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Learning Modern Algebra

From Early Attempts to Prove

Fermat’s Last Theorem

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c 2013 by The Mathematical Association of America (Incorporated)

Library of Congress Control Number: 2013940990

Print ISBN: 978-1-93951-201-7

Electronic ISBN: 978-1-61444-612-5

Printed in the United States of America

Current Printing (last digit):

10 9 8 7 6 5 4 3 2 1

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Learning Modern Algebra

From Early Attempts to Prove

Fermat’s Last Theorem

Al Cuoco

EDC, Waltham MA

and

Joseph J. Rotman

University of Illinois at Urbana–Champaign

Published and distributed by

The Mathematical Association of America

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Committee on Books

Frank Farris, Chair

MAA Textbooks Editorial Board

Zaven A. Karian, Editor

Matthias Beck Richard E. Bedient

Thomas A. Garrity

Charles R. Hampton

John Lorch

Susan F. Pustejovsky Elsa J. Schaefer

Stanley E. Seltzer

Kay B. Somers

MAA TEXTBOOKS

Bridge to Abstract Mathematics, Ralph W. Oberste-Vorth, Aristides Mouzakitis, and

Bonita A. Lawrence

Calculus Deconstructed: A Second Course in First-Year Calculus, Zbigniew H. Nitecki

Combinatorics: A Guided Tour, David R. Mazur

Combinatorics: A Problem Oriented Approach, Daniel A. Marcus

Complex Numbers and Geometry, Liang-shin Hahn

A Course in Mathematical Modeling, Douglas Mooney and Randall Swift

Cryptological Mathematics, Robert Edward Lewand

Differential Geometry and its Applications, John Oprea

Elementary Cryptanalysis, Abraham Sinkov

Elementary Mathematical Models, Dan Kalman

An Episodic History of Mathematics: Mathematical Culture Through Problem Solving,

Steven G. Krantz

Essentials of Mathematics, Margie Hale

Field Theory and its Classical Problems, Charles Hadlock

Fourier Series, Rajendra Bhatia

Game Theory and Strategy, Philip D. Straffin

Geometry Revisited, H. S. M. Coxeter and S. L. Greitzer

Graph Theory: A Problem Oriented Approach, Daniel Marcus

Knot Theory, Charles Livingston

Learning Modern Algebra: From Early Attempts to Prove Fermat’s Last Theorem, Al

Cuoco and and Joseph J. Rotman

Lie Groups: A Problem-Oriented Introduction via Matrix Groups, Harriet Pollatsek

Mathematical Connections: A Companion for Teachers and Others, Al Cuoco

Mathematical Interest Theory, Second Edition, Leslie Jane Federer Vaaler and James

W. Daniel

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Mathematical Modeling in the Environment, Charles Hadlock

Mathematics for Business Decisions Part 1: Probability and Simulation (electronic text-

book), Richard B. Thompson and Christopher G. Lamoureux

Mathematics for Business Decisions Part 2: Calculus and Optimization (electronic text-

book), Richard B. Thompson and Christopher G. Lamoureux

Mathematics for Secondary School Teachers, Elizabeth G. Bremigan, Ralph J. Bremi-

gan, and John D. Lorch

The Mathematics of Choice, Ivan Niven

The Mathematics of Games and Gambling, Edward Packel

Math Through the Ages, William Berlinghoff and Fernando Gouvea

Noncommutative Rings, I. N. Herstein

Non-Euclidean Geometry, H. S. M. Coxeter

Number Theory Through Inquiry, David C. Marshall, Edward Odell, and Michael Star-

bird

A Primer of Real Functions, Ralph P. Boas

A Radical Approach to Lebesgue’s Theory of Integration, David M. Bressoud

A Radical Approach to Real Analysis, 2nd edition, David M. Bressoud

Real Infinite Series, Daniel D. Bonar and Michael Khoury, Jr.

Topology Now!, Robert Messer and Philip Straffin

Understanding our Quantitative World, Janet Andersen and Todd Swanson

MAA Service Center

P.O. Box 91112

Washington, DC 20090-1112

1-800-331-1MAA FAX: 1-301-206-9789

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vii

Per Micky: Tutto quello che faccio, lo faccio per te.

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Contents

Preface xiii

Some Features of This Book . . . . . . . . . . . . . . . . . . . . . xiv

A Note to Students . . . . . . . . . . . . . . . . . . . . . . . . . . xv

A Note to Instructors . . . . . . . . . . . . . . . . . . . . . . . . . xv

Notation xvii

1 Early Number Theory 1

1.1 Ancient Mathematics . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Diophantus . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Geometry and Pythagorean Triples . . . . . . . . . . . . . 8

The Method of Diophantus . . . . . . . . . . . . . . . . . 11

Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . 14

Connections: Congruent Numbers . . . . . . . . . . . . . . 16

1.3 Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Greek Number Theory . . . . . . . . . . . . . . . . . . . . 21

Division and Remainders . . . . . . . . . . . . . . . . . . 22

Linear Combinations and Euclid’s Lemma . . . . . . . . . 24

Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . . 30

1.4 Nine Fundamental Properties . . . . . . . . . . . . . . . . . . 36

1.5 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . 41

Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 Induction 45

2.1 Induction and Applications . . . . . . . . . . . . . . . . . . . 45

Unique Factorization . . . . . . . . . . . . . . . . . . . . . 53

Strong Induction . . . . . . . . . . . . . . . . . . . . . . . 57

Differential Equations . . . . . . . . . . . . . . . . . . . . 60

2.2 Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . 63

Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . 69

2.3 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

An Approach to Induction . . . . . . . . . . . . . . . . . . 73

Fibonacci Sequence . . . . . . . . . . . . . . . . . . . . . 75

3 Renaissance 81

3.1 Classical Formulas . . . . . . . . . . . . . . . . . . . . . . . 82

3.2 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . 91

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x Contents

Algebraic Operations . . . . . . . . . . . . . . . . . . . . 92

Absolute Value and Direction . . . . . . . . . . . . . . . . 99

The Geometry Behind Multiplication . . . . . . . . . . . . 101

3.3 Roots and Powers . . . . . . . . . . . . . . . . . . . . . . . . 106

3.4 Connections: Designing Good Problems . . . . . . . . . . . . 116

Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Pippins and Cheese . . . . . . . . . . . . . . . . . . . . . 118

Gaussian Integers: Pythagorean Triples Revisited . . . . . . 119

Eisenstein Triples and Diophantus . . . . . . . . . . . . . . 122

Nice Boxes . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Nice Functions for Calculus Problems . . . . . . . . . . . 124

Lattice Point Triangles . . . . . . . . . . . . . . . . . . . . 126

4 Modular Arithmetic 131

4.1 Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

4.2 Public Key Codes . . . . . . . . . . . . . . . . . . . . . . . . 149

4.3 Commutative Rings . . . . . . . . . . . . . . . . . . . . . . . 154

Units and Fields . . . . . . . . . . . . . . . . . . . . . . . 160

Subrings and Subfields . . . . . . . . . . . . . . . . . . . . 166

4.4 Connections: Julius and Gregory . . . . . . . . . . . . . . . . 169

4.5 Connections: Patterns in Decimal Expansions . . . . . . . . . 177

Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . 177

Decimal Expansions of Rationals . . . . . . . . . . . . . . 179

Periods and Blocks . . . . . . . . . . . . . . . . . . . . . . 182

5 Abstract Algebra 191

5.1 Domains and Fraction Fields . . . . . . . . . . . . . . . . . . 192

5.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

Polynomial Functions . . . . . . . . . . . . . . . . . . . . 204

5.3 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . 206

Extensions of Homomorphisms . . . . . . . . . . . . . . . 213

Kernel, Image, and Ideals . . . . . . . . . . . . . . . . . . 216

5.4 Connections: Boolean Things . . . . . . . . . . . . . . . . . . 221

Inclusion-Exclusion . . . . . . . . . . . . . . . . . . . . . 227

6 Arithmetic of Polynomials 233

6.1 Parallels to Z . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

Greatest Common Divisors . . . . . . . . . . . . . . . . . 243

Unique Factorization . . . . . . . . . . . . . . . . . . . . . 248

Principal Ideal Domains . . . . . . . . . . . . . . . . . . . 255

6.2 Irreducibility . . . . . . . . . . . . . . . . . . . . . . . . . . 259

Roots of Unity . . . . . . . . . . . . . . . . . . . . . . . . 264

6.3 Connections: Lagrange Interpolation . . . . . . . . . . . . . . 270

7 Quotients, Fields, and Classical Problems 277

7.1 Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . 277

7.2 Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

Characteristics . . . . . . . . . . . . . . . . . . . . . . . . 287

Extension Fields . . . . . . . . . . . . . . . . . . . . . . . 289

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Contents xi

Algebraic Extensions . . . . . . . . . . . . . . . . . . . . 293

Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . 300

Classification of Finite Fields . . . . . . . . . . . . . . . . 305 7.3 Connections: Ruler–Compass Constructions . . . . . . . . . . 308

Constructing Regular n-gons . . . . . . . . . . . . . . . . 320

Gauss’s construction of the 17-gon . . . . . . . . . . . . . 322

8 Cyclotomic Integers 329

8.1 Arithmetic in Gaussian and Eisenstein Integers . . . . . . . . 330

Euclidean Domains . . . . . . . . . . . . . . . . . . . . . 333

8.2 Primes Upstairs and Primes Downstairs . . . . . . . . . . . . 337

Laws of Decomposition . . . . . . . . . . . . . . . . . . . 339 8.3 Fermat’s Last Theorem for Exponent 3 . . . . . . . . . . . . 349

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . 350

The First Case . . . . . . . . . . . . . . . . . . . . . . . . 351

Gauss’s Proof of the Second Case . . . . . . . . . . . . . . 354

8.4 Approaches to the General Case . . . . . . . . . . . . . . . . 359 Cyclotomic integers . . . . . . . . . . . . . . . . . . . . . 360

Kummer, Ideal Numbers, and Dedekind . . . . . . . . . . . 365

8.5 Connections: Counting Sums of Squares . . . . . . . . . . . . 371

A Proof of Fermat’s Theorem on Divisors . . . . . . . . . 373

9 Epilog 379

9.1 Abel and Galois . . . . . . . . . . . . . . . . . . . . . . . . . 379

9.2 Solvability by Radicals . . . . . . . . . . . . . . . . . . . . . 381

9.3 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 9.4 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389

9.5 Wiles and Fermat’s Last Theorem . . . . . . . . . . . . . . . 396

Elliptic Integrals and Elliptic Functions . . . . . . . . . . . 397

Congruent Numbers Revisited . . . . . . . . . . . . . . . . 400

Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . 404

A Appendices 409

A.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409

A.2 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . 420 A.3 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 424

Bases and Dimension . . . . . . . . . . . . . . . . . . . . 427

Linear Transformations . . . . . . . . . . . . . . . . . . . 435

A.4 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

A.5 Generalized Associativity . . . . . . . . . . . . . . . . . . . . 442

A.6 A Cyclotomic Integer Calculator . . . . . . . . . . . . . . . . 444 Eisenstein Integers . . . . . . . . . . . . . . . . . . . . . . 445

Symmetric Polynomials . . . . . . . . . . . . . . . . . . . 446

Algebra with Periods . . . . . . . . . . . . . . . . . . . . . 446

References 449

Index 451

About the Authors 459

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Preface

This book is designed for college students who want to teach mathematics in

high school, but it can serve as a text for standard abstract algebra courses as

well. First courses in abstract algebra usually cover number theory, groups, and commutative rings. We have found that the first encounter with groups is

not only inadequate for future teachers of high school mathematics, it is also

unsatisfying for other mathematics students. Hence, we focus here on number

theory, polynomials, and commutative rings. We introduce groups in our last

chapter, for the earlier discussion of commutative rings allows us to explain how groups are used to prove Abel’s Theorem: there is no generalization of the

quadratic, cubic, and quartic formulas giving the roots of the general quintic

polynomial. A modest proposal: undergraduate abstract algebra should be a

sequence of two courses, with number theory and commutative rings in the

first course, and groups and linear algebra (with scalars in arbitrary fields) in

the second. We invoke an historically accurate organizing principle: Fermat’s Last The-

orem (in Victorian times, the title of this book would have been Learning Mod-

ern Algebra by Studying Early Attempts, Especially Those in the Nineteenth

Century, that Tried to Prove Fermat’s Last Theorem Using Elementary Meth-

ods). To be sure, another important problem at that time that contributed to modern algebra was the search for formulas giving the roots of polynomials.

This search is intertwined with the algebra involved in Fermat’s Last Theo-

rem, and we do treat this part of algebra as well. The difference between our

approach and the standard approach is one of emphasis: the natural direction

for us is towards algebraic number theory, whereas the usual direction is to- wards Galois theory.

Four thousand years ago, the quadratic formula and the Pythagorean The-

orem were seen to be very useful. To teach them to new generations, it was

best to avoid square roots (which, at the time, were complicated to compute),

and so problems were designed to have integer solutions. This led to Pythag-

orean triples: positive integers a; b; c satisfying a2 C b2 D c2. Two thousand years ago, all such triples were found and, when studying them in the seven-

teenth century, Fermat wondered whether there are positive integer solutions

to an C bn D cn for n > 2. He claimed in a famous marginal note that there are no solutions, but only his proof of the case n D 4 is known. This problem, called Fermat’s Last Theorem, intrigued many of the finest mathematicians, but it long resisted all attempts to solve it. Finally, using sophisticated tech-

niques of algebraic geometry developed at the end of the twentieth century,

Andrew Wiles proved Fermat’s Last Theorem in 1995.

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xiv Preface

Before its solution, Fermat’s Last Theorem was a challenge to mathemati-

cians (as climbing Mount Everest was a challenge to mountaineers). There are

no dramatic applications of the result, but it is yet another triumph of human in- tellect. What is true is that, over the course of 350 years, much of contemporary

mathematics was invented and developed in trying to deal with it. The num-

ber theory recorded in Euclid was shown to have similarities with the behavior

of polynomials, and generalizations of prime numbers and unique factoriza-

tion owe their initial study to attempts at proving Fermat’s Last Theorem. But these topics are also intimately related to what is actually taught in high school.

Thus, abstract algebra is not merely beautiful and interesting, but it is also a

valuable, perhaps essential, topic for understanding high school mathematics.

Some Features of This Book

We include sections in every chapter, called Connections, in which we explic- itly show how the material up to that point can help the reader understand and

implement the mathematics that high school teachers use in their profession.

This may include the many ways that results in abstract algebra connect with

core high school ideas, such as solving equations or factoring. But it may also

include mathematics for teachers themselves, that may or may not end up “on the blackboard;” things like the use of abstract algebra to make up good prob-

lems, to understand the foundations of topics in the curriculum, and to place

the topics in the larger landscape of mathematics as a scientific discipline.

Many students studying abstract algebra have problems understanding

proofs; even though they can follow each step of a proof, they wonder how

anyone could have discovered its argument in the first place. To address such problems, we have tried to strike a balance between giving a logical develop-

ment of results (so the reader can see how everything fits together in a coherent

package) and discussing the messier kinds of thinking that lead to discovery

and proofs. A nice aspect of this sort of presentation is that readers participate

in doing mathematics as they learn it. One way we implement this balance is our use of several design features,

such as the Connections sections described above. Here are some others.

� Sidenotes provide advice, comments, and pointers to other parts of the text related to the topic at hand. What could be more fitting for a book related to

Fermat’s Last Theorem than to have large margins? � Interspersed in the text are boxed “callouts,” such as How to Think About

It, which suggest how ideas in the text may have been conceived in the first place, how we view the ideas, and what we guess underlies the formal

exposition. Some other callouts are:

Historical Note, which provides some historical background. It often helps

to understand mathematical ideas if they are placed in historical con-

text; besides, it’s interesting. The biographies are based on those in the MacTutor History of Mathematics Archive of the School of Mathemat-

ics and Statistics, University of St. Andrews, Scotland. It can be found

on the internet: its URL is

www-history.mcs.st-andrews.ac.uk

Etymology, which traces out the origin of some mathematical terms. We

believe that knowing the etymology of terms often helps to understand

the ideas they name.

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Preface xv

Etymology. The word mathematics comes from classical Greek; it

means “knowledge,” “something learned.” But in ancient Rome through

the thirteenth century, it meant “astronomy” and “astrology.” From the

Middle Ages, it acquired its present meaning.

The word arithmetic comes from the Greek word meaning “the art of

counting.” The word geometry, in classical Greek, meant “science of measuring;” it arose from an earlier term meaning “land survey.”

It is a pleasure to acknowledge those who have contributed valuable com-

ments, suggestions, ideas, and help. We thank Don Albers, Carol Baxter, Bruce

Berndt, Peter Braunfeld, Keith Conrad, Victoria Corkery, Don DeLand, Ben Conrad’s website www.math.uconn.edu/

˜kconrad/blurbs/

is full of beautiful ideas.

Fischer, Andrew Granville, Heini Halberstam, Zaven Karian, Tsit-Yuen Lam,

Paul Monsky, Beverly Ruedi, Glenn Stevens, and Stephen Ullom.

A Note to Students

The heart of a mathematics course lies in its problems. We have tried to or-

chestrate them to help you build a solid understanding of the mathematics in

the sections. Everything afterward will make much more sense if you work through as many exercises as you can, especially those that appear difficult.

Quite often, you will learn something valuable from an exercise even if you

don’t solve it completely. For example, a problem you can’t solve may show

that you haven’t fully understood an idea you thought you knew; or it may

force you to discover a fact that needs to be established to finish the solution.

There are two special kinds of exercises.

� Those labeled Preview may seem to have little to do with the section at hand; they are designed to foreshadow upcoming topics, often with numerical ex-

periments.

� Those labeled Take it Further develop interesting ideas that are connected to the main themes of the text, but are somewhat off the beaten path. They

are not essential for understanding what comes later in the text.

An exercise marked with an asterisk, such as 1.8*, means that it is either used in some proof or it is referred to elsewhere in the text. For ease of finding

such exercises, all references to them have the form “Exercise 1.8 on page 6”

giving both its number and the number of the page on which it occurs.

A Note to Instructors

We recommend giving reading assignments to preview upcoming material.

This contributes to balancing experience and formality as described above, and it saves time. Many important pages can be read and understood by students,

and they should be discussed in class only if students ask questions about them.

It is possible to use this book as a text for a three hour one-semester course,

but we strongly recommend that it be taught four hours per week.

—Al Cuoco and Joe Rotman

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Notation

.a; b; c/ 4 triangle with sides of lengths a; b; c

ABC 4 triangle with vertices A;B; C

N 21 natural numbers

Z 21 integers

a j b 21 a is a divisor of b gcd.a; b/ 24 greatest common divisor

bxc 29 greatest integer in x Q 36 rational numbers

R 36 real numbers

) 46 implies lcm.a; b/ 55 least common multiple�

n r

� 63 binomial coefficient

<.z/ 92 real part of complex number z =.z/ 92 imaginary part of complex number z C 92 complex numbers

��! PQ 93 arrow from P to Q

z 96 conjugate of z

jzj 99 modulus of z arg.z/ 100 argument of z

ez 108 complex exponential

�.n/ 111 Euler �-function

N.z/ 116 norm of z

ZŒi  119 Gaussian integers

ZŒ!/ 120 Eisenstein integers

a � b mod m 132 a is congruent to b modulom m1 � � �cmi � � �mr 147 expression withmi deleted

Œa 154 congruence class of integer a

Zm 154 integers mod m

ZŒ� 157 cyclotomic integers

RS 157 ring of functionsR ! S C.X/ 157 ring of continuous functionsX ! R

xvii

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xviii Notation

Fun.R/ 157 ring of functionsR ! R F4 165 field with 4 elements

2X 167 Boolean ring of subsets of set X

j.m/ 172 calendar month function

Frac.D/ 194 fraction field of domain D

a=b 195 element of Frac.D/

deg.f / 198 degree of polynomial f

RŒŒx 198 all power series over R

RŒx 198 all polynomials over R

x 200 indeterminate inRŒx

f 0.x/ 202 derivative of f .x/ 2 RŒx f # 204 associated polynomial function of f

Poly.R/ 204 all polynomials functions over R

k.x/ 205 field of rational functions over k

Fq 205 finite field with exactly q elements

RŒx1; : : : ; xn 205 polynomials in several variables over R

D.x1; : : : ; xn/ 206 rational functions in several variables over

domain D

R Š S 207 ringsR and S are isomorphic ker' 217 kernel of homomorphism '

im' 217 image of homomorphism '

.b1; : : : ; bn/ 218 ideal generated by b1; : : : ; bn

.a/ 218 principal ideal generated by a

.0/ 219 zero ideal D f0g IJ 220 product of ideals I and J

I C J 220 sum of ideals I and J R � S 221 direct product of rings R and S a _ b 223 binary operation in Boolean ring jAj 227 number of elements in finite set A PID 255 principal ideal domain

UFD 258 unique factorization domain

ˆd .x/ 265 cyclotomic polynomial

aC I 278 coset of element a mod ideal I a � b mod I 279 congruent mod ideal I

R=I 280 quotient ringR mod I˝ X ˛

293 subfield generated by subset X

ŒK W k 291 degree of extension field K=k k.z1 ; : : : ; zn/ 294 extension field adjoining z1; : : : ; zn to k

irr.z; k/ 296 minimal polynomial of z over k

PQ 310 line segment with endpoints P;Q

PQ 310 length of segment PQ

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Notation xix

L.P;Q/ 309 line determined by points P;Q

C.P;Q/ 309 circle with center P , radius PQ

@ 333 size function on Euclidean domain

� 348 � D 1 � ! � 350 valuation

r.n/ 371 number of non-associate z 2 ZŒi  of norm n Q1 372 first quadrant

�.s/ 374 Riemann zeta function

�.n/ 375 a multiplicative function on ZŒi 

Gal.f / 386 Galois group of polynomial f

Gal.E=k/ 387 Galois group of field extension E=k

Sn 389 symmetric group on n letters

G=N 392 quotient group

a 2 A 409 a is an element of set A 1X 411 identity function on set X

f W a 7! b 411 f .a/ D b U � V 410 U is a subset of set V U ¨ V 410 U is a proper subset of V

¿ 410 empty set

g ı f 414 composite f followed by g Œa 421 equivalence class of element a

SpanhXi 427 subspace spanned by subset X dim.V / 433 dimension of vector space V

V � 437 dual space of vector space V

A> 438 transpose of matrix A

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1 Early Number Theory Algebra, geometry, and number theory have been used for millennia. Of course,

numbers are involved in counting and measuring, enabling commerce and ar-

chitecture. But reckoning was also involved in life and death matters such as astronomy, which was necessary for navigation on the high seas (naval com-

merce flourished four thousand years ago) as well as to predict the seasons,

to apprise farmers when to plant and when to harvest. Ancient texts that have

survived from Babylon, China, Egypt, Greece, and India provide evidence for

this. For example, the Nile River was the source of life in ancient Egypt, for its banks were the only arable land in the midst of desert. Mathematics was

used by the priestly class to predict flooding as well as to calculate area (taxes

were assessed according to the area of land, which changed after flood waters

subsided). And their temples and pyramids are marvels of engineering.

1.1 Ancient Mathematics

The quadratic formula was an important mathematical tool, and so it was

taught to younger generations training to be royal scribes. Here is a problem from an old Babylonian cuneiform text dating from about 1700 BCE. We quote

from van der Waerden [35], p. 61 (but we write numbers in base 10 instead

of in base 60, as did the Babylonians). We also use modern algebraic notation

that dates from the fifteenth and sixteenth centuries (see Cajori [6]).

I have subtracted the side of the square from the area, and it is 870. What

is the side of my square?

The text rewrites the data as the quadratic equation x2 � x D 870; it then gives a series of steps showing how to find the solution, illustrating that the

Babylonians knew the quadratic formula.

Historians say that teaching played an important role in ancient mathe- matics (see van der Waerden [35], pp. 32–33). To illustrate, the coefficients

of the quadratic equation were chosen wisely: the discriminant b2 � 4ac D 1 � 4.�870/ D 3481 D 592 is a perfect square. Were the discriminant not a The number 59 may have

been chosen because

the Babylonians wrote

numbers in base 60, and

59 D 60 � 1.

perfect square, the problem would have been much harder, for finding square

roots was not routine in those days. Thus, the quadratic in the text is well- chosen for teaching the quadratic formula; a good teaching prize would not be

awarded for x2 � 47x D 210. The Babylonians were not afraid of cubics. Another of their problems from

about the same time is

1

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2 Chapter 1 Early Number Theory

Solve 12x3 D 3630,

and the answer was given. The solution was, most likely, obtained by using tables of approximations of cube roots.

A standard proof of the quadratic formula is by “completing the square.”

This phrase can be taken literally. Given a quadratic x2C bx D c with b and c positive, we can view x2 C bx as the shaded area in Figure 1.1. Complete the

x

x

Figure 1.1. Completing the Square.

figure to a square by attaching the corner square having area 1 2 b � 1

2 b D 1

4 b2;

the new square has area

c C 1 4 b2 D x2 C bx C 1

4 b2 D .x C 1

2 b/2:

Thus, x C 1 2 b D

q c C 1

4 b2, which simplifies to the usual formula giving

the roots of x2 C bx � c. The algebraic proof of the validity of the quadratic

In [35], pp. 26–35, van

der Waerden considers

the origin of proofs in

mathematics, suggesting

that they arose in Europe

and Asia in Neolithic

(late Stone Age) times,

4500 BCE–2000 BCE.

formula works without assuming that b and c are positive, but the idea of the

proof is geometric.

a2

b2

a

b

a

b

a b

a

b

c2

Figure 1.2. Pythagorean Theorem.

The Babylonians were aware of the Pythagorean Theorem. Although they

believed it, there is no evidence that the Babylonians had proved the Pythag-

orean Theorem; indeed, no evidence exists that they even saw a need for a

proof. Tradition attributes the first proof of this theorem to Pythagoras, who Exercise 1.4 on page 5

asks you to show that the

rhombus in Figure 1.2

with sides of length c is a

square.

lived around 500 BCE, but no primary documents extant support this. An ele-

gant proof of the Pythagorean Theorem is given on page 354 of Heath’s 1926

translation [16] of Euclid’s The Elements; the theorem follows from equality

of the areas of the two squares in Figure 1.2.

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1.1 Ancient Mathematics 3

Here is an ancient application of the Pythagorean Theorem. Aristarchus

(ca. 310 BCE–250 BCE) saw that the Moon and the Sun appear to be about

the same size, and he wondered how far away they are. His idea was that at the time of the half-moon, the Earth E , Moon M , and Sun S form a right

triangle with right angle †M (that is, looking up at the Moon, the line of sight seems to be perpendicular to the Sun’s rays). The Pythagorean Theorem gives

a

S M

E

Figure 1.3. Earth, Moon, and Sun.

jSEj2 D jSM j2 C jMEj2. Thus, the Earth is farther from the Sun than it is from the Moon. Indeed, at sunset, ˛ D †E seems to be very close to 90ı: if we are looking at the Moon and we wish to watch the Sun dip below the horizon,

we must turn our head all the way to the left. Aristarchus knew trigonometry;

he reckoned that cos˛ was small, and he concluded that the Sun is very much

further from the Earth than is the Moon.

Example 1.1. Next, we present a geometric problem from a Chinese collec-

tion of mathematical problems, Nine Chapters on the Mathematical Art, writ- ten during the Han Dynasty about two thousand years ago. Variations of this

problem still occur in present day calculus books!

There is a door whose height and width are unknown, and a pole whose There are similar problems from the Babylonians and

other ancient cultures. length p is also unknown. Carried horizontally, the pole does not fit by 4

ch’ihI vertically, it does not fit by 2 ch’ihI slantwise, it fits exactly. What are the height, width, and diagonal of the door?

p p – 2

p – 4

Figure 1.4. Door Problem.

The data give a right triangle with sides p � 4, p � 2, and p, and the Py- thagorean Theorem gives the equation .p � 4/2 C .p � 2/2 D p2, which simplifies to p2�12pC20 D 0. The discriminant b2�4ac is 144�80 D 64, a perfect square, so that p D 10 and the door has height 8 and width 6 (the other root of the quadratic is p D 2, which does not fit the physical data). The sides of the right triangle are 6, 8, 10, and it is similar to the triangle with

sides 3; 4; 5. Again, the numbers have been chosen wisely. The idea is to teach

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4 Chapter 1 Early Number Theory

students how to use the Pythagorean Theorem and the quadratic formula. As

we have already remarked, computing square roots was then quite difficult, so

that the same problem for a pole of length p D 12 would not have been very bright because there is no right triangle with sides of integral length that hasThe word hypotenuse

comes from the Greek verb

meaning to stretch. hypotenuse 12. N

Are there right triangles whose three sides have integral length that are not

similar to the 3; 4; 5 triangle? You are probably familiar with the 5; 12; 13 tri-

angle. Let’s use 4.a; b; c/ (lower case letters) to denote the triangle whose sides have length a, b, and c; if 4.a; b; c/ is a right triangle, then c denotes the length of its hypotenuse, while a and b are its legs. Thus, the right trian-

gle with side-lengths 5, 12, 13 is denoted by 4.5; 12; 13/. (We use the usual notation, 4ABC , to denote a triangle whose vertices are A;B; C .)

Definition. A triple .a; b; c/ of positive integers with a2 C b2 D c2 is called a Pythagorean triple.

If .a; b; c/ is a Pythagorean triple, then the triangles 4.a; b; c/ and 4.b; a; c/ are the same. Thus, we declare that the Pythagorean triples .a; b; c/ and .b; a; c/ are the same.

Historical Note. Pythagorean triples are the good choices for problems teach-

ing the Pythagorean Theorem. There are many of them: Figure 1.5 shows a

Babylonian cuneiform tablet dating from the dynasty of Hammurabi, about

1800 BCE, whose museum name is Plimpton 322, which displays fifteen Pythagorean triples (translated into our number system).

b a c

120 119 169

3456 3367 4825

4800 4601 6649

13500 12709 18541

72 65 97

360 319 481

2700 2291 3541

960 799 1249

600 481 769

6480 4961 8161

60 45 75

2400 1679 2929

240 161 289

2700 1771 3229

90 56 106

Figure 1.5. Plimpton 322.

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1.1 Ancient Mathematics 5

It is plain that the Babylonians had a way to generate large Pythagorean

triples. Here is one technique they might have used. Write

a2 D c2 � b2 D .c C b/.c � b/:

If there are integers m and n with

c C b D m2

c � b D n2;

then

a D p .c C b/.c � b/ D mn: (1.1)

We can also solve for b and c:

b D 1 2

� m2 � n2

� (1.2)

c D 1 2

� m2 C n2

� : (1.3)

Summarizing, here is what we call the Babylonian method. Choose odd num-

bers m and n (forcing m2 C n2 and m2 � n2 to be even, so that b and c are integers), and define a, b, and c by Eqs. (1.1), (1.2), and (1.3). For example, if m D 7 and n D 5, we obtain 35, 12, 37. If we choose m D 179 and n D 71, we obtain 13500, 12709, 18541, the largest triple on Plimpton 322.

The Babylonian method does not give all Pythagorean triples. For example,

.6; 8; 10/ is a Pythagorean triple, but there are no odd numbers m > n with

6 D mn or 8 D mn. Of course, .6; 8; 10/ is not signifcantly different from .3; 4; 5/, which arises from 3 > 1. In the next section, we will show, follow-

ing Diophantus, ca. 250 CE, how to find all Pythagorean triples. But now we

should recognize that practical problems involving applications of pure math-

ematics (e.g., surveying) led to efforts to teach this mathematics effectively,

which led to more pure mathematics (Pythagorean triples) that seems at first to After all, what practi-

cal application does

the Pythagorean triple

.13500; 12709; 18541/

have?

have no application outside of teaching. The remarkable, empirical, fact is that pure mathematics yields new and valuable applications. For example, we shall

see in the next section that classifying Pythagorean triples leads to simplifying

the verification of some trigonometric identities as well as the solution of cer-

tain integration problems (for example, we will see a natural way to integrate

sec x).

Exercises

1.1 Prove the quadratic formula for the roots of ax2Cbx Cc D 0 whose coefficients a, b, and c may not be positive.

1.2 Give a geometric proof that .a C b/2 D a2 C 2ab C b2 for a; b positive. 1.3 * Let f .x/ D ax2C bx C c be a quadratic whose coefficients a; b; c are rational.

Prove that if f .x/ has one rational root, then its other root is also rational.

1.4 *

(i) Prove that the rhombus with side lengths c in the left square of Figure 1.2 is The book by Loomis [20] contains 370 different

proofs of the Pythagorean

Theorem, by the author’s

count.

a square.

(ii) Prove the Pythagorean Theorem in a way suggested by Figure 1.2.

(iii) Give a proof of the Pythagorean Theorem different from the one suggested

by Figure 1.2.

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6 Chapter 1 Early Number Theory

1.5 Here is another problem from Nine Chapters on the Mathematical Art. A pond is

10 ch’ih square. A reed grows at its center and extends 1 ch’ih out of the water.

If the reed is pulled to the side of the pond, it reaches the side precisely. What are

the depth of the water and the length of the reed?

Answer: Depth = 12 ch’ih and length = 13 ch’ih.

1.6 *

(i) Establish the algebraic identity

� a C b

2

�2 � �

a � b 2

�2 D ab:

(ii) Use (i) to establish the Arithmetic–Geometric Mean Inequality: if a and b

are positive reals, then

p ab � 12 .a C b/:

When is there equality?

(iii) Show how to dissect an a � b rectangle so that it fits inside a square with side-length .a C b/=2. How much is “left over?”

Hint: Try it with numbers. Cut an 8 � 14 rectangle to fit inside an 11 � 11 square.

(iv) Show that a rectangle of maximum area with fixed perimeter is a square.

(v) The hyperbolic cosine is defined by

cosh x D 12 .e x C e�x/:

Prove that cosh x � 1 for all real numbers x, while coshx D 1 if and only if x D 0.

(vi) Use Figure 1.6 to give another proof of the Arithmetic-Geometric Mean In-

equality.

a b

Figure 1.6. Arithmetic–Geometric Mean Inequality.

1.7 * Prove that there is no Pythagorean triple .a; b; c/ with c D 12.

1.8 * Let .a; b; c/ be a Pythagorean triple.

(i) Prove that the legs a and b cannot both be odd.

(ii) Show that the area of 4.a; b; c/ is an integer.

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1.2 Diophantus 7

1.9 * Show that 5 is not the area of a triangle whose side-lengths form a Pythagorean

triple.

1.10 * Let .a; b; c/ be a Pythagorean triple. If m is a positive integer, prove that

.ma; mb; mc/ is also a Pythagorean triple.

1.11 .Converse of Pythagorean Theorem/: * Let 4 D 4.a; b; c/ be a triangle with sides of lengths a; b; c (positive real numbers, not necessarily integers). Prove that

if a2 C b2 D c2, then 4 is a right triangle.

Hint: Construct a right triangle 40 with legs of lengths a; b, and prove that 40 is congruent to 4 by side-side-side.

1.12 * Prove that every Pythagorean triple .a; b; c/ arises from a right triangle 4.a; b; c/ having sides of lengths a; b; c.

1.13 If P D .a; b; c/ is a Pythagorean triple, define r.P / D c=a. If we label the Py- thagorean triples on Plimpton 322 as P1; : : : ; P15 , show that r.Pi / is decreasing:

r.Pi / > r.PiC1/ for all i � 14.

1.14 * If .a; b; c/ is a Pythagorean triple, show that .a=c; b=c/ is a point on the graph

of x2 C y2 D 1. What is the graph of x2 C y2 D 1?

1.15 Preview: Let L be the line through .�1; 0/ with slope t . (i) If t D 12 , find all the points where L intersects the graph of x

2 C y2 D 1.

Answer: .35 ; 4 5 /.

(ii) If t D 32 , find all the points where L intersects the graph of x2 C y2 D 1.

Answer: . �513 ; 12 13

/.

(iii) Pick a rational number t , not 12 or 3 2 , and find all the points where L intersects

the graph of x2 C y2 D 1. (iv) Suppose ` is a line that contains .�1; 0/ with slope r . If r is a rational number,

show that ` intersects the graph of x2 C y2 D 1 in two points, each of which has rational number coordinates.

1.16 Preview: A Gaussian integer is a complex number a C bi where both a and b are integers. Pick six Gaussian integers r C si with r > s > 0 and square them. State something interesting that you see in your results.

1.17 Preview: Consider a complex number z D q C ip, where q > p are positive integers. Prove that

.q2 � p2; 2qp; q2 C p2/

is a Pythagorean triple by showing that jz2j D jzj2.

If z is a complex number,

say z D aC bi , then we define jzj D

p a2 C b2.

1.18 Preview: Show, for all real numbers m and n, that

h 1 2 .m C n/ C

1 2 .m � n/i

i2 D mn C 12 .m

2 � n2/i:

1.2 Diophantus

We are going to classify Pythagorean triples using a geometric method of Dio-

phantus that describes all Pythagorean triples.

Historical Note. We know very little about the life of Diophantus. He was

a mathematician who lived in Alexandria, Egypt, but his precise dates are

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8 Chapter 1 Early Number Theory

unknown; most historians believe he lived around 250 CE. His extant work

shows systematic algebraic procedures and notation, but his leaps of intuition

strongly suggest that he was thinking geometrically; indeed, Newton called Diophantus’s discussion of Pythagorean triples the chord method (see Fig-

ure 1.7). Thus, geometry (the Pythagorean Theorem) and applied problems

(teaching) suggested an algebraic problem (find all Pythagorean triples), and

we now return to geometry to solve it. Here is evidence that the distinction

between algebra and geometry is an artificial one; both are parts of the same subject.

Geometry and Pythagorean Triples

Before we get into the technicalities of Diophantus’s classification of Pythag-

orean triples, let’s note that geometry is lurking nearby. Exercise 1.14 above

makes a natural observation: if .a; b; c/ is a Pythagorean triple, then

�a c

�2 C � b

c

�2 D 1;

a point on the unit circle, the circle having radius 1, center the origin, and

equation x2 C y2 D 1. Dividing through by c2 is a good idea. For example, .6; 8; 10/ is a “duplicate” of .3; 4; 5/, and both of these Pythagorean triples

determine the same point, .3=5; 4=5/, on the unit circle. Here is the main idea of Diophantus. Even though those points arising from

Pythagorean triples are special (for example, they lie in the first quadrant and

both their coordinates are rational numbers), let’s parametrize all the pointsP

on the unit circle. Choose a point on the unit circle “far away” from the first

quadrant; the simplest is .�1; 0/, and let ` D `.P / be the line joining it to P . We shall see that the slopes of such lines parametrize all the points on the

unit circle. In more detail, any line ` through .�1; 0/ (other than the tangent)

(–1,0)

P x, y= ( )

slope = t

l

Figure 1.7. Geometric Idea of Diophantus.

intersects the unit circle in a unique second point, P D .x; y/; let t be the slope of `. As t varies through all real numbers, �1 < t < 1, the intersection points P of ` and the unit circle trace out the entire circle (except for .�1; 0/).

Proposition 1.2. The points P on the unit circle .other than .�1; 0// are parametrized as

P D � 1 � t2 1C t2 ;

2t

1C t2

� ; where � 1 < t < 1:

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1.2 Diophantus 9

Proof. The line through points .a; b/ and .c; d / has equation y�b D t.x�a/, where t D .d � b/=.c � a/, so the line ` through .�1; 0/ and a point P D .x; y/ on the unit circle has an equation of the form y D t.x C 1/, so that x D .y � t/=t . Thus, .x; y/ is a solution of the system

y D t.x C 1/ x2 C y2 D 1:

An obvious solution of this system is .�1; 0/, because this point lies on both the line and the circle. Let’s find x and y in terms of t . If the slope t D 0, then ` is the x-axis and the other solution is .1; 0/. To find the solutions when

t ¤ 0, eliminate x: the equations

y � t t

D x and x2 C y2 D 1

give

� y � t t

�2 C y2 D 1:

Expanding and simplifying, we obtain

y � .1 C t2/y � 2t

� D 0:

We knew at the outset that y D 0 makes this true. If y ¤ 0, then canceling gives

y D 2t 1C t2 ;

and solving for x gives

x D y � t t

D 2t 1Ct2 � t

t D 1 � t

2

1C t2 :

In Exercise 1.12 on page 7, we saw that every Pythagorean triple .a; b; c/

arises from a right triangle 4.a; b; c/ having sides of integral lengths a; b; c. Conversely, the Pythagorean Theorem says that every right triangle 4.a; b; c/ whose sides have integral length gives the Pythagorean triple .a; b; c/. Thus,

Pythagorean triples and certain right triangles are merely two ways to view the

same idea, one algebraic, one geometric. At any given time, we will adopt that

viewpoint which is most convenient.

We have already run across distinct Pythagorean triples that are essentially

the same; Exercise 1.10 on page 7 shows that if .a; b; c/ is a Pythagorean triple, then so is .ma;mb;mc/, where m is a positive integer. The right trian-

gles 4.a; b; c/ and 4.ma;mb;mc/ determined by these Pythagorean triples are similar, for their sides are proportional. More generally, the Pythagorean

triples .6; 8; 10/ and .9; 12; 15/ are not really different, for each arises from

.3; 4; 5/; however, neither .6; 8; 10/ nor .9; 12; 15/ is obtained from the other by multiplying its terms by some integer m.

Definition. Two Pythagorean triples .a; b; c/ and .u; v; z/ are similar if their

right triangles 4.a; b; c/ and 4.u; v; z/ are similar triangles.

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10 Chapter 1 Early Number Theory

The method of Diophantus will give a formula for certain special Pythago-

rean triples, and it will then show that every Pythagorean triple is similar to a

special one.

Definition. A point .x; y/ in the plane is a rational point if both x and y are rational numbers.

A Pythagorean point is a rational point in the first quadrant, lying on the

unit circle, and above the diagonal line with equation y D x.

Remember that we regard Pythagorean triples .a; b; c/ and .b; a; c/ as the

same. Recall some analytic geometry: if a ¤ b, P D .a; b/, and Q D .b; a/, then the diagonal is the perpendicular bisector of the segment PQ. (The line

through P and Q has equation y D �x C a C b; it is perpendicular to the diagonal for the product of their slopes is �1; the line intersects the diagonal in the point .aCb

2 ; aCb 2 /, which is equidistant from P and Q. If a � b, then P

is above the diagonal and Q is below.)

Proposition 1.3. A triple .a; b; c/ of integers is a Pythagorean triple if and

only if .a=c; b=c/ is a Pythagorean point.

Proof. Let .a; b; c/ be a Pythagorean triple. Dividing both sides of the defining

equation a2 C b2 D c2 by c2 gives .a=c/2 C .b=c/2 D 1;

so that the triple gives an ordered pair of positive rational numbers .x; y/ D .a=c; b=c/ with x2 C y2 D 1. Thus, the rational point P D .x; y/ lies in the first quadrant. As both .a; b; c/ and .b; a; c/ are the same Pythagorean triple,

we may assume that

x D a=c � b=c D y;

so that .x; y/ lies above the diagonal line with equation y D x. Hence, .x; y/ is a Pythagorean point.

Conversely, let’s now see that a Pythagorean point .x; y/ gives rise to a

Pythagorean triple. Write the rational numbers x � y with the same denomi- nator, say, x D a=c and y D b=c, where a; b; and c are positive integers and a � b � c. Now

1 D x2 C y2 D a 2

c2 C b

2

c2 ;

so that a2 C b2 D c2 and hence .a; b; c/ is a Pythagorean triple. In summary, the problem of finding all Pythagorean triples corresponds to

the problem of finding all Pythagorean points. This is exactly what the geo-

metric idea of Diophantus does. In fact, a Pythagorean point .x; y/ gives rise

to infinitely many Pythagorean triples. Write the coordinates with another de-

nominator, say x D u=z and y D v=z. The calculation at the end of the proof of Proposition 1.3 shows that .u; v; z/ is another Pythagorean triple arising from .x; y/.

Etymology. Here are sources of some common words of mathematics.

� Proposition: From Latin, meaning a statement or something pictured in the mind.

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1.2 Diophantus 11

� Theorem: From the Greek word meaning “spectacle” or “contemplate.” Re- lated words are “theory” and “theater.” Theorems are important propositions.

� Corollary:From the Latin word meaning “flower.” In ancient Rome, it meant a “gratuity;” flowers were left as tips. In mathematics, corollaries follow eas-

ily from theorems; they are gifts bequeathed to us.

� Lemma: From Greek; it meant something taken for granted. In mathematics nowadays, it is usually a technical result, a minor theorem, which can be

used in the course of proving a more important theorem.

� Proof: From Medieval French, meaning an argument from evidence estab- lishing the truth. The adage, “The exception proves the rule,” uses the word in the sense of testing: it originally meant a kind of indirect proof. We test

whether a rule is true by checking whether an exception to it leads to a con-

tradiction. Nowadays, this adage seems to have lost its meaning.

The Method of Diophantus

Proposition 1.2 parametrizes all the points P on the unit circle other than

.�1; 0/. We are now going to see which values of t produce Pythagorean points: rational points on the unit circle lying in the first quadrant above the

diagonal line with equation y D x.

Theorem 1.4. Let P D .x; y/ ¤ .�1; 0/ be a point on the unit circle, and let t be the slope of the line ` joining .�1; 0/ and P .

(i) The slope t is a rational number if and only if P is a rational point.

(ii) The point P is a Pythagorean point if and only if t is a rational number

satisfying p 2 � 1 < t < 1.

Proof. (i) The parametrization P D .x; y/ gives a pair of equations:

x D 1 � t 2

1C t2 and y D 2t

1C t2 :

Clearly, if t is rational, then both x and y are rational. Conversely, if

P D .x; y/ is a rational point, then the slope t of ` is t D y�0 x�.�1/ D

y xC1 ,

and so t is a rational number.

(ii) Pythagorean points correspond to rational points on the unit circle that lie

in the first quadrant above the line y D x. Points on the circle lying in the first quadrant arise from lines having slope t with 0 < t < 1. The point The slope of the line joining

.�1; 0/ to .0; 1/ is 1.in the first quadrant that is the intersection of the unit circle and the line y D x is .

p 2 2 ;

p 2 2 /, and the slope of the line joining .�1; 0/ to .

p 2 2 ;

p 2 2 /

is

t D

p 2 2

1C p 2 2

D p 2 � 1 � :414:

Therefore, Pythagorean points correspond to the lines ` through .�1; 0/ having rational slope t satisfying

p 2 � 1 < t < 1.

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12 Chapter 1 Early Number Theory

Let’s look at this more closely. If t D p=q is a rational number betweenp 2� 1 and 1, then the Pythagorean point it gives can be expressed in terms of

p and q:

� 1 � t2 1C t2 ;

2t

1C t2

� D

0 B@ 1 �

� p q

�2

1C � p q

�2 ; 2 � p q

�

1C � p q

�2

1 CA

D � q2 � p2 q2 C p2 ;

2qp

q2 C p2

� : (1.4)

Theorem 1.5 (Diophantus). Every Pythagorean triple .a; b; c/ is similar to a

Pythagorean triple of the form

.2qp; q2 � p2; q2 C p2/;

where p and q are positive integers with q > p > p 2 � 1.

Proof. Since .a; b; c/ is a Pythagorean triple, P D .a=c; b=c/ is a Pythago- rean point. By Eq. (1.4),

�a c ; b

c

� D � 1 � t2 1C t2 ;

2t

1C t2 �

D � q2 � p2 q2 C p2 ;

2qp

q2 C p2 � :

It follows that 4.a; b; c/ is similar to 4.2pq; q2�p2; q2Cp2/, because their sides are proportional. Therefore, the Pythagorean triple .a; b; c/ is similar to .2qp; q2 � p2; q2 C p2/, as claimed.

How to Think About It. The strategy of Diophantus is quite elegant. The problem of determining all Pythagorean triples is reduced from finding three

unknowns, a; b; and c, to two unknowns, x D a=c and y D b=c, to one unknown, t D p=q. In effect, all Pythagorean triples are parametrized by t ; that is, as t varies over all rational numbers between

p 2�1 and 1, the formulas

involving t vary over all Pythagorean points and hence over all Pythagorean triples.

We can now show that the Babylonians had, in fact, found all Pythagorean

triples. We are tacitly using a

technique of proof called

Infinite Descent. If, for a

given positive integer n

with certain properties,

there always exists a

strictly smaller positive

integer n1 having the

same properties, then

there are infinitely many

such integers. But this is

impossible; there are only

finitely many integers with

n > n1 > n2 > � � � > 0.

Corollary 1.6. Every Pythagorean triple is similar to one arising from the

Babylonian method.

Proof. By Theorem 1.5, every Pythagorean triple is similar to one of the form

.2qp; q2�p2; q2Cp2/, where q > p are positive integers. If both q and p are even, then we can replace q > p by 1

2 q > 1

2 p, obtaining a Pythagorean triple

.1 4 2qp; 1

4 .q2�p2/; 1

4 .q2Cp2// similar to the original one. If both parameters

of the new triple are still even, replace 1 2 q > 1

2 p by 1

4 q > 1

4 p. Eventually, we

arrive at a Pythagorean triple .2rs; r2 � s2; r2 C s2/, similar to the original triple, that arises from parameters r > s, at least one of which is odd.

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1.2 Diophantus 13

There are two possibilities. If r and s have different parity, definem D rCs and n D r � s. Both m and n are odd, and the Pythagorean triple given by the Babylonian method fromm > n is

B D � mn; 1

2 .m2 � n2/; 1

2 .m2 C n2/

� :

Substitute:

mn D .rCs/.r�s/ D r2�s2; 1 2 .m2�n2/ D 2rs; and 1

2 .m2�n2/ D r2Cs2:

Thus, the Pythagorean tripleB is similar to .2rs; r2 � s2; r2 C s2/. If both r and s are odd, then the Pythagorean triple given by the Babylo-

nian method from r > s is .rs; 1 2 .r2 � s2/; 1

2 .r2 C s2// which is similar to

.2rs; r2 � s2; r2 C s2/.

Not every Pythagorean triple .a; b; c/ is equal to .2qp; q2 � p2; q2 C p2/ for some q > p, nor does the theorem say that it is; the theorem asserts only

that .a; b; c/ is similar to a Pythagorean triple arising from the formula. For example, let us show that .9; 12; 15/ is not of this form. Since the leg 9 is odd,

the even leg 12 must be 2qp, so that qp D 6, and the only possible parameters are 3 > 2 or 6 > 1. But 3 > 2 gives .5; 12; 13/ and 6 > 1 gives .12; 35; 37/,

neither of which is similar to .9; 12; 15/. However, .9; 12; 15/ is similar to

.3; 4; 5/, and .3; 4; 5/ arises from 2 > 1. A Pythagorean triple .a; b; c/ is primitive if there is no integer d > 1 that

is a divisor of a; b, and c. Thus, .3; 4; 5/ is primitive but .9; 12; 15/ is not.

In Theorem 1.25, we’ll give a rigorous proof that every Pythagorean triple is

similar to exactly one primitive Pythagorean triple.

Exercises

1.19 Find q and p in Theorem 1.5 for each of the following Pythagorean triples.

(i) .7; 24; 25/.

Answer: q D 5 and p D 3. (ii) .129396;261547;291805/:

Answer: q D 526 and p D 123.

1.20 * Show that every Pythagorean triple .x; y; z/ with x; y; z having no common

factor d > 1 is of the form

.r2 � s2; 2rs; r2 C s2/

for positive integers r > s having no common factor > 1; that is,

x D r2 � s2; y D 2rs; z D r2 C s2:

1.21 A line in the plane with equation y D mx C c is called a rational line if m and c are rational numbers. If P and Q are distinct rational points, prove that the line

joining them is a rational line.

1.22 A lattice point is a point in the plane whose coordinates are integers. Let P D .x; y/ be a Pythagorean point and ` the line through P and the origin. Prove that

if Q D .a; b/ is a lattice point on ` and c is the distance from Q to the origin, then .a; b; c/ is a Pythagorean triple.

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14 Chapter 1 Early Number Theory

1.23 * Let P D .x0; y0/ be a Pythagorean point and L the line joining P and the origin (so the equation of L is y D mx, where m D y0=x0). Show that if .a=c; b=c/ is a rational point on L, then .a; b; c/ is a Pythagorean triple.

1.24 Does every rational point in the plane correspond to a Pythagorean point? If so,

prove it. If not, characterize the ones that do.

Answer: No. For example, .12 ; 1 2 / does not correspond.

1.25 * Prove the identity � x2 C y2

�2 D � x2 � y2

�2 C .2xy/2: 1.26 *

(i) Show that the same number can occur as a leg in two nonsimilar Pythagorean

triangles.

(ii) Prove that the area of 4.a; b; c/, a right triangle with integer side lengths, is an integer.

(iii) A Heron triangle is a triangle with integer side lengths and area. Find a Heron

triangle that is not a right triangle.

Hint: Use parts (i) and (ii).

1.27 Show that every integer n � 3 occurs as a leg of some Pythagorean triple.

Hint: The cases n even and n odd should be done separately.

1.28 Distinct Pythagorean triples can have the same hypotenuse: both .33;56; 65/ and

.16; 63; 65/ are Pythagorean triples. Find another pair of distinct Pythagorean

triples having the same hypotenuse.

1.29 * If .cos �; sin �/ is a rational point, prove that both cos.� C30ı/ and sin.� C30ı/ are irrational.

Fermat’s Last Theorem

About fourteen centuries after Diophantus, Fermat (1601–1665) proved that

there are no positive integers a; b; c with a4 C b4 D c4. He was studying his copy of Diophantus’s Arithmetica, published in 1621, and he wrote in its

margin,

: : : it is impossible for a cube to be written as a sum of two cubes or a

fourth power to be written as a sum of two fourth powers or, in general,

for any number which is a power greater than the second to be written

as a sum of two like powers. I have discovered a truly marvelous demon-

stration of this proposition which this margin is too narrow to contain.

Fermat never returned to this problem (at least, not publicly) except for hisFermat was not the first mathematician to write a

marginal note in a copy

of Diophantus. Next to

the same problem, the

Byzantine mathematician

Maximus Planudes wrote,

Thy soul, Diophantus, be

with Satan because of the

difficulty of your theorems.

proof of the case n D 4, which we give below. The statement: If n > 2, there are no positive integers a; b; c with an C bn D cn, was called Fer- mat’s Last Theorem, perhaps in jest. The original text in which Fermat wrote

his famous marginal note is lost today. Fermat’s son edited the next edition

of Diophantus, published in 1670; this version contains Fermat’s annotations,

including his famous “Last Theorem;” it contained other unproved assertions as well, most true, some not. By the early 1800s, only Fermat’s Last Theorem

remained undecided. It became a famous problem, resisting the attempts of

mathematicians of the highest order for 350 years, until it was finally proved,

in 1995, by Wiles. His proof is very sophisticated, and most mathematicians

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1.2 Diophantus 15

believe that Fermat did not have a correct proof. The quest for a proof of Fer-

mat’s Last Theorem generated much beautiful mathematics. In particular, it led

to an understanding of complex numbers, factorization, and polynomials. We’ll see, in the Epilog, that extending the method of Diophantus from quadratics to

cubics involves elliptic curves, the study of which is the setting for Wiles’

proof of Fermat’s Last Theorem.

Fermat proved the next theorem (which implies the case n D 4 of Fermat’s Last Theorem) because he was interested in the geometric problem of deter- mining which right triangles having all sides of rational length have integer

area (we’ll soon discuss this problem in more detail).

Theorem 1.7 (Fermat). There is no triple .x; y; z/ of positive integers with

x4 C y4 D z2: (1.5) This proof is not difficult,

but it uses several elemen-

tary divisibility results we’ll

prove later. Since we feel

that this is the appropriate

place for this theorem, we’ll

just refer to the needed

things.

Proof. The proof will be by infinite descent (Fermat invented infinite descent

for this very problem). Given a triple of positive integers .x; y; z/ satisfying Eq. (1.5), we’ll show there is another triple .u; v; w) of the same sort with

w < z, and so repeating this process leads to a contradiction.

Let’s say that integers x and y are relatively prime if there is no integer

d > 1 dividing both of them; that is, it’s not true that x D da and y D db. We can assume that x and y are relatively prime, for otherwise a common

factor of x and y would also be a factor of z, and we could divide it out. It

follows (and we’ll prove it in the next chapter) that x2 and y2 are also relatively

prime. And note that x4 C y4 D z2 implies that � x2 �2 C

� y2 �2 D z2

so that .x2; y2; z/ form a Pythagorean triple.

We also observe that x2 and y2 can’t both be odd; if x2 D 2k C 1 and y2 D 2j C 1, then

.2k C 1/2 C .2j C 1/2 D z2:

Expanding and collecting terms gives z2 D 4hC2 for some integer h. But you can check that the square of any integer is either of the form 4h or 4hC 1.

We can now assume that .x2; y2; z/ is a Pythagorean triple in which x and

y are relatively prime, x is odd, and y is even. By Exercise 1.20 on page 13,

there are relatively prime integers r and s with r > s > 0 such that

x2 D r2 � s2; y2 D 2rs; and z D r2 C s2:

The first equation says that x2 C s2 D r2; that is, .x; s; r/ is another Pythago- rean triple with x odd. Moreover, x and s have no common factor (why?), so

that Exercise 1.20 gives relatively prime integers a and b such that

x D a2 � b2; s D 2ab; and r D a2 C b2:

Now,

y2 D 2rs D 2.a2 C b2/.2ab/ D 4ab.a2 C b2/:

Since y is even, we have an equation in integers:

�y 2

�2 D ab.a2 C b2/: (1.6)

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16 Chapter 1 Early Number Theory

As a and b are relatively prime (no common factor d > 1), each pair from

the three factors on the right-hand side of Eq. (1.6) is relatively prime. Since

the left-hand side .y=2/2 is a square, each factor on the right is a square (Ex- ercise 2.12 on page 59). In other words, there are integers u, v, and w such

that

a D u2; b D v2; and a2 C b2 D w2:

And, since a and b are relatively prime, so, too, are u and v relatively prime.

Hence, we have

u4 C v4 D w2:

This is our “smaller” solution to Eq. (1.5), for

0 < w < w2 D a2 C b2 D r < r2 < r2 C s2 D z:

We can now repeat this process on .u; v; w/. By infinite descent, there is no

solution to Eq. (1.5).

Corollary 1.8 (Fermat’s Last Theorem for Exponent 4). There are no pos-

itive integers x; y; z with

x4 C y4 D z4:

Proof. If such a triple existed, we’d have

x4 C y4 D � z2 �2 ;

and that’s impossible, by Theorem 1.7.

Call an integer n � 2 good if there are no positive integers a; b; c with an C bn D cn. If n is good, then so is any multiple nk of it. Otherwise, there are positive integers r; s; t with rnk C snk D tnk , and this gives the contradic- tion an C bn D cn, where a D rk , b D sk , and c D tk . For example, Corol- lary 1.8 shows that that any positive integer of the form 4k is good. Since every

n � 2 is a product of primes, it follows that Fermat’s Last Theorem would be true if every odd prime is good.

Connections: Congruent Numbers

Fermat’s motivation for Theorem 1.7 came, not from a desire to prove there

are no non-trivial integer solutions to x4 C y4 D z4, but from a problem in the intersection of arithmetic and geometry. In more detail, suppose that

4 D 4.a; b; c/ is the right triangle arising from a Pythagorean triple .a; b; c/. Since 4 is a right triangle, the leg a is an altitude and the area of 4.a; b; c/ is 1 2 ab; since .a; b; c/ is a Pythagorean triple, the area is an integer (Exercise 1.8

on page 6). Tipping this statement on its head, we ask which integers are areas

of right triangles having integer side-lengths. Certainly 6 is, because it’s the

area of 4.3; 4; 5/. But 5 is not the area of such a triangle (Exercise 1.9 on page 7).

However, we claim that 5 is the area of a right triangle whose side-lengths

are rational numbers. Consider the Pythagorean triple .9; 40; 41/; its right tri-

angle 4 D 4.9; 40; 41/ has area 1 2 .9 � 40/ D 180. Now 180 D 36 � 5. Scaling

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1.2 Diophantus 17

the side-lengths of 4 by 1 6

scales the area by 1 36

, so that 4 � 3 2 ; 20 3 ; 41 6

� has area

180=36 D 5. So, the question arises: “Is every integer the area of a right triangle with

rational side-lengths?” Fermat showed that 1 and 2 are not, and his proof for 2

involved Eq. (1.5).

Theorem 1.9. There is no right triangle with rational side-lengths and area 2.

Proof. Suppose, on the contrary, that the rational numbers r; s; t are the lengths

of the sides of a right triangle with area 2. Then we have two equations:

r2 C s2 D t2 1 2 rs D 2:

Multiply the first equation by r2 to obtain

r4 C .rs/2 D .rt/2; so that (since rs D 4),

r4 C 24 D .rt/2:

Write the rational numbers r and t as fractions with the same denominator:

r D a=c and t D b=c. When we clear denominators, we get a4 C z4c4 D t2, an equation in integers x; y; z of the form

x4 C y4 D z2:

This is Eq. (1.5), and Theorem 1.7 says that this cannot occur.

So, not every positive integer is the area of a right triangle with rational

side-lengths.

Definition. A congruent number is a positive integer n that is the area of a

right triangle having rational side-lengths.

Theorem 1.9 says that 2 is not a congruent number. Using similar ideas,

Fermat showed that 1 is not a congruent number (Exercise 1.31 below).

One way to generate congruent numbers is to scale a Pythagorean triple using the largest perfect square that divides its area. For example, the area of We have already used this

method on the Pythag-

orean triple .9; 40; 41/

when we showed that 5 is

a congruent number.

4.7; 24; 25/ is 84 D 22 � 21. Since 4 D 22 is the largest perfect square in 84, scaling the sides by 2 will produce a triangle of area 21, so that 21 is the area of

4.7 2 ; 12; 25

2 / and, hence, 21 is a congruent number. More generally, we have

Proposition 1.10. Let .a; b; c/ be a Pythagorean triple. If its right triangle

4.a; b; c/ has area m2n, where n is squarefree, then n is a congruent number. Moreover, every squarefree congruent number is obtained in this way.

Proof. Since .a; b; c/ is a Pythagorean triple, 4 D 4.a; b; c/ is a right trian- gle. Now area.4/ D m2n D 1

2 ab, so that

area � 4. a

m ; b m ; c m / �

D 1 2

� a m b m

� D m2n

m2 D n;

and so n is a congruent number.

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18 Chapter 1 Early Number Theory

Conversely, if n is a square-free congruent number, then there are rational

numbers r , s, and t so that

r2 C s2 D t2 1 2 rs D n:

Clearing denominators, we find integers a, b, c, and m so that

a2 C b2 D c2 1 2 ab D m2n:

The first few congruent numbers are

5; 6; 7; 13; 14; 15; 20; 21; 22; 23:

In light of Exercise 1.33 on page 20, we now have a method for determin-

ing all congruent numbers: generate the areas of all Pythagorean triangles (we

know how to do that), and then divide out its largest perfect square factor: case closed.

Not quite. The trouble with this method is that you have no idea how many

triangle areas to calculate before (if ever) you get to an area of m2n for a

particular n. For some congruent numbers, it takes a long time. For example,

157 is a congruent number, but the smallest rational right triangle with area

157 has side lengths

This triangle was found

by Don Zagier, using

sophisticated techniques

investigating elliptic curves,

and using a substantial

amount of computer power

(see [19] for more details).

224403517704336969924557513090674863160948472041

8912332268928859588025535178967163570016480830 ;

6803294847826435051217540

411340519227716149383203 ;

411340519227716149383203

21666555693714761309610 :

A method for effectively determining whether or not an integer is a congru-

ent number is an unsolved problem (this problem is at least a thousand years

old, for historians have found it in manuscripts dating from the late tenth cen- tury). A detailed discussion of the Congruent Number Problem is in [19].A readable account of

the congruent prob-

lem, with more exam-

ples than we provide

here, can be found at

www.math.uconn.edu/

˜kconrad/blurbs/

How to Think About It. Proposition 1.10 shows that every squarefree con-

gruent number n is the area of a scaled Pythagorean triangle. But there might

be more than one Pythagorean triangle whose area has n as its squarefree

part. The search for more than one rational right triangle with the same area

leads to some fantastic calculations. For example, we saw that 5 is the area of

4 � 3 2 ; 20 3 ; 41 6

� , which comes from the Pythagorean triangle 4.9; 40; 41/whose

area is 5 � 62. But 5 is also the area of

4 � 1519

492 ; 4920

1519 ; 3344161

747348

� ;

and this comes from the Pythagorean triangle 4.2420640; 2307361; 3344161/ whose area is 5 � 7473482.

As usual, this isn’t magic; in Chapter 9, we’ll show how to find infinitely

many rational right triangles with the same congruent number as area.

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1.2 Diophantus 19

There’s a surprising connection between congruent numbers and 3-term

arithmetic sequences of perfect squares of rational numbers: positive rationals

s2 < t2 < u2 with u2 � t2 D t2 � s2, like 1; 25; 49 (common difference 24) and

961

36 ;

1681

36 ;

2401

36

(common difference 20). Note that 24 D 4 � 6 and 20 D 4 � 5. So, for these examples, at least, the common difference is 4 times a congruent number. This

suggests that something’s going on. One approach is due to Fibonacci.

Our two equations

a2 C b2 D c2

ab D 2n

might lead us to think that we could find a and b by finding their sum and product, for this would lead to a quadratic equation whose roots are a and b.

Well, we know ab, and

.aC b/2 D a2 C b2 C 2ab D c2 C 4n:

So, aCb D p c2 C 4n (why can we take the positive square root?), and hence

a and b are roots of the quadratic equation

x2 � p c2 C 4n x C 2n:

The quadratic formula gives us a and b:

a D p c2 C 4nC

p .c2 C 4n/� 4.2n/ 2

D p c2 C 4nC

p c2 � 4n

2

and

b D p c2 C 4n�

p .c2 C 4n/ � 4.2n/ 2

D p c2 C 4n�

p c2 � 4n

2 :

But we want a and b to be rational, so we want c2˙4n to be perfect squares. That produces an arithmetic sequence of three perfect squares:

c2 � 4n; c2; c2 C 4n:

There are details to settle, but that’s the gist of the proof of the following theo-

rem.

Theorem 1.11. An integer n is a congruent number if and only of there is a

3-term arithmetic sequence of perfect squares whose common difference is 4n.

Exercises

1.30 * Show that 1 is not a congruent number.

1.31 Show that there are no positive rational numbers x and y so that

x4 ˙ 1 D y2:

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20 Chapter 1 Early Number Theory

1.32 Show that if n is a congruent number and m is an integer, then m2n is also a

congruent number.

1.33 Show that there are no right triangles with rational side-lengths whose area is a

perfect square or twice a perfect square.

1.34 Show that 7 and 14 are congruent numbers.

1.35 Take It Further. Show that 13 is a congruent number.

1.36 * Prove Theorem 1.11.

1.3 Euclid

Euclid of Alexandria (ca. 325 BCE–ca.265 BCE) is one of the most prominent

mathematicians of antiquity. He is best known for The Elements, his treatise consisting of thirteen books: six on plane geometry, four on number theory,

and three on solid geometry. The Elements has been used for over two thou-

sand years, which must make Euclid the leading mathematics teacher of all

time. We do not know much about Euclid himself other than that he taught in

Alexandria, Egypt around 270 BCE. We quote from Sir Thomas Heath [16],

the great translator and commentator on The Elements.

It is most probable that Euclid received his mathematical training in

Athens from the students of PlatoI for most of the geometers who could have taught him were of that school ... Pappus says ... such was .Euclid’s/Pappus (ca. 290 CE–

ca. 350 CE), was one

of the last great classic

geometers.

scrupulous fairness and his exemplary kindliness towards all who could

advance mathematical science to however small an extentI .he was/ in no wise contentious and, though exact, yet no braggart.

Eight hundred years after Euclid, Proclus (412 CE–485 CE) wrote:

Not much younger than these .pupils of Plato/ is Euclid, who put to-

gether The Elements, collecting many of Eudoxus’s theorems, perfecting

many of Theaetetus’s, and also bringing to irrefragable demonstration

the things which were only somewhat loosely proved by his predecessors.

This man lived in the time of the first Ptolemy .323 BCE � 283 BCE/. For Archimedes, who came immediately after the first Ptolemy makes men-

tion of EuclidI and further they say that Ptolemy once asked him if there were a shorter way to study geometry than The Elements, to which he

replied that there was no royal road to geometry. He is therefore younger

than Plato’s circle, but older than Eratosthenes and ArchimedesI for these were contemporaries, as Eratosthenes somewhere says.

The Elements is remarkable for the clarity with which its theorems are stated

and proved. The standard of rigor was a goal (rarely achieved!) for the inven-

tors of calculus centuries later. As Heath writes in the preface to the second

edition of his translation [16] of The Elements,

. . . so long as mathematics is studied, mathematicians will find it neces-

sary and worthwhile to come back again and again . . . to the twenty-two-

centuries-old book which, notwithstanding its imperfections, remains the

greatest elementary textbook in mathematics that the world is privileged

to possess.

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1.3 Euclid 21

More than one thousand editions of The Elements have been published since

it was first printed in 1482. In the Encyclopedia Britannica, van der Waerden

wrote,

Almost from the time of its writing and lasting almost to the present,

The Elements has exerted a continuous and major influence on human

affairs. It was the primary source of geometric reasoning, theorems, and

methods at least until the advent of non-Euclidean geometry in the 19th century. It is sometimes said that, next to the Bible, The Elements may

be the most translated, published, and studied of all the books produced

in the Western world.

Greek Number Theory

In spite of the glowing reviews of The Elements, we must deviate a bit from

Euclid, for the Greeks, and Euclid in particular, recognized neither negative numbers nor zero.

Notation. The natural numbers is the set

N D f0; 1; 2; 3; : : :g:

The set of all integers, positive, negative, and 0, is denoted by The set of integers is denoted by Z because the

German word for numbers

is Zahlen.

Z D f˙n W n 2 Ng:

We are going to assume that the set N of natural numbers satisfies a certain

property—a generalized version of Infinite Descent.

Definition. The Least Integer Axiom (often called the Well-Ordering Axiom)

states that every nonempty collectionC of natural numbers contains a smallest Note that the set of positive rationals QC does not satisfy an analogous

property: the nonempty

subset fx 2QC W x2 > 2g contains no smallest

element.

element; that is, there is a number c0 2 C with c0 � c for all c 2 C .

This axiom is surely plausible. If 0 2 C , then c0 D 0. If 0 … C and 1 2 C , then c0 D 1. If 0; 1 … C and 2 2 C , then c0 D 2. Since C is not empty, you will eventually bump into C , and c0 is the first number you’ll meet.

We now define some familiar terms.

Definition. If a and b are integers, then a divides b, denoted by

a j b;

if there is an integer c with b D ca. We also say that a is a divisor (or a factor) of b, and that b is a multiple of a.

Example 1.12. Consider some special cases. Every number a divides itself, Note that 0 divides itself:

0 j 0 is true. Do not confuse the notation a j b, which is the relation “a is a divisor

of b” with a=b, which is a

number. In particular, we

are not saying that 0=0 is a

number.

for a D a � 1; similarly, 1 divides every number. Every number a divides 0: taking c D 0, we have 0 D a�0. On the other hand, if 0 divides b, then b D 0, for b D 0� c D 0. Note that 3 j 6, because 6 D 3� 2, but 3 − 5 (that is, 3 does not divide 5): even though 5 D 3 � 5

3 , the fraction 5

3 is not an integer. N

Lemma 1.13. If a and b are positive integers and a j b, then a � b.

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22 Chapter 1 Early Number Theory

Proof. There is a positive integer c with b D ca; note that 1 � c, for 1 isInequalities are discussed in Appendix A. 4. the smallest positive integer. Multiplying by the positive number a, we have

a � ac D b.

Every integer a has 1;�1; a;�a as divisors. A positive integer a ¤ 1 having only these divisors is called prime.

Definition. An integer a is prime if a � 2 and its only divisors are ˙1 and ˙a; if a � 2 has other divisors, then it is called composite.

The first few primes are 2, 3, 5, 7, 11, 13; : : : : We will soon see that there

are infinitely many primes.

The reason we do not consider 1 to be a prime is that theorems about primes

would then require special cases treating the behavior of 1. For example, we

will prove later that every positive integer a � 2 has exactly one factorization of the form a D p1p2 � � �pt , where p1 � p2 � � � � � pt are primes. This statement would be more complicated if we allowed 1 to be a prime.

Proposition 1.14. Every integer a � 2 is a product of primes.We allow products having only one factor; it’s okay to

say that a single prime is a

product of primes. Proof. Let C be the set of all natural numbers a � 2 that are not products of primes. If the proposition is false, then C is nonempty, and the Least Integer

Axiom gives a smallest such integer, say, c0. Since c0 2 C , it is not prime; hence, it factors, say, c0 D ab, where a; b ¤ 1. As a j c0, we have a � c0, by Lemma 1.13; but a ¤ c0, lest b D 1, so that a < c0. Therefore, a … C , for c0 is the smallest number in C , and so a is a product of primes: a D p1 � � �pm for m � 1. Similarly, b is a product of primes: b D q1 � � �qn. Therefore, c0 D ab D p1 � � �pmq1 � � �qn is a product of primes, a contradiction, and so C is empty.

Division and Remainders

Dividing an integer b by a positive integer a gives

b=a D q C r=a;

where q is an integer and 0 � r=a < 1. If we clear denominators, we get the statement b D qaC r which involves only integers. For example, 225 D 4C

2 5

becomes 22 D 4 � 5C 2.

b

a a a a r

b a r= 4 +

A A A A

B

Figure 1.8. Division Algorithm.

Euclid viewed division geometrically, as in Figure 1.8. Suppose B is a line

segment of length b, and thatA is a shorter segment of length a. Lay off copies

of A along B as long as possible. If there’s nothing left over, then a is a divisor

of b; if some segment of length, say r , is left over, then r is the remainder.

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1.3 Euclid 23

Theorem 1.15 (Division Algorithm). If a and b are positive integers, then

there are unique .i.e.,exactly one/ integers q .the quotient/ and r .the remain-

der/ with

b D qaC r and 0 � r < a:

Proof. We first prove that q and r exist; afterward, we’ll prove their unique- The hypothesis of Theo-

rem 1.15 can be weakened

to a; b 2 Z and a ¤ 0; the inequalities for the remain-

der now read 0 � r < jaj.

ness.

If b < a, set q D 0 and r D b. Now b D qa C r D 0 � a C b, while 0 � b < a. Hence, we may assume that b � a; that is, b�a � 0. Consider the sequence b > b�a > b�2a > b�3a > � � � . There must be an integer q � 1 with b�qa � 0 and b�.qC1/a < 0 (this is just Infinite Descent, described on page 12; in more down-to-earth language, there can be at most b steps before this sequence becomes negative). If we define r D b � qa, then b D qa C r . If a � b, the quotient q is

the largest multiple qa with

qa � b. This is very much the way young children are

taught to find the integer

quotient in division when a

and b are small.

We also have the desired inequalities. Clearly, 0 � r . If r D b � qa � a, then b � qa � a � 0; that is, b � .q C 1/a � 0, contradicting the definition of q.

Let’s prove uniqueness. If there are integers Q and R with b D Qa C R and 0 � R < a, then qa C r D b D Qa CR and

.Q � q/a D r �R:

If Q ¤ q, there is no loss in generality in assuming that Q > q; that is, 0 < Q � q. By Lemma 1.13, a � .Q � q/a D r � R. But r < a and R � 0 gives r �R < r < a. Therefore, a � .Q � q/a D r � R < a; that is, a < a, a contradiction. Hence Q D q. It follows that R D r , and we are done.

For example, there are only two possible remainders after dividing by 2,

namely, 0 and 1. An integer b is even if the remainder is 0; b is odd if the

remainder is 1. Thus, either b D 2q or b D 2qC 1. The equation b D qa C r is of no value at all without the restriction on

the remainder r . For example, the equations 1000 D 3 � 25C 925 and 1000 D 2 � 53C 894 are true and useless.

How to Think About It. We have been trained to regard the quotient q as

more important than the remainder; r is just the little bit left over. But our viewpoint now is just the reverse. Given a and b, the important question for us

is whether a is a divisor of b. The remainder is the obstruction: a j b if and only if r D 0. This will be a common strategy: to see whether a j b, use the Division Algorithm to get b D qaC r , and then try to show that r D 0.

The next result shows that there is no largest prime. The proof shows, given

any finite set of primes, that there always exists another one.

Corollary 1.16. There are infinitely many primes.

Proof. .Euclid/ Suppose, on the contrary, that there are only finitely many primes. If p1; p2; : : : ; pk is the complete list of all the primes, define

M D .p1 � � �pk/C 1:

By Proposition 1.14, M is a product of primes. But M has no prime divisor

pi , for dividing M by pi gives remainder 1 and not 0. For example, dividing

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24 Chapter 1 Early Number Theory

M by p1 givesM D p1.p2 � � �pk/C 1, so that the quotient and remainder are q D p2 � � �pk and r D 1; dividingM by p2 givesM D p2.p1p3 � � �pk/C 1, so that q D p1p3 � � �pk and r D 1; and so forth. The assumption that there are only finitely many primes leads to a contradiction, and so there must be an

infinite number of them.

Linear Combinations and Euclid’s Lemma

The greatest common divisor of two integers is a fundamental tool in studying

factorization.

Definition. A common divisor of integers a and b is an integer c with c j a and c j b. The greatest common divisor of a and b, denoted by gcd.a; b/ (or, more briefly, by .a; b/ ), is defined by

gcd.a; b/ D ( 0 if a D 0 D b the largest common divisor of a and b otherwise.

We saw, in Lemma 1.13, that if a and m are positive integers with a j m, then a � m. It follows, if at least one of a; b is not zero, that gcd’s exist: there are always common divisors (1 is always a common divisor), and there are

only finitely many positive common divisors � maxfjaj; jbjg.

Lemma 1.17. If p is a prime and b is an integer, then

gcd.p; b/ D ( p if p j b 1 otherwise.

Proof. A common divisor c of p and b is, in particular, a divisor of p. But the only positive divisors of p are p and 1, and so gcd.p; b/ D p or 1; it is p if p j b, and it is 1 otherwise.

If b � 0, then gcd.0; b/ D b (why?).

Definition. A linear combination of integers a and b is an integer of the form

saC tb;

where s; t 2 Z (the numbers s; t are allowed to be negative).

Example 1.18. The equation b D qa C r in the Division Algorithm displays b as a linear combination of a and r (for b D qaC 1 � r ). Note that 0 is a linear combination of any pair of integers: 0 D 0 �aC 0 �b. There are infinitely many linear combinations of 12 and 16, each of which is divisible by 4 (why?). It

follows that 5, for example, is not such a linear combination. N

The next result is one of the most useful properties of gcd’s. See Exercise 1.47 on

page 30. Theorem 1.19. If a and b are integers, then gcd.a; b/ is a linear combination

of a and b.

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1.3 Euclid 25

Proof. We may assume that at least one of a and b is not zero (otherwise,

the gcd is 0 and the result is obvious). Consider the set I of all the linear

combinations of a and b:

I D fsaC tb W s; t 2 Zg:

Both a and b are in I (take s D 1 and t D 0 or vice versa). It follows that The proof of Theorem 1.19 contains an idea that

will be used again, as in

Exercise 1.49 on page 30.

I contains positive integers (if a < 0, then �a is positive, and I contains �a D .�1/aC 0b); hence, the set C of all those positive integers lying in I is nonempty. By the Least Integer Axiom, C contains a smallest positive integer,

In other words, d is the

smallest positive linear

combination of a and b.

say, d ; we claim that d is the gcd.

Since d 2 I , it is a linear combination of a and b: there are integers s and t with

d D saC tb:

We’ll show that d is a common divisor by trying to divide each of a and b

by d . The Division Algorithm gives integers q and r with a D qd C r , where 0 � r < d . If r > 0, then

r D a � qd D a � q.saC tb/ D .1 � qs/aC .�qt/b 2 C;

contradicting d being the smallest element of C . Hence r D 0 and d j a; a similar argument shows that d j b.

Finally, if c is a common divisor of a and b, then Exercise 1.46 on page 29

shows that c divides every linear combination of a and b; in particular, c j d . By Lemma 1.13, we have c � d .

If d D gcd.a; b/ and if c is a common divisor of a and b, then c � d , by Lemma 1.13. The next corollary shows that more is true: c is a divisor of d ;

that is, c j d for every common divisor c.

Corollary 1.20. Let a and b be integers. A nonnegative common divisor d is In some treatments of num- ber theory, Corollary 1.20

is taken as the definition

of gcd. Later, we will want

to define greatest common

divisor in other algebraic

structures. It often will not

make sense to say that one

element of such a structure

is greater than another, but

it will make sense to say

that one element divides

another. Corollary 1.20

will allow us to extend the

notion of gcd.

their gcd if and only if c j d for every common divisor c of a and b.

Proof. Necessity (the implication )). We showed that any common divisor of a and b divides gcd.a; b/ at the end of the proof of Theorem 1.19.

Sufficiency (the implication (). Let d D gcd.a; b/, and let D � 0 be a common divisor of a and b with c j D for every common divisor c of a and b. Now D is a common divisor, so that d j D, by hypothesis; hence, d � D, by Lemma 1.13. But the definition of gcd (d is the greatest common divisor) givesD � d , and so D D d .

The next theorem is of great interest: not only is it very useful, but it also

characterizes prime numbers.

Theorem 1.21 (Euclid’s Lemma). If p is a prime and p j ab for integers a; b, then p j a or p j b. Conversely, if m � 2 is an integer such that m j ab always impliesm j a orm j b, thenm is a prime.

Proof. ()): Suppose that p j ab and that p − a; that is, p does not divide a; we must show that p j b. Since gcd.p; a/ D 1 (by Lemma 1.17), Theorem 1.19 gives integers s and t with 1 D sp C ta. Hence,

b D spb C tab:

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26 Chapter 1 Early Number Theory

Now p divides both expressions on the right, for p j ab, and so p j b, by Exercise 1.46 on page 29.The contrapositive of a

statement P implies

Q is “not Q” implies

“not P ”. For example, the

contrapositive of “If I live

in Chicago, then I live in

Illinois” is “If I don’t live

in Illinois, then I don’t live

in Chicago.” A statement

and its contrapositive are

either both true or both

false. Thus, to prove a

statement, it suffices to

prove its contrapositive.

((): We prove the contrapositive. If m is composite, then m D ab, where a < m and b < m. Now m j m D ab, but m − a and m − b, by Lemma 1.13. Thus, m divides a product but it divides neither factor.

To illustrate: 6 j 12 and 12 D 4� 3, but 6 − 4 and 6 − 3. Of course, 6 is not prime.

We will generalize Euclid’s Lemma in the next chapter. Theorem 2.8 says

that if p is a prime and p j a1 � � �an for integers a1; : : : ; an, where n � 2, then p j ai for some i .

Definition. Call integers a and b relatively prime if their gcd is 1.

Thus, a and b are relatively prime if their only common divisors are ˙1. For example, 2 and 3 are relatively prime, as are 8 and 15.

Here is a generalization of Euclid’s Lemma having the same proof.

Corollary 1.22. Let a, b, and c be integers. If c and a are relatively prime and

c j ab, then c j b.

Proof. Theorem 1.19 gives integers s and t with 1 D sc C ta. Hence, b D scb C tab. Now c divides both expressions on the right, for c j ab, and so c j b, by Exercise 1.46 on page 29.

How to Think About It.

We have just seen one reason why it is important to know proofs: Corol-

lary 1.22 does not follow from the statement of Euclid’s Lemma, but it does follow from its proof. See Exercise 1.54 on page 34 for another example of

this.

Proposition 1.23. Let a and b be integers.

(i) gcd.a; b/ D 1 .that is, a and b are relatively prime/ if and only if 1 is a linear combination of a and b.

(ii) If d D gcd.a; b/ ¤ 0, then the integers a=d and b=d are relatively prime.

Proof. (i) By Theorem 1.19, the gcd d is a linear combination; here, d D 1. Conversely, if 1 D sa C tb and c is a common divisor of a and b, then c j 1, by Exercise 1.46 on page 29. Hence, c D ˙1.

(ii) There are integers s and t with d D sa C tb. Divide both sides by d D gcd.a; b/:

1 D s � a d

� C t

� b

d

� D � s d

� aC

� t

d

� b:

Since d is a common divisor, both a=d and b=d are integers, and part (i)

applies.

Definition. An expression a=b for a rational number (where a and b are inte-

gers and b ¤ 0) is in lowest terms if a and b are relatively prime.

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1.3 Euclid 27

Proposition 1.24. Every nonzero rational number a=b has an expression in

lowest terms.

Proof. If d D gcd.a; b/, then a D a0d , b D b0d , and a b

D a 0d

b0d D a

0

b0 . But

a0 D a d

and b0 D b d

, so gcd.a0; b0/ D 1 by Lemma 1.23.

We can now complete our discussion of Pythagorean triples.

Definition. A Pythagorean triple .a; b; c/ is primitive if a; b; c have no com-

mon divisor d � 2; that is, there is no integer d � 2 which divides each of a; b, and c.

Theorem 1.25 (Diophantus). Every Pythagorean triple .a; b; c/ is similar to

a unique primitive Pythagorean triple.

Proof. We show first that .a; b; c/ is similar to a primitive Pythagorean triple.

If d is a common divisor of a; b; c, then a D du, b D dv, and c D dz, and .u; v; z/ is a Pythagorean triple similar to .a; b; c/ (why?). If d is the largest

common divisor of a; b; c, we claim that .u; v; z/ is primitive. Otherwise, there

is an integer e � 2 with u D eu0, v D ev0, and z D ez0; hence, a D du D deu0, b D dv D dev0, and c D dz D dez0. Thus, de > d is a common divisor of a; b; c, contradicting d being the largest such.

To prove uniqueness, suppose that .a; b; c/ is similar to two primitive Py-

thagorean triples, say .u; v; z/ and .r; s; t/. It follows that the right triangles

4.u; v; z/ and 4.r; s; t/ are similar, and so their sides are proportional, so there is some positive number h with

u D hr; v D hs; and z D ht:

Since the side lengths are integers, h is rational, say h D m=`, and we may assume that it is in lowest terms; that is, gcd.m; `/ D 1. Cross multiply:

mu D `r; mv D `s and mz D `t:

By Corollary 1.22, ` is a common divisor of u, v, and z and m is a common

divisor of r , s, and t . Since both .u; v; z/ and .r; s; t/ are primitive, ` D 1 D m, and so .u; v; z/ D .r; s; t/.

This next result is significant in the history of mathematics.

Proposition 1.26. There is no rational number a=b whose square is 2.

Proof. Suppose, on the contrary, that .a=b/2 D 2. We may assume that a=b is An indirect proof or proof by contradiction has the

following structure. We

assume that the desired

statement is false and

reach a contradiction. We

conclude that the original

statement must be true.

in lowest terms; that is, gcd.a; b/ D 1. Since a2 D 2b2, Euclid’s Lemma gives 2 j a, and so 2m D a. Hence, 4m2 D a2 D 2b2, and 2m2 D b2. Euclid’s Lemma now gives 2 j b, contradicting gcd.a; b/ D 1.

It follows that the legs of a Pythagorean triple .a; b; c/ cannot be equal, for

if a D b, then a2 C a2 D c2, which implies that 2 D .c=a/2.

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28 Chapter 1 Early Number Theory

Proposition 1.26 is often stated as “ p 2 is irrational,” which is a stronger

statement than what we’ve just proved. We can assert that p 2 is irrational only

if we further assume that there exists a number u with u2 D 2. Our proof can be made more elementary; we need assume only that at least

one of a; b is odd. Also, see Exercise 1.75 on page 41.

Historical Note. The ancient Greeks defined number to mean “positive inte- ger.” Rationals were not viewed as numbers but, rather, as ways of comparing

two lengths. They called two segments of lengths a and b commensurable

if there is a third segment of length c with a D mc and b D nc for posi- tive integers m and n. That

p 2 is irrational was a shock to the Pythagoreans

(ca. 500 BCE); given a square with sides of length 1, its diagonal and side are not commensurable; that is,

p 2 cannot be defined in terms of numbers

(positive integers) alone. Thus, there is no numerical solution to the equation

x2 D 2, but there is a geometric solution.To bridge the gap between numbers and geometric

magnitudes, Eudoxus

(408 BCE–355 BCE) intro-

duced the sophisticated

notion of proportions (this

idea, discussed in The Ele-

ments, is equivalent to our

contemporary definition of

real numbers).

By the time of Euclid, around 270 BCE, this problem had been resolved by

splitting mathematics into two disciplines: number theory and geometry.

In ancient Greece, algebra as we know it did not really exist. Euclid and the Greek mathematicians did geometric algebra. For simple ideas, e.g.,

.a C b/2 D a2 C 2ab C b2 or completing the square, geometry clarifies al- gebraic formulas (for example, see the right-hand part of Figure 1.2 on page 2

without the dashed lines). For more difficult ideas, say equations of higher de-

gree, the geometric figures involved are very complicated, so that geometry is no longer clarifying. As van der Waerden writes in [34], p. 266,

one has to be a mathematician of genius, thoroughly versed in trans-

forming proportions with the aid of geometric figures, to obtain results

by this extremely cumbersome method. Anyone can use our algebraic

notation, but only a gifted mathematician can deal with the Greek theory

of proportions and with geometric algebra.

The problem of defining number has arisen several times since the classical

Greek era. Mathematicians had to deal with negative numbers and with com-

plex numbers in the 1500s after the discovery of the Cubic Formula, because

that formula often gives real roots of a cubic polynomial, even integer roots, in

unrecognizable form (see Chapter 3). The definition of real numbers generally accepted today dates from the late 1800s. But there are echos of ancient Athens

in our time. Kronecker (1823–1891) wrote,

Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Men-

schenwerk. (God created the integers; everything else is the work of

Man.)

Even today some logicians argue for a new definition of number.

Exercises

1.37 True or false, with reasons. Of course, it is important to get the right answer, but

most attention should be paid to your reasoning.

(i) 6 j 2. Answer: False. (ii) 2 j 6. Answer: True. (iii) 6 j 0. Answer: True. (iv) 0 j 6. Answer: False. (v) 0 j 0. Answer: True.

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1.3 Euclid 29

1.38 True or false, with reasons.

(i) gcd.n;n C 1/ D 1 for every natural number n. Answer: True. (ii) gcd.n;n C 2/ D 2 for every natural number n. Answer: False.

(iii) 113 is a sum of distinct powers of 2. Answer: True.

(iv) If a and b are natural numbers, there there are natural numbers s and t with

gcd.a; b/ D sa C tb. Answer: False. (v) If an integer m is a divisor of a product of integers ab, then m is a divisor of

either a or b (or both). Answer: False.

1.39 Prove, or disprove and salvage if possible. “Disprove” here means “give a concrete counterex-

ample.” “Salvage” means

“add a hypothesis to make

it true.”

(i) gcd.0; b/ D b (ii) gcd.a2; b2/ D gcd.a; b/2

(iii) gcd.a; b/ D gcd.a; b C ka/ for all k 2 Z (iv) gcd.a; a/ D a (v) gcd.a; b/ D gcd.b; a/

(vi) gcd.a; 1/ D 1 (vii) gcd.a; b/ D � gcd.�a; b/

(viii) gcd.a; 2b/ D 2 gcd.a; b/

1.40 * If x is a real number, let bxc denote the greatest integer n with n � x. (For example, 3 D b�c and 5 D b5c.) If q is the quotient in Theorem 1.15, show that q D bb=ac.

1.41 *

(i) Given integers a and b (possibly negative) with a ¤ 0, prove that there exist unique integers q and r with b D qa C r and 0 � r < jaj.

Hint: Use the portion of the Division Algorithm that has already been proved.

(ii) If b and a are positive integers, do b and �b have the same remainder after dividing by a? Answer: No.

1.42 For each of the following pairs a; b, find the largest nonnegative integer n with

n � b=a < n C 1. (i) a D 4 and b D 5. Answer: n D 1.

(ii) a D 5 and b D 4. Answer: n D 0. (iii) a D 16 and b D 36. Answer: n D 2. (iv) a D 36 and b D 124. Answer: n D 3. (v) a D 124 and b D 1028. Answer: n D 7.

1.43 Let p1; p2; p3; : : : be the list of the primes in ascending order: p1 D 2, p2 D 3, p3 D 5, and so forth. Define fk D 1 C p1p2 � � � pk for k � 1. Find the smallest k for which fk is not a prime.

Hint: 19 j f7, but 7 is not the smallest k.

1.44 What can you say about two integers a and b with the property that a j b and b j a? What if both a and b are positive?

1.45 * Show that if a is positive and a j b, then gcd.a; b/ D a. Why do we assume that a is positive?

1.46 *.Two Out of Three/: Suppose that m, n, and q are integers and m D n C q. If c is an integer that divides any two of m; n; q, show that c divides the third one as

well.

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30 Chapter 1 Early Number Theory

1.47 *

(i) For each a and b, give the smallest positive integer d that can be written as

sa C tb for integers s and t :Allow for positive and negative values of s and t.

� a D 12 and b D 16. Answer: d D 4. � a D 12 and b D 17. Answer: d D 1. � a D 12 and b D 36. Answer: d D 12. � a D 0 and b D 4. Answer: d D 4. � a D 4 and b D 16. Answer: d D 4. � a D 16 and b D 36. Answer: d D 4. � a D 36 and b D 124. Answer: d D 4. � a D 124 and b D 1028. Answer: d D 4.

(ii) How is “smallest positive integer d expressible as saC tb” related to a and b in each case? Is d a divisor of both a and b?

1.48 * Show that the set of all linear combinations of two integers is precisely the set

of all multiples of their gcd.

1.49 * Let I be a subset of Z such that

(i) 0 2 I (ii) if a; b 2 I , then a � b 2 I

(iii) if a 2 I and q 2 Z, then qa 2 I . Prove that there is a nonnegative integer d 2 I with I consisting precisely of all the multiples of d .

1.50 How might one define the gcd.a;b; c/ of three integers? When applied to a prim-

itive Pythagorean triple .a; b; c/, your definition should say that gcd.a;b; c/ D 1.

Euclidean Algorithm

Our discussion of gcd’s is incomplete. What is gcd.12327; 2409/? To ask the question another way, is the expression 2409/12327 in lowest terms? The next

result enables us to compute gcd’s efficiently. We first prove another lemma

from Greek times.

Lemma 1.27. Let a and b be integers.

(i) If b D qa C r , then gcd.a; b/ D gcd.r; a/. (ii) If b � a, then gcd.a; b/ D gcd.b � a; a/.

Proof. (i) In light of Corollary 1.20, it suffices to show that an integer c is a

common divisor of a and b if and only if it is a common divisor of a and r . Since b D qaC r , this follows from Exercise 1.46 on page 29.

(ii) This follows from part (i) because b D 1 � aC .b � a/.

The hypothesis b � a in part (ii) of Lemma 1.27 is not necessary; it is there only to put you in the mood to accept the next example showing a method

the Greeks probably used to compute gcd’s. This method of computation is

nowadays called the Euclidean Algorithm; it is Theorem 1.29.

Example 1.28. In this example, we will abbreviate gcd.b; a/ to .b; a/. Com-

puting .b; a/ is simple when a and b are small. If b � a, then Lemma 1.27

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1.3 Euclid 31

allows us to replace .b; a/ by .b � a; a/; indeed, we can continue replacing numbers, .b�2a; a/, .b�3a; a/; : : : ; .b�qa; a/ as long as b�qa > 0. Since the natural numbers b�a; b�2a; : : : ; b�qa are strictly decreasing, the Least Integer Axiom (or Infinite Descent) says that they must reach a smallest such

integer: r D b�qa; that is, 0 < r < a. Now .b; a/ D .r; a/. (We see the proof of the Division Algorithm in this discussion.) Since .r; a/ D .a; r/ and a > r , they could continue replacing numbers: .a; r/ D .a�r; r/ D .a�2r; r/ D � � � (remember that the Greeks did not recognize negative numbers, so it was nat- ural for them to reverse direction). This process eventually ends, computing

gcd’s; we call it the Euclidean Algorithm. The Greek term for this method is

antanairesis, a free translation of which is “back and forth subtraction.” Let us

implement this idea before we state and prove the Euclidean Algorithm.

Antanairesis computes gcd.326; 78/ as follows:

.326; 78/ D .248; 78/ D .170; 78/ D .92; 78/ D .14; 78/:

So far, we have been subtracting 78 from the other larger numbers. At this

point, we now start subtracting 14 (this is the reciprocal, direction-changing, aspect of antanairesis), for 78 > 14.

.78; 14/ D .64; 14/ D .50; 14/ D .36; 14/ D .22; 14/ D .8; 14/:

Again we change direction:

.14; 8/ D .6; 8/:

Change direction once again to get .8; 6/ D .2; 6/, and change direction one last time to get

.6; 2/ D .4; 2/ D .2; 2/ D .0; 2/ D 2:

Thus, gcd .326; 78/ D 2. The Division Algorithm and Lemma 1.27(i) give a more efficient way of

performing antanairesis. There are four subtractions in the passage from

.326; 78/ to .14; 78/; the Division Algorithm expresses this as

326 D 4 � 78C 14:

There are then five subtractions in the passage from .78; 14/ to .8; 14/; the

Division Algorithm expresses this as

78 D 5 � 14C 8:

There is one subtraction in the passage from .14; 8/ to .6; 8/:

14 D 1 � 8C 6:

There is one subtraction in the passage from .8; 6/ to .2; 6/:

8 D 1 � 6C 2;

and there are three subtractions from .6; 2/ to .0; 2/ D 2:

6 D 3 � 2: N

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32 Chapter 1 Early Number Theory

Theorem 1.29 (Euclidean Algorithm I). If a and b are positive integers,

there is an algorithm computing gcd.a; b/.

Proof. Let us set b D r0 and a D r1, so that the equation b D qa C r reads r0 D q1aC r2. There are integers qi and positive integers ri such that

The beginning of the proof

of the theorem gives the

algorithm.

b D r0 D q1a C r2; r2 < a a D r1 D q2r2 C r3; r3 < r2

r2 D q3r3 C r4; r4 < r3 :::

:::

rn�3 D qn�2rn�2 C rn�1; rn�1 < rn�2 rn�2 D qn�1rn�1 C rn; rn < rn�1 rn�1 D qnrn

(remember that all qj and rj are explicitly known from the Division Algo-

rithm). There is a last remainder: the procedure stops (by Infinite Descent!)

because the remainders form a strictly decreasing sequence of nonnegative in-

tegers (indeed, the number of steps needed is less than a).

Lamé (1795–1870) proved

that the number of steps

in the Euclidean Algorithm

cannot exceed 5 times

the number of digits in the

smaller number (see [26],

p. 49).

We now show that the last remainder rn is the gcd.

b D q1aC r2 ) gcd.a; b/ D gcd.a; r2/ a D q2r2 C r3 ) gcd.a; r2/ D gcd.r2; r3/ r2 D q3r3 C r4 ) gcd.r2; r3/ D gcd.r3; r4/

:::

rn�2 D qn�1rn�1 C rn ) gcd.rn�2; rn�1/ D gcd.rn�1; rn/ rn�1 D qnrn ) gcd.rn�1; rn/ D rn:

All the implications except the last follow from Lemma 1.27. The last one

follows from Exercise 1.45 on page 29.

Let’s rewrite the previous example in the notation of the proof of Theo- rem 1.29. The passage from one line to the line below it involves moving the

boldface numbers “southwest.”

326 D 4 � 78 C 14 (1.7) 78 D 5 � 14 C 8 (1.8) 14 D 1 � 8 C 6 (1.9) 8 D 1 � 6 C 2 (1.10) 6 D 3 � 2:

The Euclidean Algorithm also allows us to find a pair of integers s and t

expressing the gcd as a linear combination.

Theorem 1.30 (Euclidean Algorithm II). If a and b are positive integers,

there is an algorithm computing a pair of integers s and t with gcd.a; b/ D saC tb.

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1.3 Euclid 33

Proof. It suffices to show, given equations

b D qaC r a D q0r C r 0

r D q00r 0 C r 00;

how to write r 00 as a linear combination of b and a (why?). Start at the bottom, and write

r 00 D r � q00r 0:

Now rewrite the middle equation as r 0 D a � q0r , and substitute:

r 00 D r � q00r 0 D r � q00.a � q0r/ D .1 � q00q0/r � q00a:

Now rewrite the top equation as r D b � qa, and substitute:

r 00 D .1 � q00q0/r � q00a D .1 � q00q0/.b � qa/ � q00a:

Thus, r 00 is a linear combination of b and a.

We use the equations to find coefficients s and t expressing 2 as a linear

combination of 326 and 78. Work from the bottom up.

2 D 8 � 1 � 6 by Eq. (1.10) D 8 � 1 � .14 � 1 � 8/ by Eq. (1.9) D 2 � 8 � 1 � 14 D 2 � .78 � 5 � 14/ � 1 � 14 by Eq. (1.8) D 2 � 78 � 11 � 14 D 2 � 78 � 11 � .326 � 4 � 78/ by Eq. (1.7) D 46 � 78 � 11 � 326:

Thus, s D 46 and t D �11.

How to Think About It. The algorithm produces one pair of coefficients that

works. However, it’s not the only pair. For example, consider gcd.2; 3/ D 1. A moment’s thought gives s D �1 and t D 1; but another moment’s thought gives s D 2 and t D �1 (see Exercise 1.57 on page 35). However, the Eu- clidean Algorithm always produces a specific pair of coefficients; assuming

that no mistakes in arithmetic are made, two people using the algorithm al-

ways come up with the same s and t .

Students usually encounter greatest common divisors in elementary school, sometimes as early as the fifth grade, when they learn how to add fractions

and put the sum in lowest terms. As we have seen, putting a fraction in lowest Putting a fraction in lowest terms is not always wise.

For example,

2 3 C 1

5 D 10

15 C 3

15

D 13 15

:

terms involves the gcd of numerator and denominator. The preferred method

of finding gcd’s in early grades involves prime factorization, for if integers a

and b are small, then it is easy to factor them into primes: after several cancel- lations, the expression a=b is in lowest terms. Pedagogically, this may be the

right choice, but finding gcd’s using prime factorization is practical only when

numbers are small; can you put the fraction 167291=223377 in lowest terms

using prime factorization?

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34 Chapter 1 Early Number Theory

How to Think About It. In calculating gcd’s with the Euclidean Algorithm,

many students get confused keeping track of the divisors and remainders. We

illustrate one way to organize the steps that has been effective with high school

students. Arrange the steps computing gcd.124; 1028/ as on the left:

8

124

� 1028

992 3

36

� 124

108 2

16

� 36

32 4

4

� 16

16

0

4 D 36 � 2 � 16 &

D 36 � 2 � .124 � 3 � 36/ D �2 � 124C 7 � 36

& D �2 � 124C 7 � .1028 � 8 � 124/ D 7 � 1028� 58 � 124

The last nonzero remainder is the gcd, so gcd.124; 1028/ D 4. This arrange- ment can be used to read off coefficients s and t so that 4 D 124sC128t . Start at the next to last division and solve for each remainder.

Exercises

1.51 If a and b are positive integers, then gcd.a; b/ D sa C tb. Prove that either s or t is negative.

1.52 * Use Infinite Descent to prove that every positive integer a has a factorization

a D 2km, where k � 0 and m is odd. Now prove that p

2 is irrational using this

fact instead of Euclid’s Lemma.

1.53 Prove that if n is squarefree (i.e., n > 1 and n is not divisible by the square of

any prime), then there is no rational number x with x2 D n. Hint: Adapt the proof of Proposition 1.26.

1.54 * Assuming there is a real number x with x3 D 2, prove that x is irrational.

1.55 (i) Find d D gcd.326; 78/, find integers s and t with d D 326s C 78t , and put the expression 326/78 in lowest terms.

Answer: d D 2, s D �11, t D 46, and 163 39

.

(ii) Find d D gcd.12327;2409/, find integers s and t with d D 12327s C 2409t , and put the expression 2409/12327 in lowest terms.

Answer: d D 3, s D 299, t D �1530, and 8034109 . (iii) Find d D gcd.7563;526/, and express d as a linear combination of 7563 and

526.

Answer: d D 1, s D �37, t D 532. (iv) Find d D gcd.73122; 7404621/ and express d as a linear combination of

73122 and 7404621.

Answer: d D 21, s D 34531, t D �7404621.

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1.3 Euclid 35

1.56 * Prove that if gcd.r; m/ D 1 and gcd.r 0 ; m/ D 1, then gcd.rr 0 ; m/ D 1. Con- clude that if both r and r 0 are relatively prime to m, then so is their product rr 0 .

Hint: If ar C bm D 1 and sr 0 C tm D 1, consider .ar C bm/.sr 0 C tm/.

1.57 * Let a; b, and d be integers. If d D sa C tb, where s and t are integers, find infinitely many pairs of integers .sk ; tk/ with d D ska C tkb.

Hint: If 2s C 3t D 1, then 2.s C 3/ C 3.t � 2/ D 1.

1.58 * If a and b are relatively prime and each divides an integer n, prove that their

product ab also divides n.

Hint: Use Corollary 1.22.

1.59 If m > 0, prove that m gcd.b; c/ D gcd.mb; mc/. (We must assume that m > 0 lest m gcd.b; c/ be negative.)

Hint: Show that if k is a common divisor of mb and mc, then k j m gcd.b; c/.

1.60 Write d D gcd.a; b/ as a linear combination of a and b. (i) a D 4 and b D 16.

Answer: d D 4 D 5 � 4 C .�1/ � 16 (or, 4 D 1 � 4 C 0 � 16). (ii) a D 16 and b D 36.

Answer: d D 4 D .�2/ � 16 C 1 � 36. (iii) a D 36 and b D 124.

Answer: d D 4 D 7 � 36 C .�2/ � 124. (iv) a D 124 and b D 1028.

Answer: d D 4 D .�58/ � 124 C 7 � 1028.

1.61 Given integers a, b, and c with c j a and c j b, prove that c divides every linear combination sa C tb.

1.62 Is anything wrong with this calculation? Explain your answer.

4

7 �

37 28

9

1.63 Given integers b, c, d , and e satisfying b D 7c C 2 and d D 7e C 4, (i) What’s the remainder when b C d is divided by 7?

Answer: 6.

(ii) What’s the remainder when bd is divided by 7?

Answer: 1.

(iii) Explain your answers.

1.64 A lattice point is a point .x; y/ in the plane with both x and y integers.

(i) Which lattice points are on the line whose equation is 4x C 6y D 24? (ii) Which lattice points are on the line whose equation is 3x C 6y D 24?

(iii) Find a line whose equation has integer coordinates but that never passes

through a lattice point.

(iv) Explain how to tell whether the line with equation y D axCb contains lattice points.

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36 Chapter 1 Early Number Theory

1.65 Consider the calculation of gcd.124; 1028/ on page 34. Show that the integer pairs

.124; 1028/; .36; 124/; .16; 36/; .4; 16/; .0; 4/

have the same greatest common divisor.

1.66 Most calculators have functions computing quotients and remainders. Let r.b; a/

denote the remainder when b is divided by a, and let q.b; a/ denote the quotient.

Find r.b; a/ and q.b; a/ if

(i) a D 12, b D 16. Answer: q.16;12/ D 1; r.16; 12/ D 4. (ii) a D 16, b D 12. Answer: q.12; 16/ D 0; r.12; 16/ D 12.

(iii) a D 124, b D 1028. Answer: q.1028;124/ D 8; r.1028; 24/ D 36. (iv) a D 78, b D 326. Answer: q.326;78/ D 4; r.326; 78/ D 14.

1.67 Preview. Using the notation in Exercise 1.66, consider the pair of recursivelyThe functions can be pro- grammed into a calculator. defined functions on N:

s.a; b/ D (

0 a D 0 t .r.b; a/; a/ � q.b; a/ � s .r.b; a/; a/ a > 0

t.a; b/ D (

1 a D 0 s .r.b; a/; a/ a > 0:

Find s.a; b/ and t .a; b/ if

(i) a D 124; b D 1028. Answer: s.124; 1028/ D �58; t .124; 1028/ D 7. (ii) a D 36; b D 124. Answer: s.36; 124/ D 7; t .36;124/ D �2.

(iii) a D 78; b D 326. Answer: s.78; 326/ D 46; t .78; 326/ D �11. (iv) a D 12327;b D 2409. Answer: s.1237;2409/ D 1186; t .1237;2409/ D

�609. (v) a D 7563; b D 526. Answer: s.7563;526/ D �37; t .7563;526/ D 532.

(vi) a D 167291;b D 223377. Answer: s.167291;223377/ D �4, t .167291;223377/ D 3.

1.4 Nine Fundamental Properties

We now focus on a small number (nine) of properties of arithmetic, for it turns

out that many of the usual rules follow from them. This obviously simplifies

things, making explicit what we are allowed to assume. But we have an ulterior

motive. The properties will eventually be treated as axioms that will describe

addition and multiplication in other systems, such as complex numbers, poly- nomials, and modular arithmetic; these systems lead naturally to their common

generalization, commutative rings.

Notation. The set of all rational numbers is denoted by Q, and the set of all

real numbers is denoted by R.

We begin by stating some basic properties of real numbers (of course, inte-

gers and rationals are special cases). These properties undergird a great deal of high school algebra; they are essential for the rest of this book and, indeed, for

abstract algebra.

Addition and multiplication are functions R � R ! R, namely, .a; b/ 7!Functions are discussed in Appendix A.1. a C b and .a; b/ 7! ab. The Laws of Substitution say that if a; a0; b; b0 are

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1.4 Nine Fundamental Properties 37

real numbers with a D a0 and b D b0, then

a C b D a0 C b0 and ab D a0b0:

The Laws of Substitution are used extensively (usually tacitly) when solving

equations or transforming expressions, and they merely say that addition and

multiplication are single-valued. For example, since �5C 5 D 0, we have

.�5C 5/ � .�1/ D 0 � .�1/ D 0:

Here are the properties of addition we are emphasizing.

Addition W For all real numbers a; b, and c,

(i) Commutativity: aC b D b C a, (ii) 0C a D a,

(iii) there is a number �a, called the negative of a (or its additive inverse), with �a C a D 0,

(iv) Associativity: a C .b C c/ D .a C b/C c.

Let’s say a bit more about associativity. Addition is defined as an operation

performed on two numbers at a time, but it’s often necessary to add three or

more numbers. Associativity says that, when evaluating, say 2 C 5 C 3, we Given associativity for the sum or product of

3 numbers, generalized

associativity is also true:

we don’t need parentheses

for the sum or product of

n � 3 numbers. A proof is in Appendix A.5.

can first add 2 and 5, giving 7 C 3 D 10, or we can first add 5 and 3, giving 2 C 8 D 10. In other words, we don’t need parentheses: writing 2 C 5 C 3 is unambiguous because .2 C 5/ C 3 D 2 C .5 C 3/. This is not the case with subtraction. What is 8� 3� 2? If we first subtract 8 � 3, then the answer is 5 � 2 D 3. However, if we evaluate 8 � .3 � 2/ D 8 � 1, we obtain a different answer. Thus, subtraction R2 ! R, defined by .a; b/ 7! a�b, is not associative, and we do need parentheses for it.

Here are the properties of multiplication that we are emphasizing; note that

they are, formally, the same as those for addition: just replace “plus” by “times”

(we usually denote the product of numbers a and b by ab, although we will

occasionally write a � b or a � b).

Multiplication W For all real numbers a; b, and c,

(i) Commutativity: ab D ba, (ii) 1 � a D a,

(iii) If a ¤ 0, there is a number a�1, called its (multiplicative) inverse (or its reciprocal) with a � a�1 D 1, Why do we assume that

a ¤ 0? Read on. (iv) Associativity: a.bc/ D .ab/c.

Finally, we highlight a property involving both addition and multiplication.

Distributivity: a.b C c/ D ac C ab:

Reading from left to right, distributivity says that we can “multiply a through;”

reading from right to left, distributivity says that we can “factor a out.”

Aside from the two Laws of Substitution, one for addition and one for mul-

tiplication, we have now listed nine properties of addition and multiplication.

Subtraction and division are defined as follows.

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38 Chapter 1 Early Number Theory

Definition. If a and b are numbers, define subtraction by

b � a D b C .�a/;

where �a is the negative of a; that is, �a is the number which, when added to a, gives 0.

Quotient (or division) is defined similarly.

Definition. If a and b are numbers with b ¤ 0, then the quotient of a by b is ab�1, where b�1 is the number which, when multiplied by b, gives 1. We often denote ab�1 by a=b.

The word quotient is used here in a different way than in the Division Algo-

rithm, where it is bb=ac, the integer part of b=a (see Exercise 1.40 on page 29).

How to Think About It. Almost all the properties just listed for the set R

of real numbers also hold for the set Z of integers—these properties are “in-

herited” from R because integers are real numbers. The only property that Z

doesn’t inherit is the existence of multiplicative inverses. While every nonzero integer does have an inverse in R, it may not be an integer; in fact, the only

nonzero integers whose inverses also lie in Z are 1 and �1. There are other familiar algebraic systems that are more like Z than R in the sense that multi-

plicative inverses may not exist in the system. For example, all polynomials in

one variable with rational coefficients form such a system, but the multiplica-

tive inverse 1=x of x is not a polynomial.

Other familiar “rules” of arithmetic are easy consequences of these fundamen-

tal ones. Here are some of them.

Proposition 1.31. For every number a, we have 0 � a D 0.

Proof. By Addition Rule (ii), we have 0C 0 D 0. Therefore,

0 � a D .0C 0/a D .0 � a/C .0 � a/:

Now subtract 0 � a from both sides to obtain 0 D 0 � a.

What is the meaning of division by 5? When we say that 20 � 5 D 4, we mean that 20=5 is a number (namely 4), and that .20=5/ � 5 D 20. Dividing is the “opposite” of multiplying: dividing by 5 undoes multiplying by 5. This agrees with our formal definition. The inverse of 5 is 5�1, and 20 � 5�1 D 20=5 D 4.Multiplication by 5 is a

bijection, and we are

saying that division by 5

is its inverse function.

See Example A.10 in

Appendix A. 1.

Can we divide by zero? If so, then 1=0would be a number with 0� .1=0/ D 1. But we have just seen that 0 � a D 0 for any number a. In particular, 0 � .1=0/ D 0, giving the contradiction 1 D 0. It follows that 1=0 D 0�1 is not a number; we cannot divide by 0.

Here is another familiar consequence of the nine fundamental properties.

Proposition 1.32. For a number a, we have

.�a/ � .�1/ D a:

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1.4 Nine Fundamental Properties 39

In particular,

.�1/ � .�1/ D 1:

Proof. The distributive law gives The Law of Substitution allows us to replace 0 by

�1C 1.0 D 0 � .�a/ D .�1 C 1/.�a/ D .�1/ � .�a/C .�a/:

Now, add a to both sides to get a D .�1/.�a/.

How to Think About It.

Even though its proof is very simple, Proposition 1.32 is often presented to

high school students as something mysterious and almost magical. We can only guess at a reason. From Euclid’s time until the 1500s, numbers were always

positive; either negative numbers were not recognized at all or, if they did

appear, they were regarded with suspicion, as not being bona fide (the complex

numbers, which came on the scene around the same time, were also suspected

of witchcraft). In the proof of Proposition 1.32, we treated negative numbers

without prejudice, and we assumed that they obey the same elementary rules as positive numbers do. And we have reaped a reward for clear thinking.

Addition Rule (iii) states that every real number has a negative, an additive

inverse. Can a number a have more than one negative? Intuition tells us no,

and this can be proved using the nine fundamental properties.

Proposition 1.33. Negatives in R are uniqueI that is, for a 2 R, there is exactly one number b in R with b C a D 0.

Multiplicative inverses of nonzero real numbers are uniqueI that is, for nonzero c 2 R, there is exactly one real number d with cd D 1.

Proof. Suppose b is a number with aC b D 0. Add �a to both sides:

�aC .a C b/ D �a:

We can now use associativity to calculate, like this.

�aC .a C b/ D �a .�a C a/C b D �a

0C b D �a b D �a:

This argument can be adapted to prove uniqueness of multiplicative in- verses; merely replace + by � and “additive inverse” by “multiplicative in- verse.”

Uniqueness theorems like Proposition 1.33 are useful because they show

that certain objects are characterized by their behavior. For example, to show

that a number b is equal to �a, add b to a and see if you get 0. This is the strategy in the next proof.

Corollary 1.34. For every real number a, we have �a D .�1/a. Similarly, if b ¤ 0, then .b�1/�1 D b.

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40 Chapter 1 Early Number Theory

Proof. We add .�1/a to a and see if we get 0.

.�1/a C a D .�1/a C 1 � a D a.�1C 1/ D a � 0 D 0:

We do get 0, and so Proposition 1.33 guarantees that �a D .�1/a. To prove the second statement, interpret the equation bb�1 D 1 as saying

that b is an element which, when multiplied by b�1, gives 1.

We can now prove the distributive law for subtraction.

Corollary 1.35. If a; b; c are real numbers, then a.b � c/ D ab � ac.

Proof. By definition, b � c D b C .�c/. But b � c D b C .�1/c, by Corol- lary 1.34. Therefore, distributivity gives

a.b � c/ D a � b C .�1/c

�

D ab C a.�1/c D ab C .�1/.ac/ D ab � ac:

We have just displayed some properties of addition and multiplication of real numbers following from the nine fundamental properties. The proofs fol-

low only from the nine properties; we did not use any other properties of R,

such as decimal expansions or inequalities. Hence, if we show, for example,

that addition and multiplication of complex numbers or of polynomials satisfy

the nine properties, then each of these systems satisfy the “other properties,”

Propositions 1.31, 1.32, and 1.33, as well.

Exercises

1.68 (i) Prove the additive cancellation law using only the nine properties: if a; b; c

are real numbers with a C c D b C c, then a D b. (ii) Prove the multiplicative cancellation law for real numbers using only the nine

properties: if a; b; c are real numbers with ac D bc and c ¤ 0, then a D b.

1.69 Suppose that b ¤ 0. Show that a=b is the unique real number whose product with b is a.

1.70 (i) Prove that a real number a is a square if and only if a � 0. (ii) Prove that every complex number is a square.

1.71 * Let a; b; c be numbers.

(i) Prove that �ac, the negative of ac, is equal to .�a/c; that is, acC.�a/c D 0. (ii) In the proof of Corollary 1.35, we stated that

ab C a.�1/c D ab � ac:

Prove this.

Hint: Evaluate a.0 C 0/ in two ways.

1.72 * Suppose that e and f are integers and let m D minfe; f g and M D maxfe; f g. Show that

m C M D e C f:

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1.5 Connections 41

1.73 *

(i) If a is a positive real number such that an D 1 for an integer n � 1, prove that a D 1.

(ii) If a is a real number such that an D 1 for an integer n � 1, prove that a D ˙1.

1.74 The Post Office has only 5 and 8 cent stamps today. Which denominations of

postage can you buy?

1.75 * Later in this book, we’ll prove Theorem 2.10: every integer can be factored into

primes in essentially only one way. You may use this theorem here.

(i) If a 2 Z, prove that every prime p that divides a2 shows up with even expo- nent; that is, if p j a2, then p2 j a2.

(ii) Show that there are no integers a and b so that 2a2 D b2. (iii) Use part(ii) to show that there is no rational number x with x2 D 2.

1.76 Use Euclid’s idea of a geometric Division Algorithm (see Figure 1.8 on page 22)

to give a geometric version of the Euclidean Algorithm that uses repeated geo-

metric division. Apply your geometric algorithm to

(i) two segments of length 12 and 90.

(ii) the diagonal and the side of a square.

1.5 Connections

This section applies the method of Diophantus to trigonometry and to calculus.

Trigonometry

The formulas x D .1 � t2/=.1 C t2/ and y D 2t=.1 C t2/, where t is a real number, parametrize all the points on the unit circle except .�1; 0/. But we know that if A D .x; y/ is a point on the unit circle, then x D cos � and y D sin � , where � D †DOA (see Figure 1.9).

O D

A = (cos , sin )q q

1

q

Figure 1.9. Cosine and Sine.

If � D 30o, then .cos �; sin �/ D . p 3 2 ; 1 2 /; one coordinate is irrational and

one is rational. Are there any acute angles � with both cos � and sin � rational?

If .x; y/ is the Pythagorean point arising from .3; 4; 5/, then x D cos � D 3 5

and y D sin � D 4 5

. With a little more work, we can prove that there are

infinitely many angles � with both cos � and sin � rational (is it obvious that

Pythagorean triples arising from distinct Pythagorean points are not similar?)

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42 Chapter 1 Early Number Theory

and also infinitely many angles with both cosine and sine irrational (see Exer-

cise 1.29 on page14).

The parametrization of the unit circle in Proposition 1.2,

cos � D 1 � t 2

1C t2 and sin � D 2t

1C t2 ; �1 < t < 1;

enables us to prove some trigonometric identities. For example, let’s prove the

identity

1C cos � C sin � 1C cos � � sin � D sec � C tan �:

First, rewrite everything in terms of sin � and cos � . The left-hand side is fine; the right-hand side is .1= cos �/ C .sin �= cos �/. Now replace these by their formulas in t . The left-hand side is

1C 1�t2 1Ct2 C

2t 1Ct2

1C 1�t2 1Ct2 �

2t 1Ct2

;

and this simplifies to a rational function of t (that is, a quotient of two poly-

nomials). Similarly, the right-hand side is also a rational function of t , for

sec � D 1 cos �

D 1C t 2

1 � t2 and tan � D 2t

1 � t2 . Thus, verifying whether the trigonometric identity is true is the same thing as verifying whether one ra-

tional expression is equal to another. This problem involves no ingenuity at

all. Just cross multiply and check whether the polynomials on either side are

equal; that is, check whether the monomials on either side having the same

degree have the same coefficients.

Integration

The parametrization of the unit circle is useful for certain integration problems.

In Figure 1.10, we see that 4AOB is isosceles, for two sides are radii; thus, the

q

O

B = (cos , sin )q q

(–1,0) = A

Figure 1.10. Tangent Half-Angle.

We denote the line de-

termined by points A

and B by L.A; B/, but

the notation ! AB is the

convention in geometry

and precalculus books.

base angles are equal. But the exterior angle � is their sum, and so †BAO D �=2. Therefore,

t D sin � 1C cos � D slope L.A; B/ D tan.�=2/I

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1.5 Connections 43

t D tan.�=2/ is called the tangent half-angle formula. Now

� D 2 arctan t and d� D 2 dt 1C t2

:

Let’s apply this substitution. In most calculus courses, the indefinite integralR sec � d� D log j sec � C tan � j is found by some unmotivated trick, but this

integration is quite natural when we use the method of Diophantus.

Z sec � d� D

Z d�

cos � D Z 1C t2 1 � t2

� 2 dt 1C t2

D Z

2 dt

1 � t2 :

Since

2

1 � t2 D 1 1C t

C 1 1 � t

;

we have Z

2 dt

1 � t2 D Z

dt

1C t C Z

dt

1 � t D log j1C t j � log j1� t j:

The hard part is now done;

log j1C t j � log j1 � t j D log ˇ̌1C t 1 � t

ˇ̌ ;

and it is cosmetic to rewrite, using the formula relating t and �

1C t 1 � t D

.1 C t/2 1 � t2 D

1C 2t C t2 1 � t2 D

1C t2 1 � t2 C

2t

1 � t2 D sec � C tan �:

Other integrands can also be integrated using the tangent half-angle formula

(see Exercise 1.78 below). Similar parametrizations of other conic sections also

lead to integration formulas (see Exercises 1.80–1.82 below and [28, pp. 86–

97]).

Exercises

1.77 Verify the following trigonometric identities.

(i) 1 C csc � D cos � cot � 1 � sin �

.

(ii) 1

csc � � cot � � 1

csc � C cot � D 2 cot � .

(iii) cot4 � C cot2 � D csc4 � � csc2 �:

1.78 Integrate the following using the tangent half-angle formula.

(i)

Z sin �

2 C cos � d� .

Answer: ln ˇ̌ ˇ 1Ct2 3Ct2

ˇ̌ ˇ, where t D sin �

1Ccos � .

(ii)

Z sin � � cos � sin � C cos � d� .

Answer: ln ˇ̌ ˇ 1Ct2 1C2tDt2

ˇ̌ ˇ, which leads to � ln j cos � C sin � j.

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44 Chapter 1 Early Number Theory

1.79 * Preview.

(i) Sketch the graph of x2 � xy C y2 D 1. (ii) Find a “sweeping lines” parametrization for the points on the graph of

x2 � xy C y2 D 1. (iii) Find a scalene triangle with integer side lengths and a 60ı angle.

1.80 Take It Further.

(i) Find a “sweeping lines” parametrization for the points on the graph of the

parabola x D y2, using lines joining A D .0; 0/ to points P D .x; y/ on the parabola.

(ii) Use this parametrization to evaluate

Z dx

1 C p

x .

1.81 Take It Further. Show that a “sweeping lines” parametrization for the points

on the ellipse x2=a2 C y2=b2 D 1, using lines joining A D .�a; 0/ to points P D .x; y/ on the ellipse, is

x D a.b 2 � a2t2/

b2 C a2t2 and

2ab2t

b2 C a2t2 :

1.82 Take It Further. Show that a “sweeping lines” parametrization for the points on

the hyperbola x2=a2 � y2=b2 D 1, using lines joining A D .�a; 0/ to points P D .x; y/ on the hyperbola, is

x D a.b 2 C a2t2/

b2 � a2t2 and y D 2ab

2t

b2 � a2t2 :

1.83 * Take It Further. Most high school texts derive the quadratic formula by “com-

pleting the square,” a method we’ll discuss and generalize in Chapter 3. Here’s

another way to derive the formula.

(i) Show that if r and s are the roots of x2 C bx C c D 0, then

r C s D �b and rs D c

(ii) If r C s D �b and rs D c, show that

.r � s/2 D b2 � 4c;

so that r � s D ˙ p

b2 � 4c. (iii) Solve the system

r C s D �b

r � s D ˙ p

b2 � 4c

for r and s.

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2 Induction In Chapter 1, we proved some basic theorems of ordinary arithmetic: Division

Algorithm; Euclidean Algorithm; prime factorization. We are now going to

prove the Fundamental Theorem of Arithmetic: any two people writing an in- teger as a product of primes always get the same factors. We need a very useful We shall see later that

many interesting number

systems do not have

unique factorization.

Indeed, not recognizing

this fact is probably

responsible for many false

“proofs” of Fermat’s Last

Theorem.

tool in order to do this, and so we interrupt our historical account to introduce

mathematical induction, a method of proof that finds application throughout

mathematics. We’ll go on here to use induction to discuss the Binomial Theo-

rem and some combinatorics.

2.1 Induction and Applications

The term induction has two meanings. The most popular one is inductive rea-

soning: the process of inferring a general law from the observation of partic-

ular instances. For example, we say that the Sun will rise tomorrow morning

because, from the dawn of time, the Sun has risen every morning. Although this notion of induction is used frequently in everyday life, it is not adequate

for mathematical proofs, as we now show.

Consider the assertion: “f .n/ D n2 � n C 41 is prime for every positive integer n.” Evaluating f .n/ for n D 1; 2; 3; : : : ; 40 gives the numbers

41; 43; 47; 53; 61; 71; 83; 97; 113; 131;

151; 173; 197; 223; 251; 281; 313; 347; 383; 421;

461; 503; 547; 593; 641; 691; 743; 797; 853; 911;

971; 1033; 1097; 1163; 1231; 1301; 1373; 1447; 1523; 1601:

It is tedious, but not very difficult (see Exercise 2.2 on page 52), to show that every one of these numbers is prime. Inductive reasoning leads you to expect

that all numbers of the form f .n/ are prime. But the next number, f .41/ D 1681, is not prime, for f .41/ D 412 � 41 C 41 D 412, which is obviously composite.

An even more spectacular example of the failure of inductive reasoning is given by the harmonic series 1 C 1

2 C 1

3 C � � � C 1

n C � � � , which diverges

(first proved by Oresme (ca.1320–1382)), and so its partial sums get arbitrarily

large. Given a number N , there is a partial sum

†m D mX

nD1

1

n D 1C 1

2 C � � � C 1

m

45

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46 Chapter 2 Induction

with †m > N . A high school student, unaware of this, playing with his cal-

culator and seeing that†315 � 6:33137, would probably make the reasonable guess that †m < 100 for all m. But he’s wrong; the series diverges! It is known that if m < 1:5 � 1043, then †m < 100. The most generous estimate of the age of the Earth is ten billion (10,000,000,000) years, or 3:65 � 1012 days, a number insignificant when compared to 1:5� 1043. Therefore, starting from the Earth’s very first day, if the statement †m < 100 was verified on the

mth day, then there would be today as much evidence of the general truth of these statements as there is that the Sun will rise tomorrow morning. And yet

most statements †m < 100 are false!

Inductive reasoning is valuable in mathematics, as it is in natural science,

because seeing patterns in data often helps us guess what may be true in general

(see Exercise 2.1 on page 52, for example). However, merely checking whether the first few (or first few trillion) statements are true is not enough. We have

just seen that checking the first 1:5� 1043 statements is inadequate to establish a general rule.

Let’s now discuss mathematical induction. Suppose we are given a se-

quence of statements

S.1/; S.2/; S.3/; : : : ; S.n/; : : : :

For example, the formula 2n > n for all n � 1 can be viewed as the sequence of statements

21 > 1; 22 > 2; 23 > 3; : : : ; 2n > n; : : : :

Mathematical induction is a technique for proving that all the statements are

true.

The key idea is just this. Imagine a stairway to the sky. We claim that if

its bottom step is white and the next step above any white step is also white,

then all the steps of the stairway are white. Here’s our reasoning. If some steps aren’t white, walk up to the first non-white step; call it Fido. Now Fido can’t

be at the bottom, for the bottom step is white, and so there is a step just below

Fido. This lower step must be white, because Fido is the first non-white one.

But Fido, being the next step above a white step, must also be white. This is a

contradiction; there is no Fido. All the steps are white.

To sum up, given a list of statements, we are claiming that if

(i) the first statement is true, and

(ii) whenever a statement is true, so is the next one,

then all the statements on the list are true.

Let’s apply this idea to the list of inequalities S.n/W 2n > n. Now S.1/ is true, for 21 D 2 > 1. Suppose we believe, for every n > 1, that the implication 2n�1 > n � 1 ) 2n > n is true. Since S.1/ is true and S.1/ ) S.2/ is true,The symbol) means

implies. we have S.2/ true; that is, if 21 > 1 and 21 > 1 ) 22 > 2 are both true, then 22 > 2. Since 22 > 2 is true and 22 > 2 ) 23 > 3 is true, we have 23 > 3; since 23 > 3 is true and 23 > 3 ) 24 > 4 is true, we have 24 > 4; and so forth. Mathematical induction replaces the phrase and so forth with statement (ii), which guarantees, for every n, that there is never an obstruction

in the passage from the truth of any statement S.n � 1/ to the truth of the next one S.n/. We will prove 2n > n for all n � 1 in Proposition 2.2.

Here is the formal statement of mathematical induction.

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2.1 Induction and Applications 47

Theorem 2.1 (Mathematical Induction). Let k be an integer. If S.k/;

S.k C 1/; S.k C 2/; : : : is a sequence of statements such that

(i) Base StepW S.k/ is true, and (ii) Inductive StepW If, for n > k, S.n � 1/ being true implies S.n/ true,

then the statements S.n/ are true for all n � k. Many people prefer to

write the inductive step as

S.n/) S.nC 1/ instead of S.n � 1/ ) S.n/ as we do. The difference is

cosmetic; the important

thing is the passage from

one statement to the next

one.

We’ll prove this in Theorem 2.17 (you’ll see then that the proof is our story about Fido), but let’s use the theorem now to prove some interesting results.

We start by completing our argument that 2n > n.

Proposition 2.2. 2n > n for all n � 1.

Proof. Here k D 1, and the statements are

S.1/ W 21 > 1; : : : ; S.n � 1/ W 2n�1 > n � 1; S.n/ W 2n > n; : : :

Base Step: If n D 1, then 21 D 2 > 1, so S.1/ is true. Inductive Step: We need to show that if n > 1 and S.n� 1/ is true, then S.n/ is true. It is always a good idea to write the statements out so that we can see

what needs to be proved. Here, we must show that if the inductive hypothesis

S.n � 1/ W 2n�1 > n � 1

is true, then so is S.n/; that is, 2n�1 > n � 1 implies 2n > n. Multiply both sides of the inequality S.n � 1/ by 2: if 2n�1 > n� 1, then

2n D 2 � 2n�1 > 2.n� 1/ D .n � 1/C .n � 1/ � .n � 1/C 1 D n

(the last inequality holds because n > 1 implies n � 1 � 1). Thus, if 2n�1 > n� 1 is true, then 2n > n is also true.

Since both the base step and the inductive step hold, Theorem 2.1 says that

all the statements are true: 2n > n for all n � 1.

Etymology. The word induction comes from the Latin word meaning to lead

into or to influence. It is used here because, as we have just seen, the truth of the nth statement arises from the truth of the previous statement.

Usually the base step in an inductive proof occurs when k D 1, although Define S.0/ W 20 > 0. Suppose we had taken

the base step in Propo-

sition 2.2 at k D 0. Can you write out a proof that

S.0/) S.1/?

many proofs occur when k D 0 (see Exercise 2.4 on page 52). Here is an example of an induction whose base step occurs when k D 5. Consider the statements

S.n/ W 2n > n2:

This is not true for small values of n: if n D 2 or 4, then there is equality, not inequality; if n D 3, the left side, 8, is smaller than the right side, 9. However, S.5/ is true: 32 > 25.

Proposition 2.3. 2n > n2 for all integers n � 5.

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48 Chapter 2 Induction

Proof. We have just checked the base step S.5/. Suppose that n > 5 and that

2n�1 > .n � 1/2: (2.1)

Can we use this to show that 2n > n2? Multiply both sides of inequality (2.1)

by 2 to obtain

2n > 2.n� 1/2:

We’ll be done if we show, for n > 5, that 2.n� 1/2 > n2. Now

2.n� 1/2 D .n � 1/2 C .n � 1/.n � 1/ > .n � 1/2 C 4.n � 1/ since n� 1 > 4 � � n2 � 2nC 1

� C 4.n � 1/

D n2 C 2n� 3:

But 2n� 3 is positive, because n > 5, and so n2 C 2n� 3 > n2.

We now use induction to prove a geometric result.

Definition. A polygon P in the plane is convex if, for every pair of distinct

pointsA;B on its perimeter, the line segment AB lies inside of P .

For example, every triangle is convex, but there are quadrilaterals that are

not convex. For example, the shaded quadrilateral in Figure 2.1 is not convex,

for the line segment joining boundary pointsA and B is not wholly inside it.

A B

Figure 2.1. Non-convex polygon.

Proposition 2.4. Let P be a convex polygon with vertices V1; : : : ; Vn. If �i is

the .interior/ angle at Vi , then

�1 C � � � C �n D .n� 2/180ı:

Proof. The proof is by induction on n � 3. For the base step n D 3, the polygon is a triangle, and it is well known that the sum of the interior angles

is 180ı. For the inductive step n > 3, let P be a convex polygon with vertices V1; : : : ; Vn. Since P is convex, the segment joining V1 and Vn�1 lies wholly inside P ; it divides P into the triangle D V1VnVn�1 and the polygon P 0 having vertices V1; : : : ; Vn�1. Now P 0 is convex (why?), so that the inductive hypothesis says that the sum of its interior angles is .n � 3/180ı. Figure 2.2 shows that the sum of the interior angles of P is the sum of the angles of 

and those of P 0, which is 180ı C .n � 3/180ı D .n� 2/180ı.

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2.1 Induction and Applications 49

P¢

D

Vn

Vn–1

Vn–2

V1 V2

V3

qn–2

q3

q2 q1

qn–1

qn

Figure 2.2. Convex polygon.

In any proof by induction, we must verify both the base step and the induc-

tive step; verification of only one of them is insufficient. For example, consider the statements S.n/Wn2 D n. The base step, S.1/, is true, but the inductive step is false; of course, these statements S.n/ are false for all n > 1. Another

example is given by the statements S.n/Wn D nC 1. It is easy to see that the inductive step is true: if S.n�1/ is true, i.e., if n�1 D .n�1/C1, then adding 1 to both sides gives n D .n � 1/C 2 D n C 1, which is the next statement, S.n/. But the base step is false; of course, all these statements S.n/ are false.

How to Think About It. When first seeing induction, many people sus-

pect that the inductive step is circular reasoning. Why are you allowed to use

statement S.n � 1/, which you don’t know is true, to prove that S.n/ is true? Isn’t the truth of S.n � 1/ essentially what you are supposed to be proving? A closer analysis shows that this is not at all what is happening. The inductive step, by itself, does not prove that S.n/ is true. Rather, it says that if S.n � 1/ is true, then S.n/ is also true. In other words, the inductive step proves that the

implication “If S.n � 1/ is true, then S.n/ is true” is correct. The truth of this implication is not the same thing as the truth of its conclusion. For example,

consider the two statements: “Your grade on every exam is 100%” and “Your

grade for the course is A.” The implication “If all your exams are perfect, then you will get the highest grade for the course” is true. Unfortunately, this does

not say it is inevitable that your grade for the course will be A. Here is a math-

ematical example: the implication “If n � 1 D n, then n D nC 1” is true, but the conclusion “n D nC 1” is false.

From now on, we usually abbreviate mathematical induction to induction.

Here is the first example of a proof by induction often given in most texts.

Proposition 2.5. For every integer n � 1, we have

1C 2C � � � C n D 1 2 n.nC 1/:

Proof. The proof is by induction on n � 1. Base step. If n D 1, then the left-hand side is 1 and the right-hand side is

1 2 1.1C 1/ D 1, as desired.

Inductive step. The .n � 1/st statement is

S.n � 1/W 1C 2C � � � C .n � 1/ D 1 2 .n � 1/n;

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50 Chapter 2 Induction

and we must show

1C 2C � � � C n D � 1C 2C � � � C .n � 1/

� C n:

By the inductive hypothesis, the right-hand side is

1 2 .n � 1/nC n:

But 1 2 .n � 1/n C n D 1

2 n.n C 1/. By induction, the formula holds for all

n � 1.

Historical Note. Here is one version of a popular story. As a 7-year old

prodigy, Gauss was examined by two mathematicians to evaluate his mathe-

matical ability. When asked to add up all the numbers from 1 to 100, he thought

a moment and then said the answer was 5050. Gauss let s denote the sum of all the numbers from 1 to 100: s D 1 C 2 C � � � C 99 C 100. Of course, s D 100C 99C � � � C 2C 1. Arrange these nicely

s D 1C 2C � � � C 99C 100 s D 100C 99C � � � C 2C 1

and add

2s D 101C 101C � � � C 101C 101;

the sum 101 occurring 100 times. We now solve: s D 1 2 .100 � 101/ D 5050.

This argument is valid for any number n in place of 100 (and there is no obvi-

ous use of induction!). Not only does this give a new proof of Proposition 2.5,

it shows how the formula could have been discovered.

Example 2.6. Another proof of the formula in Proposition 2.5 comes from an

analysis of the square in Figure 2.3.

Figure 2.3. Pn kD1 k.

Imagine an .nC 1/ � .nC 1/ square. It contains .nC 1/2 small unit squares.We have n D 7 in Fig- ure 2.3. Since there are nC 1 unit squares on the diagonal, there are

.nC 1/2 � .nC 1/ D n2 C n

unit squares off the diagonal. Half of them, 1 2 .n2 Cn/, are above the diagonal.

But, if you count by rows, there are

1C 2C � � � C n

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2.1 Induction and Applications 51

unit squares above the diagonal. Hence

1C 2C � � � C n D 1 2 .n2 C n/: N

How to Think About It. Proposition 2.5 illustrates a common problem stu- dents have when learning induction. Induction itself is a technique of proof

(involving just two steps), but it is not a method of discovery. However, the

two notions of proof and discovery are often intertwined. For example, merely

applying mathematical induction, as we did in the proof of Proposition 2.5,

is straightforward. But many beginning students get confused because, at the same time as they are following the steps of the proof, they are also wonder-

ing where the formula for the sum comes from. In contrast, neither Gauss’s

proof nor the proof using the .n C 1/ � .n C 1/ square is confusing, for the ideas of these proofs and their techniques of proof are separate. In Section 2.3,

we’ll describe a method for introducing mathematical induction to high school

students that usually minimizes this confusion.

Aside from proving statements, induction can also be used to define terms.

For example, here is an inductive definition of factorial. Inductive definitions are often called recursive

definitions.Definition. Define 0Š D 1 and, if n > 0, define nŠ D n � .n � 1/Š. In other words, nŠ is defined by Defining 0Š D 1 is con-

venient, as we shall see

in the next section when

we discuss the Binomial

Theorem.

nŠ D (

1 if n D 0 n � .n � 1/Š if n > 0:

Induction allows us to define the powers of a number.

Definition. If a 2 R, define the powers of a, for n � 0, by induction:

an D (

1 if n D 0 an�1a if n > 0:

If a D 0, we have defined 00 D 1.

Etymology. The terminology x square and x cube for x2 and x3 is, of

course, geometric in origin. Usage of the term power in this context arises

from a mistranslation of the Greek dunamis (from which the word dynamo

comes) as used by Euclid. The standard European rendition of dunamis was

“power;” for example, the first English translation of Euclid’s Elements by

H. Billingsley in 1570, renders a sentence of Euclid as “The power of a line is the square on the same line” (which doesn’t make much sense to us). However,

contemporaries of Euclid, e.g., Plato and Aristotle, used dunamis to mean “am-

plification.” This seems to be a more appropriate translation, for Euclid was

probably thinking that a one-dimensional line segment can sweep out a two-

dimensional square. We thank Donna Shalev for informing us of the classical usage of dunamis.

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52 Chapter 2 Induction

Proposition 2.7 (Laws of Exponents). Let a 2 R andm; n � 0 be integers. (i) amCn D aman.

(ii) .am/n D amn.

Proof. (i) The statement is true for all n � 0 when m D 0. We prove amCn D aman is true for all n by induction on m � 1. The base step says that aan D anC1, which is just the definition of powers. For the inductive step,

amCn D amCn�1a definition of powers D am�1Cna D am�1ana inductive hypothesis D am�1anC1 definition of powers D am�1CnC1 inductive hypothesis D amCn:

(ii) The statement is true for all m � 0 when n D 0. We prove .am/n D amn is true for all m by induction on n � 1. The base step says that .am/1 D am1 D am, which is obvious. For the inductive step,

.am/n D .am/n�1am definition of powers D am.n�1/am inductive hypothesis D am.n�1/Cm part (i) D amn:

Historical Note. The earliest known occurrence of mathematical induction is in Sefer ha-Mispar (also called Maaseh Hoshev, whose Hebrew title means

practical and theoretical calculating), written by Levi ben Gershon in 1321

(he is also known as Gersonides or as RaLBaG, the acronym for Rabbi Levi

ben Gershon). Induction appears later in Arithmeticorum libri duo, written by

Maurolico in 1557, and also in Traité du Triangle Arithmetique, written by Pascal around 1654 (in which Pascal discusses the Binomial Theorem).

Exercises

2.1 * Guess a formula for 1 C P n jD1 j Šj , and use mathematical induction to prove

that your formula is correct.

2.2 * Prove that if m � 2 is an integer not divisible by any prime p with p � p

m,

then m is a prime. Use this to prove that the numbers n2 � n C 41 are prime for all n � 40.

2.3 * Let m1; m2; : : : ; mn be integers such that gcd.mi ; mj / D 1 for all i ¤ j . If each mi divides an integer k, prove that their product m1m2 � � � mn also divides k. Hint: Use Exercise 1.58 on page 35.

2.4 * If a is positive, give two proofs that

1 C a C a2 C � � � C an�1 D a n � 1

a � 1 ;

by induction on n � 0 and by multiplying the left-hand expression by .a � 1/.

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2.1 Induction and Applications 53

2.5 Let x > �1 be a real number. Prove that .1 C x/n � 1 C nx for all n � 1.

2.6 What is the smallest value of k so that 2n > n3 for all n � k? Why?

2.7 Assuming the product rule for derivatives, .fg/0 D f 0g C fg0, prove that

.xn/0 D nxn�1 for all integers n � 1:

2.8 In high school, nŠ is usually defined as 1 � 2 � 3 � � � � � n. Show that this agrees with the definition on page 51 for all n � 1.

2.9 (Double Induction) Let k; k0 be integers, and let S.m; n/ be a doubly indexed family of statements, one for each pair of integers m � k and n � k0. Suppose that

(i) S.k; k0/ is true,

(ii) if S.m � 1; k0/ is true, then S.m; k0/ is true, (iii) if S.m; n � 1/ is true for all m � k, then S.m; n/ is true for all m � k. Prove that S.m; n/ is true for all m � k and n � k0.

2.10 Prove that .m C 1/n > mn for all m; n � 1.

2.11 Prove the Laws of Exponents by Double Induction.

Unique Factorization

Induction is useful in number theory. As a simple example, we generalize Eu-

clid’s Lemma to more than two factors.

Theorem 2.8 (Euclid’s Lemma). If p is a prime and p j a1a2 � � �an, where n � 2, then p j ai for some i .

Proof. The proof is by induction on n � 2. The base step is Theorem 1.21. To “The proof is by induction on n � 2” not only indicates the base step, it

also tells which variable

will be changing in the

inductive step.

prove the inductive step, suppose that p j a1a2 � � �an. We may group the fac- tors on the right side together so there are only two factors: .a1a2 � � �an�1/an . By Theorem 1.21, either p j a1a2 � � �an�1 or p j an. In the first case, the in- ductive hypothesis gives p j ai for some i � n � 1 and we are done. In the second case, p j an, and we are also done.

This proof illustrates an empirical fact. It is not always the case, in an in-

ductive proof, that the base step is very simple. In fact, all possibilities can

occur: both steps can be easy, both can be difficult, or one can be harder than

the other.

Here is an amusing inductive proof (due to Peter Braunfeld) of the existence of the quotient and remainder in the Division Algorithm.

Proposition 2.9. If a and b are positive integers, then there are integers q

and r with b D qaC r and 0 � r < a.

Proof. We do induction on b � 1. The base step: b D 1. Now a � 1, because it is a positive integer. If a D 1,

choose q D 1 and r D 0; if a > 1, choose q D 0 and r D 1. Let’s prove the inductive step. The inductive hypothesis is b � 1 D qaC r ,

where 0 � r < a. It follows that b D qaCrC1. Now r < a implies rC1 � a. If r C 1 < a, we are done. If r C 1 D a, then b D qaC .r C 1/ D qaC a D .q C 1/a, and we are done in this case as well.

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54 Chapter 2 Induction

We now use induction to prove unique factorization into primes.

Theorem 2.10 (Fundamental Theorem of Arithmetic). Every integer a � 2 is a product of primes. Moreover, ifWe often use the word

product even when there

is only one factor. Thus,

a prime is a product of

primes.

a D p1 � � �pm and a D q1 � � �qn;

where the p’s and q’s are primes, then n D m and the q’s can be re-indexed so that qi D pi for all i .

Proof. The existence of a factorization is Theorem 1.14. To prove uniqueness,

we may assume that m � n, and we use induction on m � 1. The base step is obvious: if m D 1, then n D 1 and the given equations are a D p1 D q1. For the inductive step, the equation

p1 � � �pm D q1 � � �qn

gives pm j q1 � � �qn. By Euclid’s Lemma, there is some i with pm j qi . But qi , being a prime, has no positive divisors other than 1 and itself, so that

qi D pm. Re-indexing, we may assume that qn D pm. Canceling, we have p1 � � �pm�1 D q1 � � �qn�1. By the inductive hypothesis, n � 1 D m � 1 (so that n D m) and the q’s may be re-indexed so that qi D pi for all i � m.

Corollary 2.11. If a � 2 is an integer, then there are distinct primes pi and integers ei > 0 with

a D pe11 � � �penn :

Moreover, if there are distinct primes qj and integers fj > 0 with

p e1 1 � � �penn D q

f1 1 � � �qfmm ;

then m D n, qi D pi and fi D ei for all i .after re-indexing the q0s/.

Proof. Just collect like terms in a prime factorization.

The Fundamental Theorem of Arithmetic says that the exponents e1; : : : ; en in the prime factorization a D pe11 � � �p

en n are well-defined integers determined

by a. It would not make sense to speak of the exponent of q dividing a if the

Fundamental Theorem were false; if an integer a had two factorizations, say,

a D p2q5r6 and a D p2q3s7, where p; q; r; s are distinct primes, would its q-exponent be 5 or 3?

It is often convenient to allow factorizations p e1 1 � � �p

en n having some ex-

ponents ei D 0, because this allows us to use the same set of primes when factoring two numbers. For example, 168 D 233171 and 60 D 223151 may be rewritten as 168 D 23315071 and 60 D 22315170.

Lemma 2.12. Let positive integers a and b have prime factorizations

a D pe11 � � �penn and b D p f1 1 � � �pfnn ;

where p1; : : : ; pn are distinct primes and ei , fi � 0 for all i . Then a j b if and only if ei � fi for all i .

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2.1 Induction and Applications 55

Proof. If ei � fi for all i , then b D ac, where c D pf1�e11 � � �p fn�en n . Now c

is an integer, because fi � ei � 0 for all i , and so a j b. Conversely, if b D ac, let the prime factorization of c be c D pg11 � � �p

gn n ,

where gi � 0 for all i . It follows from the Fundamental Theorem of Arithmetic that ei C gi D fi for all i , and so fi � ei D gi � 0 for all i ; that is, ei � fi for all i .

Definition. A common multiple of integers a and b is an integerm with a j m and b j m. The least common multiple, denoted by

lcm.a; b/;

is the smallest positive common multiple if both a; b ¤ 0, and it is 0 otherwise.

The following proposition describes gcd’s in terms of prime factorizations. This is, in fact, the method usually taught to students in elementary school for

putting fractions into lowest terms.

Proposition 2.13. Let a D pe11 � � �p en n and b D pf11 � � �p

fn n , where p1; : : : ; pn

are distinct primes and ei � 0, fi � 0 for all i . Then

gcd.a; b/ D pm11 � � �pmnn and lcm.a; b/ D p M1 1 � � �pMnn ;

where mi D minfei ; fig, andMi D maxfei ; fi g.

Proof. Define d D pm11 � � �p mn n . Lemma 2.12 shows that d is a (positive)

common divisor of a and b; moreover, if c is any (positive) common divisor,

then c D pg11 � � �p gn n , where 0 � gi � minfei ; fig D mi for all i . Therefore,

c j d . A similar argument shows that D D pM11 � � �p

Mn n is a common multiple

that divides every other such.

Computing the gcd for small numbers a and b using their prime factoriza-

tions is more efficient than using Euclidean Algorithm I. For example, since

168 D 23315071 and 60 D 22315170, we have gcd.168; 60/ D 22315070 D 12 and lcm.168; 60/ D 23315171 D 840. However, finding the prime factor- ization of a large integer is very inefficient, even with today’s fanciest com-

puters; it is so inefficient that this empirical fact is one of the main ingredients

in public key cryptography, the basic reason you can safely submit your credit

card number when buying something online.

Corollary 2.14. If a and b are positive integers, then

lcm.a; b/ gcd.a; b/ D ab:

Proof. The result follows from Proposition 2.13 and Exercise 1.72 on page 40: Notice how a computa- tional inquiry has given a

theorem.mi CMi D ei C fi ;

where mi D minfei ; fi g and Mi D maxfei ; fig.

Since gcd’s can be computed by Euclidean Algorithm I, this corollary al- lows us to compute lcm’s:

lcm.a; b/ D ab= gcd.a; b/:

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56 Chapter 2 Induction

Example 2.15. Sudoku is a popular puzzle. One starts with a 9 � 9 grid of cells, some filled with numbers. The object is to insert numbers in the blank

cells so that every row, every column, and every heavily bordered 3 � 3 box contains the digits 1 through 9 exactly once.

96

5

3

34

1

3

8

7

5

4

5 2

8 3

7 6

3

5

8

1

1

5

3 6

9

7

6 7

Figure 2.4. Sudoku.

KenKen is a variation of Sudoku. As in Sudoku, the object is to fill an n�nKenKenr is a registered trademark of Nextoy, LLC,

2012, KenKen Puzzle LLC.

All rights reserved.

grid with digits: 1 through 4 for a 4 � 4 grid, 1 through 5 for a 5 � 5 grid, etc., so that no digit appears more than once in any row or column. That the cells

in Sudoku are filled with 1, 2, . . . , 9 is not important; one could just as well use the first nine letters a, b, . . . , i instead. In contrast, KenKen uses arithmetic.

KenKen grids are divided into heavily bordered groups of cells, called cages,

and the numbers in the cells in each cage must produce a target number when

combined using a specified mathematical operation—either addition, subtrac-

tion, multiplication or division. Here is a 5� 5KenKen puzzle and its solution.

4– 12´ 3–

12+ 1–

2¸ 4– 3

60´ 5+ 2¸

Figure 2.5. KenKenr puzzle.

4– 12´ 3–

12+ 1–

2¸ 4– 3

60´ 5+ 2¸

5 3 2 1 4

1 2 3 4 5

2 1 4 5 3

4 5 1 3 2

3 4 5 2 1

Figure 2.6. KenKenr solution.

The difficulty in solving a KenKen puzzle arises from there being too many

ways to fill in each cage. Sometimes, the Fundamental Theorem of Arithmetic can help. Let’s start solving the puzzle in Figure 2.5. We view the grid as a

5 � 5 matrix, and we’ll abbreviate “target-operation” to T-O. Consider the L- shaped cage consisting of 4 cells whose target operation is 60�. There are two possibilities: its cells are filled with an arrangement either of 2, 2, 3, 5 or of

1, 3, 4, 5. Assume the first possibility holds. Since we cannot have both 2s in the same row or the same column, one 2 is in position .4; 3/; the other 2 is

either in position .5; 1/ or .5; 2/. Suppose 2 sits in the .5; 1/ position. There is

a cage in the first column with T-O 4�; its cells must contain 1 and 5. Hence, the other cage, with T-O 2�, must contain 3 and 5; it cannot. Thus, 2 sits in

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2.1 Induction and Applications 57

position .5; 2/. There is a cage in the second column with T-O 4�, and its cells must contain 1 and 5. This says that the top two cells in the second column

contain 3 and 4. But the L-shaped cage with T-O 12� must now have 1 in position .1; 3/. This forces the column one cage, with T-O 4�, to have 5 in position .1; 1/, because it can’t be 1. Thus, the last cage in the first row cannot

involve 1 or 5. But the only ways to fill in a 2-cell cage with T-O 3� are with 1 and 4 or with 2 and 5. Conclusion: The 4-cell cage with T-O 60� must be an arrangement of 1, 3, 4, 5. The full solution is given in Figure 2.6. N

Strong Induction

Certain situations call for a variant of induction, called Strong Induction (or the Second Form of Induction).

Definition. Given integers k and n � k, the predecessors of n are the integers ` with k � ` < n, namely, k; k C 1; : : : ; n� 1 (k has no predecessor).

Theorem 2.16 (Strong Induction). Let k be an integer. If S.k/, S.k C 1/, S.k C 2/; : : : is a sequence of statements such that

(i) Base StepW S.k/ is true, and (ii) Inductive StepW If, for n > k, S.`/ being true for all predecessors ` of n

implies S.n/ true,

then the statements S.n/ are true for all n � k.

How to Think About It. Let’s compare the two forms of induction. Both

start by verifying the base step, and both have an inductive step to prove

S.n/. The inductive hypothesis in the first form is that S.n � 1/ is true; the inductive hypothesis in Strong Induction is that all the preceding statements

S.k/; : : : ; S.n � 1/ are true. Thus, Strong Induction has a stronger inductive hypothesis (actually, each of Theorems 2.1 and 2.16 implies the other).

We are going to prove Theorem 2.16 and Theorem 2.1 simultaneously (we

haven’t yet proved the latter theorem). But first we need an easy technical

remark. The Least Integer Axiom says that every nonempty subset C of the natural numbers N contains a smallest number; that is, there is some c0 2 C with c0 � c for all c 2 C . This axiom holds, not only for N, but for any subset

Nk D fn 2 Z W n � kg

as well, where k is a fixed, possibly negative, integer. If k � 0, then Nk � N, and there is nothing to prove; if k < 0, then we argue as follows. Let C � Nk be a nonempty subset. If C contains no negative integers, then C � N, and the Least Integer Axiom applies; otherwise, keep asking, in turn, whether

k; k C 1; : : : ;�1 are in C , and define c0 to be the first one that lies in C . We have alreaady seen the basic idea of the next proof, involving Fido, on

page 46. Here are the two statements again.

Theorem 2.17 (= Theorems 2.1 and 2.16). Let k be an integer. If S.k/,

S.k C 1/, S.k C 2/; : : : is a sequence of statements such that

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58 Chapter 2 Induction

Base StepW S.k/ is true, and Inductive StepW either

(i) if, for n > k, S.n � 1/ being true implies S.n/ true, or

(ii) if, for n > k, S.`/ being true for all predecessors ` of n implies S.n/ true,

then the statements S.n/ are true for all n � k.

Proof. Let Nk D fn 2 Z W n � kg. We show that there are no integers n 2 Nk for which S.n/ is false. Otherwise, the subset C � Nk , consisting of all n for which S.n/ is false, is nonempty, and so C has a smallest element, say, c0. The

truth of the base step says that k < c0, so that c0�1 lies in Nk and hence there is a statement S.c0 � 1/.

Case 1. As S.c0/ is the first false statement, S.c0�1/must be true. Assuming inductive step (i), S.c0/ D S ..c0 � 1/C 1/ is true, and this is a contradiction. Case 2. As S.c0/ is the first false statement, all the statements S.`/, where ` is

a predecessor of c0, are true. Assuming inductive step (ii), the strong version,

we again reach the contradiction that S.c0/ is true.

In either case, C D ¿ (i.e., C is empty), which says that every S.n/ is true.

Here’s a second proof that prime factorizations exist.

Proposition 2.18 (= Proposition 1.14). Every integer n � 2 is a product of primes.

Proof. The base step S.2/ is true because 2 is a prime. We prove the inductive

step. If n � 2 is a prime, we are done. Otherwise, n D ab, where 2 � a < n and 2 � b < n. As a and b are predecessors of n, each of them is a product of primes:

a D pp0 � � � and b D qq0 � � � :

Hence, n D pp0 � � �qq0 � � � is a product of (at least two) primes.

The reason why strong induction is more convenient here is that it is more natural to use S.a/ and S.b/ than to use S.n � 1/; indeed, it is not at all clear how to use S.n � 1/.

The next result says that we can always factor out a largest power of 2 from

any integer. Of course, this follows easily from the Fundamental Theorem of

Arithmetic, but we prove the proposition to illustrate further situations in which strong induction is more appropriate than the first form.

Proposition 2.19. Every integer n � 1 has a unique factorization n D 2km, where k � 0 and m � 1 is odd.

Proof. We use strong induction on n � 1 to prove the existence of k and m. If n D 1, take k D 0 and m D 1. For the inductive step n � 1, we distinguish

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2.1 Induction and Applications 59

two cases. If n is odd, take k D 0 and m D n. If n is even, then n D 2b. Why isn’t the first form of induction convenient here?Since b < n, it is a predecessor of n, and so the inductive hypothesis allows us

to assume b D 2`m, where ` � 0 and m is odd. The desired factorization is n D 2b D 2`C1m.

To prove uniqueness (induction is not needed here), suppose that 2km D n D 2tm0, where both k and t are nonnegative and both m and m0 are odd. We may assume that k � t . If k > t , then canceling 2t from both sides gives 2k�tm D m0. Since k � t > 0, the left side is even while the right side is odd; this contradiction shows that k D t . We may thus cancel 2k from both sides, leaving m D m0.

Exercises

2.12 (i) Prove that an integer a � 2 is a perfect square if and only if whenever p is prime and p j a, then p2 j a.

(ii) Prove that if an integer z � 2 is a perfect square and d4 j z2 , then d2 j z.

2.13 Let a and b be relatively prime positive integers. If ab is a perfect square, prove

that both a and b are perfect squares.

2.14 * Let a; b; c; n be positive integers with ab D cn. Prove that if a and b are relatively prime, then both a and b are nth powers; that is, there are positive

integers k and ` with a D kn and b D `n.

2.15 * For any prime p and any positive integer n, denote the highest power of p

dividing n by Op.n/. That is, Corollary 2.11 guarantees that Op is well-defined.

Op.n/ D e;

where pe j n but peC1 − n. If m and n are positive integers, prove that (i) Op.mn/ D Op.m/ C Op.n/

(ii) Op.m C n/ � min ˚ Op.m/;Op.n/

. When does equality occur?

There is a generalization of Exercise 1.6 on page 6. Using a (tricky) inductive

proof (see FCAA [26], p. 11), we can prove the Inequality of the Means: if n � 2 and a1; : : : ; an are positive numbers, then

n p

a1 � � � an � 1n .a1 C � � � C an/:

2.16 (i) Using the Inequality of the Means for n D 3, prove, for all triangles having a given perimeter, that the equilateral triangle has the largest area.

Hint: Use Heron’s Formula for the area A of a triangle with sides of lengths

a; b; c: if the semiperimeter is s D 1 2

.a C b C c/, then

A2 D s.s � a/.s � b/.s � c/:

(ii) What conditions on a, b, and c ensure that Heron’s formula produces 0? In-

terpret geometrically.

2.17 Let a, b, and c be positive numbers with a > b > c, and let L D 13 .a Cb Cc/. (i) Show that either a > b > L > c or a > L > b > c.

(ii) Assume that a > b > L > c. Show that

L3 � .L � b/2c � .L � b/.L � c/c � .L � c/2L D abc:

(iii) Use part (ii) to prove the Inequality of the Means for three variables.

(iv) Show that a box of dimensions a � b � c can be cut up to fit inside a cube of side length L with something left over.

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60 Chapter 2 Induction

Differential Equations

You may have seen differential equations in other courses. If not, don’t worry;

the next example is self-contained.

Definition. A differential equation is an equation involving a function y D y.x/ and its derivatives; a solution is a function y that satisfies the equation.

Solving a differential equation generalizes indefinite integration: R f .x/ dx

is a solution to the differential equation y0 D f . There may be many solutions: for example, if y D F.x/ is an indefinite integral of

R f .x/ dx, then so is

F.x/ C c for any constant c. Assume that a differential equation has a solution y that is a power series.

Because factorials occur in the coefficients of Taylor series, let’s write a solu-

tion in the formDefining 0Š D 1 allows us to write the coefficient of

xn in Eq. (2.2) as an=nŠ

for all n � 0. y.x/ D a0 C a1x C

a2

2Š x2 C � � � C an

nŠ xn C � � � : (2.2)

We ignore questions of convergence. Of course, some power series diverge,

but we are doing algebra here!

Induction arises here because we can often find y by relating its coefficients

an�1 and an.

Example 2.20. Consider the differential equation y0 D y; that is, we seek a function equal to its own derivative (do you know such a function?). Assuming

that y is a power series, then y has an expression as in Eq. (2.2). Using term-

by-term differentiation, we see that

y0 D a1 C a2x C a3

2Š x2 C � � � C

an

.n � 1/Šx n�1 C � � � ;

so that

an

.n � 1/Šx n�1 D

an�1 .n � 1/Šx

n�1 for all n � 1I

that is, a1 D a0, a2 D a1, and, in fact, an D an�1 for all n � 1. Rewrite the equations: there is no restriction on a0 and, for small n, we see that an D a0. If this were true for all n, then

y.x/ D a0.1C x C 12x 2 C � � � C 1

nŠ xn C � � � / D a0ex:

It is true that an D a0 for all n; one proof is by induction (see Exercise 2.18 on page 62). N

Differential equations often arise with initial conditions: values of y.0/,

y0.0/, y00.0/; : : : are specified. If y is given by a power series

P .an=nŠ/x

n, then y.0/ D a0. Thus, the ini- tial condition y.0/ D 1 chooses the solutiony D ex in the preceding example.

The next example shows how strong induction can be used in solving dif-

ferential equations.

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2.1 Induction and Applications 61

Example 2.21. Consider the differential equation

y00 D y0 C 2y (2.3)

with initial conditions

y.0/ D 2 and y0.0/ D 1:

Again, let’s see if there is a power series solution

y.x/ D a0 C a1x C a2

2Š x2 C � � � :

Substituting y into Eq. (2.3) and equating like powers of x gives

an D an�1 C 2an�2:

Tabulating an for a few values shows a pattern. All the outputs an seem to be

1 away from a power of 2, either 1 more or 1 less. The first two entries record

the initial conditions.

n an

0 2

1 1

2 5 D 1C 2 � 2 3 7 D 5C 2 � 1 4 17 D 7C 2 � 5 5 31 D 17C 2 � 7 6 65

7 127

8 257

9 511

10 1025

Looking closer, the coefficients seem to satisfy an D 2nC .�1/n. Inductive reasoning suggests the conjecture, and mathematical induction is a natural way

to prove it. But there is a problem. The inductive step for an involves not only

an�1, but an�2 as well. Strong Induction to the rescue! Before dealing with the details, we show that if the formula can be proved to hold for all an, then we

can complete our discussion of the differential equation. A solution is

y.x/ D X 2n C .�1/n

nŠ xn

D X 2n

nŠ xn C

X .�1/n nŠ

xn

D X 1

nŠ .2x/n C

X 1 nŠ .�x/n

D e2x C e�x:

You can check that e2x C e�x works by substituting it into Eq. (2.3); we have solved the differential equation. N

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62 Chapter 2 Induction

The proof of the equation relating the coefficients is by Strong Induction.

Proposition 2.22. Suppose, for all n � 0, that an satisfies

an D

8 <̂

:̂

2 n D 0 1 n D 1 an�1 C 2an�2 n > 1:

Then an D 2n C .�1/n for all integers n � 0.

Proof. Because the definition has two initial values, we need to check two base

steps:

a0 D 2 D 20 C .�1/0 and a1 D 1 D 21 C .�1/1:

If n > 1 and ak D 2k C .�1/k for all the predecessors of n, 0 � k < n, then

an D an�1 C 2an�2 D � 2n�1 C .�1/n�1

� C 2

� 2n�2 C .�1/n�2

�

D � 2n�1 C .�1/n�1

� C � 2 � 2n�2 C 2 � .�1/n�2

�

D � 2n�1 C .�1/n�1

� C � 2n�1 C 2 � .�1/n�2

�

D � 2n�1 C 2n�1

� C � .�1/n�1 C 2 � .�1/n�2

�

D 2 � 2n�1 C .�1/n�2 .�1C 2/ D 2n C .�1/n�2

D 2n C .�1/n :

Exercises

2.18 * Complete the discussion in Example 2.20: show that if

y.x/ D a0 C a1x C a2

2Š x2 C � � � C an

nŠ xn C � � �

and y0 D y, then an D a0.

2.19 Assume that “term-by-term” differentiation holds for power series: if f .x/ D c0 C c1x C c2x2 C � � � C cnxn C � � � , then the power series for the derivative f 0.x/ is

f 0.x/ D c1 C 2c2x C 3c3x2 C � � � C ncnxn�1 C � � � :

(i) Prove that f .0/ D c0. (ii) Prove, for all n � 0, that if f .n/ is the nth derivative of f , then

f .n/.x/ D nŠcn C .n C 1/ŠcnC1x C x2gn.x/;

where gn.x/ is a power series.

(iii) Prove that cn D f .n/.x/.0/=nŠ for all n � 0. (This is Taylor’s formula.)Here is an instance in which it is convenient to

write 0Š D 1. This exercise shows why, in Example 2.21, that power series were denoted by a0 C a1x C .a2=2Š/x2 C .a3=3Š/x3 C � � � .

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2.2 Binomial Theorem 63

2.20 Find the solution to the differential equation

2y00 � y0 � 3y D 0:

subject to the initial conditions y.0/ D y.1/ D 1.

Answer: y D 1 5

e�x C 4 5

e3x=2 .

2.2 Binomial Theorem

We now look at a result, important enough to deserve its own section, which

involves both mathematical induction and inductive reasoning. What is the pat-

tern of the coefficients in the formulas for the powers .1C x/n of the binomial 1C x? Let

.1C x/n D c0 C c1x C c2x2 C � � � C cnxn:

Definition. The coefficients cr are called binomial coefficients: Euler introduced the notation

� n r

� , and this

symbol evolved into �

n r

� ,

which is generally used

today:

n

r

! is the coefficient cr of x

r in .1 C x/n:

The binomial coefficient � n r

� is pronounced “n choose r” because it also

arises in counting problems, as we shall soon see. Thus,

.1C x/n D nX

rD0

n

r

! xr :

For example,

.1 C x/0 D 1

.1 C x/1 D 1C 1x

.1 C x/2 D 1C 2x C 1x2

.1 C x/3 D 1C 3x C 3x2 C 1x3

.1 C x/4 D 1C 4x C 6x2 C 4x3 C 1x4:

Etymology. Binomial means aCb; trinomial means aCbCc. But monomial usually refers to a summand of a polynomial: either axe for a polynomial in

one variable, or ax e1 1 � � �x

en n for a polynomial in several variables.

The following figure, called Pascal’s triangle, displays an arrangement of the

first few coefficients.

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64 Chapter 2 Induction

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

In Pascal’s triangle, an inside number (i.e., not a 1 on the border) of the

nth row can be computed by going up to the .n � 1/st row and adding the two neighboring numbers above it. For example, the inside numbers in row 4 can

be computed from row 3 as follows:

1 3 3 1

1 4 6 4 1

(4 D 1C3, 6 D 3C3, and 4 D 3C1). Let’s prove that this observation always holds.

Lemma 2.23. For all integers n � 1 and all r with 0 � r � n, n

r

! D ( 1 if n D 0 or n D r � n�1 r�1

� C � n�1 r

� if 0 < r < n:

Proof. The nth row of Pascal’s triangle is the coefficient list for .1C x/n. The fact that the constant term and the highest degree term have coefficient 1 is Exercise 2.21 on page 67. For the inside terms, we claim that the coefficient of

xr in .1C x/n is the sum of two neighboring coefficients in .1C x/n�1. MoreYou can also prove Lemma 2.23 by induc-

tion. See Exercise 2.22 on

page 67.

precisely, we claim that if

.1C x/n�1 D c0 C c1x C c2x2 C � � � C cn�1xn�1;

and 0 < r < n, then the coefficient of xr in .1 C x/n is cr�1 C cr . We have

.1C x/n D .1C x/.1 C x/n�1 D .1 C x/n�1 C x.1C x/n�1

D .c0 C � � � C cn�1xn�1/C x.c0 C � � � C cn�1xn�1/ D .c0 C � � � C cn�1xn�1/C .c0x C c1x2 C � � � C cn�1xn/ D 1C .c0 C c1/x C .c1 C c2/x2 C � � � :

Thus � n r

� D cr�1 C cr D

� n�1 r�1 �

C � n�1 r

� .

Pascal’s triangle was known centuries before Pascal’s birth; Figure 2.7 shows a Chinese scroll from the year 1303 depicting it. Pascal’s contribution (around

1650) is a formula for the binomial coefficients.

Proposition 2.24 (Pascal). For all n � 0 and all r with 0 � r � n, n

r

! D nŠ rŠ.n� r/Š

:

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2.2 Binomial Theorem 65

Figure 2.7. Pascal’s triangle, China, 1303 CE.

Proof. The proof is by induction on n � 0. If n D 0, then

Pascal probably discovered

this formula by regarding� n r

� in a different way. We’ll

look at this in a moment. 0

0

! D 0Š 0Š0Š

D 1:

For the inductive step, note first that the formula holds when r D 0 and Here is another place showing that defining

0Š D 1 is convenient. r D n:

n

0

! D 1 D nŠ

0Š.n � 0/Š

and n

n

! D 1 D nŠ

nŠ 0Š :

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66 Chapter 2 Induction

If 0 < r < n, then n

r

! D n� 1 r � 1

! C n � 1 r

! (Lemma 2.23)

D .n � 1/Š .r � 1/Š.n� r/Š

C .n � 1/Š rŠ.n� r � 1/Š

(inductive hypothesis)

D .n � 1/Š .r � 1/Š.n� r � 1/Š

� 1 n � r C

1

r

�

D .n � 1/Š .r � 1/Š.n� r � 1/Š

� n r.n � r/

� D nŠ rŠ.n� r/Š :

Theorem 2.25 (Binomial Theorem). (i) For all real numbers x and all in-

tegers n � 0,

.1C x/n D nX

rD0

�n r

� xr D

nX

rD0

nŠ

rŠ.n� r/Š xr :

(ii) For all real numbers a and b and all integers n � 0,

.a C b/n D nX

rD0

n

r

! an�rbr D

nX

rD0

� nŠ rŠ.n� r/Š

� an�rbr :

Proof. (i) This follows from replacing � n r

� by Pascal’s formula in Proposi-

tion 2.24.

(ii) The result is trivially true when a D 0 (we have agreed that 00 D 1). If a ¤ 0, set x D b=a in part (i), and observe that

� 1C b

a

�n D �aC b

a

�n D .a C b/

n

an :

Hence,

.a C b/n D an � 1C b

a

�n D an

nX

rD0

n

r

! br

ar D

nX

rD0

n

r

! an�rbr :

There are many beautiful connections between Pascal’s triangle and number

theory. For example, while it is not generally true that n j � n r

� (for example,

4 − 6 D � 4 2

� ), this result is true when n is prime.

Proposition 2.26. If p is a prime, then p j � p r

� for all r with 0 < r < p.

Proof. By Pascal’s Theorem,

p

r

! D

pŠ

rŠ.p � r/Š D p.p � 1/ � � � .p � r C 1/

rŠ ;

and cross multiplying gives

rŠ

p

r

! D p.p � 1/ � � � .p � r C 1/I

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2.2 Binomial Theorem 67

that is, p j rŠ � p r

� . But each factor of rŠ is strictly less than p, because r < p,

so that p is not a divisor of any of them. Therefore, Euclid’s Lemma says that

p − rŠ and, hence, that p must divide � p r

� .

Example 2.27. The Binomial Theorem can be used to express the sum of the

nth powers of two variables a and b in terms of the “elementary symmetric

functions” a C b and ab. Here are some examples for n D 2; 3; 4; from

.a C b/2 D a2 C 2abC b2

we have

a2 C b2 D .a C b/2 � 2ab:

From

.aC b/3 D a3 C 3a2b C 3ab2 C b3

we conclude

a3 C b3 D .a C b/3 � 3ab.aC b/:

For n D 4,

.a C b/4 D a4 C 4a3b C 6a2b2 C 4ab3 C b4

D .a4 C b4/C 4ab.a2 C b2/ � 6.ab/2:

Hence,

a4 C b4 D .a C b/4 � 4ab.a2 C b2/C 6.ab/2

We can now replace a2Cb2 by the already computed expression .aCb/2�2ab, collect like terms, and have an expression for a4 C b4 in terms of a C b and ab.

We could proceed inductively, expressing anC bn in terms of aC b and ab for n � 5. Try a few more examples; you’ll get the sense that there’s a general method expressing anCbn in terms of aCb, ab, and other terms like akCbk with k < n. N

Exercises

2.21 * Show, without using the Binomial Theorem, that, if n � 0 is an integer, then (i) the degree of .1 C x/n is n

(ii) the leading coefficient of .1 C x/n is 1 (iii) the constant term of .1 C x/n is 1.

2.22 Prove Lemma 2.23 by induction on n � 0.

2.23 Prove that the binomial coefficients are symmetric: for all r with 0 � r � n,

n

r

! D

n

n � r

! :

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68 Chapter 2 Induction

2.24 Find a formula for the sum of the entries in the nth row of Pascal’s triangle and

prove your assertion.

2.25 If n � 1, find a formula for the alternating sum of the binomial coefficients in the nth row of Pascal’s triangle:

n

0

! �

n

1

! C

n

2

! � � � � C .�1/n

n

n

! :

Prove what you say.

Hint: Consider f .x/ D .1 C x/n when x D �1.

2.26 If n � 1, find a formula for the sum of the squares of the binomial coefficients in the nth row of Pascal’s triangle:

n

0

!2 C

n

1

!2 C

n

2

!2 � � � C

n

n

!2 :

Prove what you say.

2.27 Prove, for a given n � 1, that the sum of all the binomial coefficients �n r

� with r

even is equal to the sum of all those �n r

� with r odd.

2.28 The triangular numbers count the number of squares in a staircase of height n.

Figure 2.8 displays the staircases of height n for 1 � n � 5.

Figure 2.8. Triangular numbers.

(i) Find a formula for the nth triangular number in terms of binomial coefficients,

and prove your assertion. Compare this exercise with the discussion of the

.n C 1/ � .n C 1/ square in Example 2.6. (ii) Show that the sum of two consecutive triangular numbers is a perfect square.

2.29 Take It Further. Using the notation of Example 2.27, use the Binomial Theorem

and induction to show that an C bn can be expressed in terms of a C b and ab.

2.30 Pascal’s triangle enjoys a sort of hockey stick property: if you start at the end of

any row and draw a hockey stick along a diagonal, as in Figure 2.9, the sum of the

entries on the handle of the stick is the entry at the tip of the blade. Express the

hockey stick property as an identity involving binomial coefficients and prove the

identity.

2.31 (Leibniz) A function f W R ! R is called a C1-function if it has an nth deriva- tive f .n/.x/ for every n � 1. Prove that if f and g are C1-functions, then

.fg/.n/.x/ D nX

kD0

n

k

! f .k/.x/ � g.n�k/.x/:

In spite of the strong resemblance, there is no routine derivation of the Leibniz

formula from the Binomial Theorem (there is a derivation using an idea from

hypergeometric series).

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2.2 Binomial Theorem 69

6435 64351 15 105 455 1365 5005 11510545513655005

1 14 1

1

1

1

1

13

12

11

10

1

1

1

1

1

1

11

1

1 1

1

1

11

1

1

1

1

14

13

12

11

10

2

3 3

4 4

5 5

6 6

7 7

8 8

9 9

3432

1716 1716

924

462 462

252

126 126

70

35 35

20

2121

56 5628 28

10 10

6

15 15

1 1

1 1

84 368436

210 210 120 4512045

30033003

91

78

66

55

91

78

66

55

300330032002 2002 10011001364 364

792 792 495495

12871287 715 286715286

220220

330330 165165

Figure 2.9. Hockey sticks.

2.32 * If p is a prime and a and b are integers, prove that there is an integer c with

.a C b/p D ap C bp C pc:

Combinatorics

Binomial coefficients have a combinatorial interpretation. Given a set X with n elements, define an r-subset of X to be a subset having exactly r elements.

How many r -subsets of X are there?

Example 2.28. There are ten 3-element subsets of the 5-element set X D fA;B; C;D;Eg. Think of forming a 3-person committee from 5 people. A com- mittee either contains Elvis or doesn’t. The committees are

fA;B; C g fA;B;Dg fB; C;Dg fA; C;Dg fA;B;Eg fA; C; Eg fA;D;Eg fB; C; Eg fB;D;Eg fC;D;Eg

The first row consists of the 3-subsets that don’t contain Elvis (there are four

such); the second row displays the 3-subsets that do contain Elvis (there are

six of these). N

In general, if X has n elements and 0 � r � n, denote the number of its r -element subsets by

Œn; r I

that is, Œn; r  is the number of ways one can choose r things from a box of n

things. Note that:

When n D 0, items (i) and (ii) give the same

answer. Why does this

make sense?

(i) Œn; 0 D 1 (there’s only one 0-subset, the empty set ¿). (ii) Œn; n D 1 (there’s only one n-subset of X , X itself).

If 0 < r < n, you can compute Œn; r  using the committee idea in Exam-

ple 2.28. If X D fa1; a2; : : : ; ang and you want to build an r -subset, first choose a “distinguished” element ofX , say an, and call him Elvis. Either your

subset contains Elvis or it doesn’t.

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70 Chapter 2 Induction

Case 1. If Elvis is in your r -subset, then you must pick r � 1 elements from the remaining n� 1; by definition, there are Œn� 1; r � 1 ways to do this. Case 2. If Elvis is not in your r -subset, then you must pick all r elements from

the remaining n� 1; there are Œn� 1; r ways to do this.

It follows that Œn; r  D Œn�1; r�1C Œn�1; r. We have proved the following result.

Lemma 2.29. For all integers n � 1 and all r with 0 � r � n,

Œn; r  D (

1 if n D 0 or n D r Œn � 1; rC Œn� 1; r if 0 < r < n.

The similarity between Lemmas 2.23 and 2.29 inspires the next theorem. It is also the reason why the binomial coefficient

� n r

� is usually pronounced

“n choose r .”

Theorem 2.30 (Counting Subsets). If n and r are nonnegative integers with

0 � r � n, then n

r

! D Œn; r :

Proof. Use induction on n � 0. If n D 0, the inequality 0 � r � n forces r D 0, and

0

0

! D 1 D Œ0; 0:

Suppose the result is true for n� 1. If 0 � r � n, then n

r

! D n � 1 r � 1

! C n � 1 r

! Lemma 2.23

D Œn � 1; r � 1C Œn� 1; r inductive hypothesis

D Œn; r  Lemma 2.29.

Theorems can often be proved in several ways. The following discussion

gives another proof of Theorem 2.30 using Pascal’s formula (Proposition 2.24)

instead of Lemmas 2.23 and 2.29.

We first compute Œn; r  by considering a related question.

Definition. Given an “alphabet” with n (distinct) letters and an integer r with

1 � r � n, an r-anagram is a sequence of r of these letters with no repetitions.

For example, the 2-anagrams on the alphabet a, b, c are

ab; ba; ac; ca; bc; cb

(note that aa, bb, cc are not on this list). How many r -anagrams are there on an

alphabet with n letters? We count the number of such anagrams in two ways.

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2.2 Binomial Theorem 71

(i) There are n choices for the first letter; since no letter is repeated, there are

only n � 1 choices for the second letter, only n � 2 choices for the third letter, and so forth. Thus, the number of r -anagrams is

n.n � 1/.n � 2/ � � � .n � .r � 1// D n.n � 1/.n � 2/ � � � .n� r C 1/:

In the special case n D r , the number of n-anagrams on n letters is nŠ. (ii) Here is a second way to count the anagrams. First choose an r -subset of

the alphabet (consisting of r letters); there are Œn; r  ways to do this, for

this is exactly what the symbol Œn; r  means. For each chosen r -subset,

there are rŠ ways to arrange the r letters in it (this is the special case of

our first count when n D r ). The number of r -anagrams is thus

rŠ Œn; r :

We conclude that

rŠ Œn; r D n.n � 1/.n � 2/ � � � .n � r C 1/;

from which it follows that

Œn; r  D n.n � 1/.n � 2/ � � � .n � r C 1/ rŠ

D nŠ .n � r/Š rŠ:

Therefore, Pascal’s formula gives

Œn; r  D n

r

! :

If you piece together the results of this section, you’ll see that we have

shown that the following ways to define binomial coefficients are all equiva-

lent: starting from any one of them, you can derive the others.

Algebraic: � n r

� is the coefficient of xr in the expansion of .1C x/n.

Pascal:

n

r

! D nŠ rŠ .n� r/Š

:

Combinatorial: � n r

� is the number of r -element subsets of an n-element set.

Inductive:

n

r

! D ( 1 if n D 0 or n D r� n�1 r�1 �

C � n�1 r

� if 0 < r < n.

Example 2.31. If you replace the symbols by their definition, Theorem 2.30

says something that is far from obvious: the coefficient of xr in .1 C x/n is the same as the number of r -element subsets of an n-element set. The proof by

induction of Theorem 2.30 establishes this, but many people are left wondering

if there is a more intuitive reason why the expansion of .1 C x/n contains all the information about subsets of various sizes from an n-element set.

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72 Chapter 2 Induction

If you were going to multiply out .1 C x/5 by hand, you could view the calculation like this:

.1 C x/5 D .1C x/.1 C x/.1 C x/.1C x/.1 C x/

The expansion is carried out by taking one term (1 or x) from each binomial

factor 1C x, multiplying them together, and then collecting like powers of x. For example, you could take a “1” from each of the first three binomials and

an x from the last two. That would produce 1 � 1 � 1 � x � x D x2. But that’s not the only way to get an x2. You could have taken an x from the first and third binomials and 1 from the rest. Or an x from the first two binomials and 1

from the last three. Do this in every possible way; the coefficient of x2 in the

expansion will be the number of ways you can pick two binomials from the set

of five to be “x terms.” And there are precisely 10 D Œ5; 2 ways to do this. Generalizing, view .1 C x/n as a product of n binomials:

.1 C x/n D .1C x/.1 C x/.1 C x/ : : : .1 C x/:„ ƒ‚ … n times:

The coefficient of xr in this product is the number of ways you can choose r of

the binomials to be “x terms” (and the rest to be 1). This number is precisely

Œn; r . Hence

.1 C x/n D nX

rD0 Œn; r xr :

When combined with the definition of binomial coefficients on page 63, this

gives another proof that � n r

� D Œn; r . N

Exercises

2.33 *

(i) For each value of r , 0 � r � 4, how many r-element subsets of the set fA; B; C; Dg are there?

(ii) For each value of r , 0 � r � 5, how many r-element subsets of the set fA; B; C; D; Eg are there?

2.34 How many subsets (of any size) are there in an n-element set? Prove your asser-

tion.

2.35 Show that

2n

r

! D

rX

kD0

n

k

! n

r � k

! :

Hint: Split a 2n-element set into two equal pieces.

2.36 Show that

3n

r

! D

rX

kD0

n

k

! 2n

r � k

! :

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2.3 Connections 73

2.37 If m, n, and r are nonnegative integers, prove Vandermonde’s Identity:

m C n

r

! D

rX

kD0

m

k

! n

r � k

! :

Hint: .1 C x/mCn D .1 C x/m.1 C x/n.

2.38 Show that nX

kD0 k

n

k

! D n2n�1:

2.39 How many ways can you choose two hats from a closet containing 14 different

hats? (One of our friends does not like the phrasing of this exercise. After all,

you can choose two hats with your left hand, with your right hand, with your

teeth, : : : , but we continue the evil tradition.)

2.40 Let D be a collection of ten different dogs, and let C be a collection of ten dif-

ferent cats. Prove that there are the same number of quartets of dogs as there are

sextets of cats.

2.41 (i) What is the coefficient of x16 in .1 C x/20? (ii) How many ways are there to choose 4 colors from a palette containing paints

of 20 different colors?

2.42 A weekly lottery asks you to select 5 different numbers between 1 and 45. At the

week’s end, 5 such numbers are drawn at random, and you win the jackpot if all

your numbers match the drawn numbers. What is your chance of winning?

The number of selections of 5 numbers is “45 choose 5”, which is �45 5

� D

1; 221; 759. The odds against your winning are more than a million to one.

2.3 Connections

An Approach to Induction

Teaching mathematical induction to high school students is often tough. In

particular, many students fall into the trap we described on page 49: in spite

of all our explanations to the contrary, they think that the inductive hypothesis

assumes what it is they are supposed to be proving. In this section, we look at

a well-tested method that avoids this trap.

Suppose you ask a class to come up with a function that agrees with the table

Input Output

0 4

1 7

2 10

3 13

4 16

5 19

We’ve found that about half a high school class (beginning algebra, say) comes

up with a closed form definition, something likef .n/ D 3nC4;while the other

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74 Chapter 2 Induction

half comes up with an inductive definition—something like “start with 4, and

each output is 3 more than the previous one.” This inductive definition can be

written more formally:

g.n/ D ( 4 if n D 0 g.n � 1/C 3 if n > 0.

That inductive definitions (or recursive definitions, as they are often called)

Of course, beginning

students don’t usually

write the definition of g

using this case notation.

Technology can help get

them used to it. seem to be, in some sense, natural for students, can be exploited to help stu-

dents understand proof by induction.

The part of g ’s definition

involving n, the equation

g.n/ D g.n � 1/ C 3, is called a recurrence.

Computer algebra systems (CAS) let you model the two definitions. Build-

ing such computational models allows students to experiment with both func- tions, and it also provides an opportunity to launch some important ideas. For

example, a teacher can use the models to discuss the domain of each function—

f accepts any real number, but g will accept only nonnegative integers.

Here’s what’s most germane to this section. If you try some values in a

spreadsheet or calculator, it seems, for a while, that f and g produce the same output when given the same input. But at some point (exactly where depends

on the system), f outputs an integer but g surrenders. Suppose this happens,

for example, at 255: both functions return 766 at 254, but f .255/ D 769 while g returns an error. Is this because f and g really are not equal at 255?

Or is it because of the limitations of the technology? Many students will say

immediately that the functions are equal when n D 255; it’s just that the com- puter can’t compute the value of g there. Tell the students, “I believe you that

f .254/ D g.254/. Convince me that they are also equal at 255.” After some polishing and a little help, their argument usually goes some-

thing like this.

g.255/ D g.254/C 3 (this is how g is defined) D f .254/C 3 (the calculator said so—they both output 766) D .3 � 254C 4/C 3 (this is how f is defined) D .3 � 254C 3/C 4 (algebra) D 3.254C 1/C 4 (more algebra) D 3 � 255C 4 (arithmetic) D f .255/ (this is how f is defined).

There’s nothing special about 255 here. If you had a more powerful calculator,

one that handled inputs to g up to, say, 567 (and then crashed at 568), you

could argue that f and g were equal at 568, too.

g.568/ D g.567/C 3 (this is how g is defined) D f .567/C 3 (the powerful calculator said so) D .3 � 567C 4/C 3 (this is how f is defined) D .3 � 567C 3/C 4 (algebra) D 3.567C 1/C 4 (more algebra) D 3 � 568C 4 (arithmetic) D f .568/ (this is how f is defined).

Computing g.568/ is the same as computing g.255/. Now imagine that you

had a virtual calculator, one that showed that f and g agreed up to some integer

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2.3 Connections 75

n � 1, but then crashed when you asked for g.n/. You could show that f and g are equal at n by the same argument:

g.n/ D g.n � 1/C 3 (this is how g is defined) D f .n � 1/C 3 (the virtual calculator said so) D .3 � .n � 1/C 4/C 3 (this is how f is defined) D 3nC 4 (algebra) D f .n/ (this is how f is defined).

So, every time f and g are equal at one integer, they are equal at the next

one. Since f and g are equal at 0 (in fact, since they are equal at every integer

between 0 and 254), they are equal at every nonnegative integer.

This argument is the essence of mathematical induction. In the example, it

shows that if two functions f and g are equal at one integer, then they are equal at the next one. Coupled with the fact that they are equal at 0, it makes

sense that they are equal for all integers greater than or equal to 0; that is,

f .n/ D g.n/ for all nonnegative integers n. We have seen that induction applies in much more general situations than

this one. But this simple context is quite effective in starting students onto a

path that helps them understand induction.

Fibonacci Sequence

Many interesting investigations in high school center around the following se- quence, which describes a pattern frequently found in nature and in art.

Definition. The Fibonacci sequence is defined by:

Fn D

€ 0 n D 0 1 n D 1 Fn�1 C Fn�2 n > 1:

There are two base steps in the definition: n D 0 and n D 1. The Fibonacci sequence begins: 0; 1; 1; 2; 3; 5; 8; 13; : : :.

Historical Note. The Fibonacci sequence is related to the golden ratio, a

number mentioned in Euclid, Book 6, Proposition 30. It is said that the ancient Greeks thought that a rectangular figure is most pleasing to the eye (such rect-

angles can be seen in the Parthenon in Athens) if its edges a and b are in the

proportion

a

b D

b

aC b :

In this case, a.aCb/ D b2, so that b2�ab�a2 D 0; that is, .b=a/2�.b=a/�1 D 0. The quadratic formula gives b=a D 1

2 .1 ˙

p 5/. Therefore,

b=a D D 1 2 .1 C

p 5/ or b=a D ı D 1

2 .1 �

p 5/:

But ı is negative, and so we must have

b=a D D 1 2 .1 C

p 5/:

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76 Chapter 2 Induction

Figure 2.10. Golden rectangle.

The number D 1:61803 : : :, is called the golden ratio. Since both and ı are roots of x2 � x � 1, we have

2 D C 1 and ı2 D ı C 1: (2.4)

So, what’s the connection of the golden ratio to the Fibonacci sequence?

We discovered the closed form for the sequence cn in Proposition 2.22 by

tabulating the first few terms of the sequence and looking for regularity—it

seemed “almost” exponential, off by 1 from a power of 2. Let’s tabulate the first few ratios of consecutive terms Fn=Fn�1 of the Fi-

bonacci sequence.

F1=F0 D 1=1 D 1 F2=F1 D 2=1 D 2 F3=F2 D 3=2 D 1:5 F4=F3 D 5=3 D 1:666 F5=F4 D 8=5 D 1:6 F6=F5 D 13=8 D 1:625:

If you tabulate a few more ratios (try it), a conjecture emerges—it appears that

the ratio of two consecutive terms in the Fibonacci sequence might converge to the golden ratio � 1:61803 (if the ratios were actually constant, Fn would be a geometric sequence (why?)). This is, in fact, the case, and you’ll see, in

Exercise 2.50 on page 78, how to refine the conjecture into the statement of the

following theorem (the exercise will also help you develop a method that will

let you find closed forms for many 2-term recurrences).

Theorem 2.32. For all n � 0, the nth term of the Fibonacci sequence satisfies

Fn D 1p 5 . n � ın/;

where D 1 2 .1 C

p 5/ and ı D 1

2 .1 �

p 5/:

Proof. We use strong induction because the inductive step involves the formu-

las for both Fn�1 and Fn�2. The base steps S.0/ and S.1/ are true:

1p 5 . 0 � ı0/ D 0 D F0 1p 5 . � ı/ D 1p

5

� 1 2 .1C

p 5/ � 1

2 .1 �

p 5/ �

D 1 D F1:

If n � 2, then

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2.3 Connections 77

Fn D Fn�1 C Fn�2 D 1p

5 . n�1 � ın�1/C 1p

5 . n�2 � ın�2/

D 1p 5

� . n�1 C n�2/� .ın�1 C ın�2/

�

D 1p 5

� n�2. C 1/� ın�2.ı C 1/

�

D 1p 5

� n�2. 2/ � ın�2.ı2/

� by Eq. (2.4)

D 1p 5 . n � ın/:

Isn’t it curious that the integers Fn are expressed in terms of the irrational

number p 5?

If n D 2, then F2 D 1 D 0, and so there is equality,

not inequality. Corollary 2.33. Fn >

n�2 for all integers n � 3, where D 1 2

� 1C

p 5 � .

Proof. The proof is by induction on n � 3. The base step S.3/ is true, for F3 D 2 > � 1:618. For the inductive step, we must show that FnC1 > n�1. By the inductive hypothesis,

FnC1 D Fn C Fn�1 > n�2 C n�3

D n�3. C 1/ D n�3 2 D n�1:

Exercises

2.43 Show that the following functions agree on all natural numbers.

f .n/ D 3n C 5 and g.n/ D (

5 if n D 0 g.n � 1/ C 3 if n > 0.

2.44 Show that the following two functions agree on all natural numbers.

f .n/ D 4n and g.n/ D (

4 if n D 0 4g.n � 1/ if n > 0.

2.45 Define the function h inductively:

h.n/ D (

4 if n D 0 h.n � 1/ C 2n if n > 0

Find a polynomial function p that agrees with h on all natural numbers, and prove

that your functions are equal on N .

Answer: n2 C n C 4.

2.46 Define the function m inductively:

m.n/ D (

0 if n D 0 m.n � 1/ C n2 if n > 0

Find a polynomial function s that agrees with m on all natural numbers, and prove

that your functions are equal on N .

Answer: 2n 3C3n2Cn 6 D

n.nC1/.2nC1/ 6 .

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78 Chapter 2 Induction

2.47 Consider the two functions f and g:

f .x/ D x4 � 6x3 C 14x2 � 6x C 2

and

g.x/ D (

2 if x D 0 s.x � 1/ C 6x � 3 if x > 0.

Are f and g equal on N?

2.48 Find a formula for 02 C 12 C 22 C � � � C .n � 1/2 as a function of n, and prove your assertion.

2.49 You saw, on page 76, that the ratio of two consecutive terms seems to converge to

the golden ratio. Using only the recurrence

Fn D Fn�1 C Fn�2 for all n � 2;

show that

lim n!1

Fn

Fn�1 D 12 .1 C

p 5/:

2.50 * You saw, on page 76, that the Fibonacci sequence seems to be “almost” expo-

nential.

(i) Suppose the Fibonacci sequence actually was exponential: Fn D rn. Show that r would have to be either

D 1 C p

5

2 or ı D 1 �

p 5

2 :

(ii) Show that the sequences n and ın satisfy the recurrence

fn D fn�1 C fn�2:

(iii) If a and b are any real numbers, show that a nC bın satisfies the recurrence

fn D fn�1 C fn�2:

(iv) Without using Theorem 2.32, find a and b so that

a n C bın D Fn:

2.51 Ms. D’Amato likes to take a different route to work every day. She will quit her

job the day she has to repeat her route. Her home and work are pictured in the grid

of streets in Figure 2.11. If she never backtracks (she only travels north or east),

how many days will she work at her job?

A valid trip Another valid trip Not a valid trip

Figure 2.11. Ms. D’Amato.

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2.3 Connections 79

2.52 Find a closed form for each of the following functions and prove your assertions.

(i) f .n/ D (

4 n D 0 f .n � 1/ C 3 n > 0:

Answer: f .n/ D 3n C 4.

(ii) f .n/ D (

4 n D 0 3f .n � 1/ n > 0:

Answer: f .n/ D 4 � 3n.

(iii) f .n/ D

˚ 2 n D 0 4 n D 1 4f .n � 1/ � 3f .n � 2/ n > 1:

Answer: f .n/ D 3n C 1.

(iv) f .n/ D

˚ 4 n D 0 4 n D 1 4f .n � 1/ � 3f .n � 2/ n > 1:

Answer: f .n/ D 4.

2.53 Find a closed form for the following function and prove your assertion.

f .n/ D

„ 3 n D 0 4 n D 1 14 n D 2 4f .n � 1/ � f .n � 2/ � 6f .n � 3/ n > 2:

2.54 Take It Further. Find or generate a copy of the first 30 rows of Pascal’s triangle.

Color the odd numbers red and the even numbers black. Explain any patterns that

you see. (Alternatively, you can use a spreadsheet to generate the triangle of 0s

and 1s that are the remainders when each entry is divided by 2.) For more on this

exercise, see

ecademy.agnesscott.edu/˜lriddle/ifs/siertri/Pascalmath.htm

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3 Renaissance For centuries, the Western World believed that the high point of civilization

took place from the Greek and Roman eras through the beginning of Chris-

tianity. But this worldview began to change dramatically about five hundred years ago. The printing press was invented around 1450, by Johannes Guten-

berg, Christopher Columbus landed in North America in 1492, Martin Luther

began the Reformation in 1517, and Nicolas Copernicus published De Revolu-

tionibus in 1530.

Mathematics was also developing. A formula giving the roots of certain cubic polynomials, similar to the quadratic formula, was discovered by Scipi-

one del Ferro around 1515; by 1545, it was extended to all cubics by Fontana

(Tartaglia) and Cardano. The cubic formula contributed to the change in world-

view that was the essence of the Renaissance, for it was one of the first math-

ematical results not known to the ancients. But its impact on contemporary

mathematics was much deeper, for it introduced complex numbers. As we shall see, the cubic formula is not as useful for numerical computations as we’d like,

because it often gives roots in unrecognizable form. Its importance, however,

lies in the ideas it generated. Trying to understand the formula, searching for

generalizations of it, and studying questions naturally arising from such en-

deavors, were driving forces in the development of abstract algebra. In many high school algebra courses today, the complex numbers, usually

denoted by C, are introduced to find the roots of ax2CbxCcwhen b2�4ac < 0. That’s not how it happened. Square roots of negative real numbers occur in

the cubic formula, but not as roots; indeed, in the 16th century, complex roots

would have been considered useless. But complex numbers arose in the middle of calculations, eventually producing real numbers (we will see this explicitly

in the next section). To understand this phenomenon, mathematicians were

forced to investigate the meaning of number; are complex numbers bona fide

numbers? Are negative numbers bona fide numbers?

Section 3.1 discusses the classical formulas giving the roots of cubic and

quartic polynomials. We will look more carefully at the complex numbers themselves in Section 3.2. Although initially used in purely algebraic contexts,

C has a rich geometric and analytic structure that, when taken together with its

algebraic properties, can tie together many of the ideas in high school mathe-

matics. Indeed, C finds applications all over mathematics. Section 3.4 uses C

to solve some problems that are especially useful for teachers (and interesting for all mathematicians). Just as the method of Diophantus was used to create

Pythagorean triples, C can be used to invent problems whose solutions “come

out nice.”

81

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82 Chapter 3 Renaissance

3.1 Classical Formulas

As Europe emerged from the Dark Ages, a major open problem in mathemat-

ics was finding solutions to polynomial equations. The quadratic formula had

been known for about four thousand years and, arising from a tradition of pub-

lic mathematical contests in Pisa and Venice, formulas for the roots of cubics

and quartics had been found in the early 1500s. Let’s look at these formulas

in modern algebraic notation; we will assume for now that the complex num- bers obey the usual laws of arithmetic (neither of these simplifying steps was

available to mathematicians of the 16th century).

Historical Note. Modern arithmetic notation was introduced in the late

1500s, but it was not generally agreed upon in Europe until after the influ-

ential book of Descartes, La Géométrie, was published in 1637 (before then,

words and abbreviations were used as well as various competing notations).

The symbols +, �, and p, as well as the symbol / for division, as in 4=5, were introduced by Widman in 1486. The equality sign, D, was invented by Recorde in 1557. Designating variables by letters was invented by Viète in

1591, who used consonants to denote constants and vowels to denote vari-

ables; the modern notation, using the letters a; b; c; : : : to denote constants and

the letters x; y; z at the end of the alphabet to denote variables, was introduced

by Descartes in 1637. The exponential notation 22; 23; 24; : : : was essentially invented by Hume in 1636, who wrote 2ii, 2iii, 2iv; : : : : The symbol � for mul- tiplication was introduced by Oughtred in 1631; the symbol � for division was introduced by Rahn in 1659. See Cajori [6].

Cubics

The following familiar fact (to be proved in Chapter 6) was known and used

by Renaissance mathematicians, and we will use it in this section.

Proposition 6.15. If r is a root of a polynomial f .x/, then x � r is a factor of f .x/I that is, f .x/ D .x � r/g.x/ for some polynomial g.x/.

One of the simplest cubics is f .x/ D x3�1. Obviously, 1 is root of f , and so x3 � 1 D .x � 1/g.x/, where

g.x/ D .x3 � 1/=.x � 1/ D x2 C x C 1:

The roots of g (and, hence, also of f ) are

! D 1 2 .�1 C i

p 3/ and ! D 1

2 .�1 � i

p 3/;

by the quadratic formula. Both ! and ! are called cube roots of unity, for

!3 D 1 D !3. Note that ! D !2 D 1=!. We know that a positive number a has two square roots. By convention,

p a

denotes the positive square root, so that the two square roots are ˙ p a. Any

real number a has three cube roots. By convention, 3 p a denotes the real cube

root, so that the three cube roots are 3 p a; ! 3

p a; !2 3

p a. Thus, cube roots of

unity generalize ˙. The general cubic equation aX3 C bX2 C cX C d D 0 can be simplified

by dividing both sides by a; this procedure does not affect the roots, and so

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3.1 Classical Formulas 83

we may assume that a D 1. Thus, we seek the roots of the cubic polynomial F.X/ D X3 C bX2 C cX C d , where b; c; d 2 R. The change of variable

X D x � 1 3 b

yields a simpler polynomial, f .x/ D F.x � 1 3 b/ D x3 C qx C r , where q

and r are expressions in b, c, and d . We call f the reduced polynomial arising

from F .

Lemma 3.1. Let f .x/ D x3CqxC r be the reduced polynomial arising from F.X/ D X3 C bX2 C cX C d D 0. If u is a root of f , then u � 1

3 b is a root

Exercise 3.5 on page 89

asks you to check that

the coefficient of x2 in

F .x� 13 b/ D .x� 13 b/3C b.x� 1

3 b/2Cc.x� 1

3 b/Cd

is zero. of F .

Proof. Since f .x/ D F.x � 1 3 b/ for all x, we have 0 D f .u/ D F.u � 1

3 b/;

that is, u� 1 3 b is a root of F.X/.

We will use the following consequence of the quadratic formula.

Lemma 3.2. Given a pair of numbersM andN , there are .possibly complex/

numbers g and h with g C h D M and gh D N . In fact, g and h are roots of x2 �Mx CN .

Proof. We have

.x � g/.x � h/ D x2 � .g C h/x C gh:

Thus, the roots g; h of f .x/ D x2 � Mx C N (which exist, thanks to the quadratic formula) satisfy the given equations g C h D M and gh D N .

Let’s try to find a general method for solving cubic equations—a method that doesn’t depend on the specific values of the coefficients—by first solving

a numerical equation.

Consider the polynomial f .x/ D x3 � 18x � 35. Since the constant term 35 D 5 � 7, we check whether ˙1;˙5;˙7 are roots. It turns out that 5 is a root and, dividing by x � 5, we can find the other two roots by solving the You can check that the

other two roots are

complex. Renaissance

mathematicians would

have dismissed these as

meaningless. But stay

tuned—we’ll soon see that

they, too, can be generated

by the emerging method.

quadratic f .x/=.x � 5/ D x2 C 5x C 7. But we are looking for a general method applicable to other cubics, so let’s pretend we don’t know that 5 is a

root.

It’s natural to look for a polynomial identity having the same form as the

equation we are trying to solve. Example 2.27 provides one. From

a3 C b3 D .a C b/3 � 3ab.aC b/;

we have the identity

.a C b/3 � 3ab.aC b/� � a3 C b3

� D 0:

Thinking of a C b as a single “chunk,” say, x D a C b, the correspondence looks like this:

.a C b/3 �3ab � .a C b/ �.a3 C b3/ D 0 # # # #

�3 �3ab � � �.a3 C b3/ D 0 " " " " x3 �18 � x �35 D 0:

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84 Chapter 3 Renaissance

So, if we can find two numbers a and b such that

�3ab D �18 and a3 C b3 D 35;

then a C b will be a root of the cubic. Hence we want a and b so that ab D 6 and a3Cb3 D 35. There’s an obvious solution here, namely, a D 3 and b D 2, but we’re looking for a general method. Cubing both sides of ab D 6, we get

a3b3 D 216 and a3 C b3 D 35:

By Lemma 3.2, a3 and b3 are roots of the quadratic equation

x2 � 35x C 216 D 0:

The roots of this are 27 and 8. So we can take a3 D 27, b3 D 8; surprise! a D 3 and b D 2. Hence, 3C 2 D 5 is a root of our original cubic.

The next theorem is usually attributed to Scipione del Ferro; we’ll use com-

plex numbers and modern notation in its statement and proof, neither of which

was available at the time. In light of Lemma 3.1, we may assume that cubics

are reduced.

Theorem 3.3 (Cubic Formula). The roots of f .x/ D x3 C qx C r are

g C h; !g C !2h; and !2g C !h;

where ! D 1 2

� �1C i

p 3 �

is a cube root of unity,

g3 D �r C

p R

2 ; h D �

q

3g ; and R D r2 C

4q3

27 :

Proof. Let u be a root of f .x/ D x3CqxC r and, as in the discussion above, we try

u D g C h:

We are led to

g3 C h3 D �r gh D �1

3 q:

Cube gh D �1 3 q, obtaining the pair of equations

g3 C h3 D �r g3h3 D � 1

27 q3:

Lemma 3.2 gives a quadratic equation whose roots are g3 and h3:

x2 C rx � 1 27 q3 D 0: (3.1)

The quadratic formula gives

g3 D 1 2

� �r C

q r2 C 4

27 q3 �

D 1 2

� �r C

p R �

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3.1 Classical Formulas 85

and

h3 D 1 2

� �r �

q r2 C 4

27 q3 �

D 1 2

� �r �

p R � :

Now there are three cube roots of g3, namely, g, !g, and !2g. Because

of the constraint gh D �1 3 q, each has a “mate,” namely, �q=.3g/ D h,

�q=.3!g/ D !2h, and �q=.3!2g/ D !h. Thus, the roots of f are

g C h; !g C its mate; !2g C its mateI

that is, the roots of f are g C h, !g C !2h, and !2g C !h.

Example 3.4 (Good Example). If f .x/ D x3 � 15x � 126, then q D �15, r D �126, R D 15376, and

p R D 124. Hence, g3 D 125, so we can take

g D 5. Thus, h D �q=3g D 1. Therefore, the roots of f are

Alternatively, having found

one root to be 6, the

other two roots can be

found as the roots of the

quadratic f.x/=.x� 6/ D x2C 6x C 21.

6; 5! C !2 D �3C 2i p 3; 5!2 C ! D �3 � 2i

p 3:

For Renaissance mathematicians, this cubic would have only one root—they would have ignored the complex roots. N

But things don’t always work out as we expect, as the next surprising ex-

ample shows.

Example 3.5 (Bad Example). The cubic formula may give the roots in unrec-

ognizable form. Let

f .x/ D .x � 1/.x � 2/.x C 3/ D x3 � 7x C 6I

the roots of f are, obviously, 1; 2, and �3. But the cubic formula gives

g3 D 1 2

� �6C

q �400 27

� and h3 D 1

2

� �6 �

q �400 27

� :

It is not at all obvious that gC h is a real number, let alone an integer! N

Imagine yourself, standing in Piazza San Marco in Venice in 1520, partici-

pating in a contest. Your opponent challenges you to find a root of f .x/ D x3 � 7x C 6 (he invented the cubic, so he knows that it comes from .x � 1/.x � 2/.x C 3/). Still, you are a clever rascal; your mentor taught you the cubic formula. You do as you were taught, and triumphantly announce that

a root is gC h, where g3; h3 are the awful expressions above. Most likely, the judges would agree that your opponent, who says that 1 is a root, has defeated you. After all, f .1/ D 1 � 7C 6 D 0, so that 1 is, indeed, a root. The judges even snickered when they asked you to evaluate f .g C h/.

With head hung low, you return home. Can you simplify your answer? Why

is g C h equal to 1? Let’s pretend you have modern notation. Well,

g3 D 1 2

� �6C

q �400 27

� D �3C i 10

p 3

9 :

The first question is how to compute cube roots of “numbers” of the form

aC bi , where i2 D �1. Specifically, we want uC iv with

.uC iv/3 D �3C i 10 p 3

9 :

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86 Chapter 3 Renaissance

Hmm! Perhaps it’s smart to separate terms involving i from honest numbers.

.uC iv/3 D u3 C 3u2iv C 3u.iv/2 C .iv/3

D u3 C 3u2iv � 3uv2 � iv3

D u3 � 3uv2 C i.3u2v � v3/:

Let’s see if the separation pays off. We want numbers u; v with u3 � 3uv2 DBut see Example 3.36, or try to solve the system

u3 � 3uv2 D �3 and 3u2v � v3 D 10

p 3

9 with a

computer.

�3 and 3u2v � v3 D 10 p 3

9 . These equations are intractible! Sigh.

Cube roots are tough. Let’s simplify things; perhaps solving a simpler prob-

lem, say, finding square roots, can give a clue to finding cube roots. And this

we can do.

Proposition 3.6. Every complex number a C bi has a square root.

Proof. If b D 0, then aC ib D a. If a � 0, then p a is well-known; if a < 0,

then a D �c, where c > 0, and p a D i

p c. We can now assume that b ¤ 0,

and our task is to find uC iv with

.uC iv/2 D u2 C 2iuv � v2 D a C ibI

that is, we seek numbers u; v such that

u2 � v2 D a (3.2) and

2uv D b: (3.3)

Since b ¤ 0, Eq. (3.3) gives u ¤ 0; define v D b=2u. Substituting into Eq. (3.2), we have

u2 � .b=2u/2 D aI rewriting,

4u4 � 4au2 � b2 D 0:

This is a quadratic in u2, and the quadratic formula gives

u2 D 1 8

� 4a˙

p 16a2 C 16b2

�

D 1 2

� a˙

p a2 C b2

� :

Since a2Cb2 > 0, it has a real square root. Now 1 2

� aC

p a2 C b2

� is positive

(because b2 > 0 implies a < p a2 C b2); hence, we can find its (real) square

root u as well as v D b=2u.

For example, our proof gives a method finding a square root of i . Set a D 0 and b D 1 to obtain

i D � 1p 2 .1 C i/

�2 :

Alas, this square root success doesn’t lead to a cube root success, although it We’ll see how to find the

roots of a complex number

in Section 3.3. does give us some confidence that our manipulations may be legitimate.

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3.1 Classical Formulas 87

You can now appreciate the confusion produced by the cubic formula; a

cloud enveloped our ancestors. First of all, what are these “numbers” a C ib? Sometimes they can help. Can we trust them to always give us the truth? Is it true that we can separate terms involving i from those that don’t? When are

two complex numbers equal? Does it make sense to do arithmetic with these

guys? Do they obey the nine properties of arithmetic on page 40 that familiar

numbers do? It took mathematicians about 100 years to become comfortable

with complex numbers, and another 100 years until all was set on a firm foun- dation.

Quartics

A method for solving fourth degree equations was found by Lodovico Ferrari

in the 1540s, but we present the version given by Descartes in 1637. Consider the quartic F.X/ D X4 C bX3 C cX2 C dX C e. The change

of variable X D x � 1 4 b yields a simpler polynomial, f .x/ D F.x � 1

4 b/ D

x4 C qx2 C rx C s, whose roots give the roots of F : if u is a root of f , then See Exercise 3.6 on page 89.u� 1

4 b is a root of F . Write f as a product of two quadratics:

f .x/ D x4 C qx2 C rx C s D .x2 C jx C `/.x2 � jx Cm/;

and determine j , `, and m (note that the coefficients of the linear terms in the

quadratic factors are j and �j because f has no cubic term). Expanding and equating like coefficients gives the equations

`Cm� j 2 D q; j.m� `/ D r;

`m D s:

The first two equations give Since j.m � `/ D r , we have �`Cm D r=j .

2m D j 2 C q C r j ;

2` D j 2 C q � r j :

Substituting these values for m and ` into the third equation and simplifying

yield a degree 6 polynomial which is a cubic in j 2 (called the resolvent cubic):

.j 2/3 C 2q.j 2/2 C .q2 � 4s/j 2 � r2:

The cubic formula gives a root j 2, from which we can determinem and ` and,

hence, the roots of the quartic.

This process is an algorithm that can easily be encoded in a computer al- gebra system; it is known as the quartic formula. The quartic formula has the

same disadvantage as the cubic formula: even though it gives correct answers,

the values it gives for the roots are usually unrecognizable. But there are some

good examples.

Example 3.7. Let’s find the roots of

f .x/ D x4 � 10x2 C 1:

First, factor f :

x4 � 10x2 C 1 D .x2 C jx C `/.x2 � jx Cm/I

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88 Chapter 3 Renaissance

in our earlier notation, q D �10, r D 0, and s D 1. The quartic formula showsWhat’s going on? Why is there a choice for ` and

j 2? If the roots of f are

˛i for 1 � i � 4, then f.x/ D

Q i.x � ˛i /. A

factorization of f into

quadratics arises from

grouping these four factors

into pairs, and there is

no reason why different

groupings should give

the same quadratics.

Of course, any such

factorization gives the

same roots of f .

us how to find j; `; m. Since r D 0, we have 2` D j 2 � 10 D 2m; hence, ` D m. But `m D 1, so that either ` D 1 and j 2 D 12 or ` D �1 and j 2 D 8. Taking ` D 1 and j 2 D 12 gives

f .x/ D .x2 C p 12x C 1/.x2 �

p 12x C 1/;

and the quadratic formula gives the four roots of f :

˛ D p 2C

p 3; ˇ D �

p 2C

p 3; D

p 2 �

p 3; ı D �

p 2 �

p 3: N

The quadratic formula can be derived in a way similar to the derivations

of the cubic and quartic formulas (in Chapter 1, we derived the formula by

completing the square). The change of variable X D x � 12b replaces the polynomialF.X/ D X2CbXCc with the simpler polynomialf .x/ D x2Cq, where q D c � 1

4 b2; the roots u D ˙p�q of f .x/ give the roots u � 1

2 b of

F . Since the roots of f are

u D ˙p�q D ˙ q

�.c � 1 4 b2/ D ˙ 1

2

p b2 � 4c;

the roots of F are our old friends

˙ 12 p b2 � 4c � 1

2 b D 1

2

� � b ˙

p b2 � 4c

� :

It is now tempting, as it was for our ancestors, to try to find the roots of the

general quinticF.X/ D X5 C bX4 C cX3 C dX2 C eX C f and to express them in a form similar to those for quadratic, cubic, and quartic polynomials;

that is, using only extraction of roots, addition, subtraction, multiplication, and division (of course, our ancestors hoped to find roots of polynomials of any

degree). They began with the change of variable X D x � 1 5 b to eliminate

the X4 term. It was natural to expect that some further ingenious substitution

together with the formulas for roots of polynomials of lower degree, analogous

to the resolvent cubic, would yield the roots of F . For almost 300 years, no

such formula was found. But, in 1824, Abel proved that there is no such quintic formula.

How to Think About It. Abel’s theorem is often misquoted. It says: there is

no formula involving only extraction of roots and the four basic operations of arithmetic that expresses the roots of the general quintic polynomial in terms

of its coefficients. Succinctly, the general quintic is not solvable by radicals.

But there are other kinds of formulas giving roots of polynomials. For exam-

ple, here is a formula, due to Viète, giving the roots in terms of trigonometric

functions. If f .x/ D x3CqxC r has three real roots, then its roots are t cos � , t cos.�C120ı/, t cos.�C240ı/, where t D

p �4q=3 and cos.3�/ D �4r=t3

(there are variations using cosh and sinh when f has complex roots ([26],

p. 445–447)). You may recall Newton’s method giving the roots as limn!1 xn, where xnC1 D xn � f .xn/=f 0.xn/. Now some quintic polynomials are solv- able by radicals; for example, we’ll see in Section 3.3 that x5 � 1 is one such. Another theorem of Abel gives a class of polynomials, of any degree, which are solvable by radicals. Galois, the young wizard who was killed before hisAbel (1802–1829) also

died young. 21st birthday, characterized all the polynomials which are solvable by radi-

cals, greatly generalizing Abel’s theorem. We will look at this more closely in

Chapter 9.

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3.1 Classical Formulas 89

Exercises

3.1 For each equation, find all roots in R and in C

(i) x2 � 2x D 15 (ii) x2 � 2x D 16 (iii) x2 � 2x D �16 (iv) 6x2 C x D 15 (v) 6x2 C x D 16 (vi) 6x2 C x D �16

(vii) x2 D 1 (viii) x3 D 1 (ix) x4 D 1 (x) x3 D 8

3.2 * We know that i satisfies x2 C 1 D 0 in C (is there another solution?). (i) Show, for all n 2 Z, that the value of in is one of 1; i; �1; �i .

(ii) Use the Division Algorithm to decide which of the four values i247 will have.

3.3 Let ! D 12 .�1 C i p

3/ be a cube root of unity.

(i) Show, for every integer n, the value of !n is one of 1; !; !2 .

(ii) Use the Division Algorithm to decide, for any fixed n, which of the three

values !n will have.

3.4 Find two numbers whose

(i) sum is 5 and product is 6. (ii) sum is 0 and product is �2. (iii) sum is 3 and product is 3. (iv) sum is �1 and product is 1. (v) sum is b and product is c (in terms of b and c).(vi)

3.5 * If F.X/ D X3CbX2CcX Cd , show that the change of variable X D x � 13b produces a polynomial f with no quadratic term,

f .x/ D F.x � 1 3

b/ D x3 C qx C r:

Express q and r in terms of b, c, and d .

3.6 *

(i) Suppose that F.X/ D X4 C bX3 C cX2 C dX C e. (a) Show that the change of variable X D x � 14b produces a polynomial f

with no cubic term,

f .x/ D F.x � 14b/ D x 4 C qx2 C rx C s:

Express q, r , and s in terms of b, c, d , and e.

(b) Show that if u is a root of f , then u � 1 4

b is a root of F .

(ii) In general, let

F.X/ D Xn C an�1Xn�1 C an�2Xn�2 C � � � C a0

be a polynomial of degree n.

(a) Show that the change of variable X D x� 1 n

an�1 produces a polynomial f with no term of degree n � 1,

f .x/ D F.x � 1nan�1/ D x n C qn�2xn�2 C � � � C q0:

(b) Show that if u is a root of f , then u � 1nan�1 is a root of F:

3.7 Take It Further. Suppose that g and h are complex numbers and

! D 1 2

� �1 C i

p 3 �

:

Show that

g3 C h3 D .g C h/.!g C !2h/.!2g C !h/:

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90 Chapter 3 Renaissance

3.8 In Example 3.7, we found the roots of x4 � 10x2 C 1 by factoring it into two quadratics (which came from taking, in the notation of page 87, ` D 1 and j 2 D 12). Another choice was ` D �1 and j 2 D 8.

(i) Using the alternate choice, get a different factorization of the quartic into

quadratic factors.

(ii) Show that the two factorizations produce the same linear factors.

3.9 The following problem, from an old Chinese text, was solved by Qin Jiushao

(Ch’in Chiu-shao) in 1247. There is a circular castle (see Figure 3.1) whose di-

ameter is unknown; it is provided with four gates, and two lengths out of the

north gate there is a large tree, which is visible from a point six lengths east of the

south gate. What is the length of the diameter? (The answer is a root of a cubic

polynomial.)

S E

N C

O

T

2

r

r

r

a

6

Figure 3.1. Castle problem.

3.10 Show that there is no real number whose square is �1.

3.11 (i) Find the roots of x3 � 3x C 1 D 0. (ii) Find the roots of x4 � 2x2 C 8x � 3 D 0.

3.12 Find a complex number s so that s3 D 9 � 46i .

3.13 Find the roots of x3 � 21x C 20. (i) by finding a root and reducing the cubic to a quadratic.

(ii) by the cubic formula.

(iii) Verify that the answers are the same.

3.14 Suppose that ˛ and ˇ are roots of the quadratic equation x2 C bx C c D 0. Find expressions in terms of b and c for

(i) ˛ C ˇ (ii) ˛2 C ˇ2

(iii) ˛3 C ˇ3

(iv) .˛ � ˇ/2

(v) Use parts (i) and (iv) to derive the quadratic formula.

3.15 Suppose that ˛, ˇ, and are roots of the cubic equation x3C bx2C cx C d D 0. Show that

(i) ˛ C ˇ C D �b (ii) ˛ˇ C ˛ C ˇ D c

(iii) ˛ˇ D �d .

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3.2 Complex Numbers 91

3.16 (i) Show that a rectangle is determined by its perimeter and area.

(ii) Is a rectangular box determined by its volume and surface area? Explain.

3.17 Suppose that ˛, ˇ, and are roots of the cubic equation x3C bx2C cx C d D 0. Exercise 3.17 can be done without the cubic formula.Find, in terms of b, c, and d ,

(i) ˛2 C ˇ2 C 2

(ii) ˛3 C ˇ3 C 3

(iii) ˛2ˇ2 C ˛2ˇ 2 C ˛ˇ2 2

3.18 Take It Further.

(i) Suppose that ˛, ˇ, and are three numbers whose sum is 0. Show that

� .˛ � ˇ/.˛ � /.ˇ � /

�2 C 4.˛ˇ C ˛ C ˇ /3 C 27 � ˛ˇ

�3 D 0:

(ii) Suppose that ˛, ˇ, and are roots of x3 C qx C r . Show that

� � .˛ � ˇ/.˛ � /.ˇ � /

�2 D 27r2 C 4q3:

3.19 Take It Further. When finding the roots of x3Cqx C r with the cubic formula, you are led to Eq. (3.1): x2 C rx � 1

27 q3, whose roots are g3 and h3.

(i) Show that the discriminant ı of this quadratic is

ı D r2 C 427q 3

(ii) If ı > 0, show that the cubic has one real root and two complex conjugate

roots. The discriminant of the cubic x3CqxCr is defined to be D �4q3 � 27r2.(iii) If ı D 0, show that the cubic has two real roots, one of them with multiplic-

ity 2.

(iv) If ı < 0, show that the cubic has three distinct real roots.

3.2 Complex Numbers

Before the cubic formula, mathematicians had no difficulty in ignoring neg- ative numbers or square roots of negative numbers. For example, consider

the problem of finding the sides x and y of a rectangle having area A and

perimeter p. The equations xy D A and 2x C 2y D p give the equation 2x2 � px C 2A D 0, and the quadratic formula gives

x D 1 4

� p˙

p p2 � 16A

� :

If p2 � 16A � 0, the problem is solved. If p2 � 16A < 0, people didn’t in- vent fantastic rectangles whose sides involve square roots of negative numbers.

Instead, they merely said that there is no rectangle whose area and perimeter are so related. But the cubic formula doesn’t allow us to avoid “imaginary”

numbers, for we have just seen, in Example 3.5, that an “honest” real and posi-

tive root can appear in terms of such expressions. Complex numbers arose, not

as an attempt to get roots of equations involving square roots of negative real

numbers, but as a device to solve cubic equations having real coefficients and real roots.

The cubic formula was revolutionary. For the next 100 years, mathemati-

cians were forced to reconsider the meaning of number, calculating with strange

objects of the form a C ib (where a and b are real numbers) as if they were

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92 Chapter 3 Renaissance

actual numbers enjoying the simplification rule i2 D �1. It was during thisIn Chapter 7, using ideas of abstract algebra, we’ll

see that the naive way of

thinking about complex

numbers, as polynomials in

i obeying the rule i2 D �1, can be made precise.

time that the terms real and imaginary arose. In this section and the next, we’ll

develop complex numbers in a more careful and formal way, and we’ll see that complex numbers are as real as real numbers!

The Complex Plane

When considering expressions of the form aC bi , it is natural to separate the two summands. Geometry rears its head.

Definition. A complex number is an ordered pair z D .a; b/ of real numbers, denoted by z D aC bi . We call a the real part of z, denoting it by <.z/ D a,Right now, aC bi is just

an “alias” for .a;b/ but, in

your previous experience

with C, the summand bi

denoted the product of b

and i . We’ll soon recover

this notion.

and b the imaginary part of z, denoting it by =.z/ D b.

Both the real and the imaginary parts of a complex number are real numbers.

Moreover, equality of ordered pairs says that complex numbers z D a C bi and z0 D a0 C b0i are equal if and only if <.z/ D <.z0/ and =.z/ D =.z0/; that is, a D a0 and b D b0. Thus, one equation of complex numbers is the same as two equations of real numbers.

There is an immediate geometric interpretation of complex numbers: they

can be viewed as points in the plane. Real numbers are complex numbers z

with =.z/ D 0; that is, they correspond to points .a; 0/ on the x-axis (which is called the real axis in this context). We usually abbreviate .a; 0/ to a; thus, the

set of real numbers R is a subset of C. We denote the complex number .0; 1/You hear the following message when you call

one of our friends. “The

number you have reached

is imaginary; please rotate

your phone 90 degrees.”

by i , so that the purely imaginary complex numbers z, those with <.z/ D 0, correspond to points on the y-axis (which is called the imaginary axis in this

context). When we view points as complex numbers, the plane R2 is called the

complex plane, and it is denoted by C.

Thus, an ordered pair .a; b/ of real numbers has two interpretations: alge-

braic, as the complex number z D a C bi , and geometric, as the point P in the plane R2 having coordinates a and b. We will use both interpretations, al-

gebraic and geometric, depending on which is more convenient for the context

in which we are working.

Historical Note. Surprisingly, it took a very long time for people to embrace

the idea of representing the elements of C as points in the plane. It wasn’t until

Wessel presented a paper in 1797 to the Royal Danish Academy of Sciences, A bijection is a one-to-

one correspondence. See

Appendix A.1, page 416,

for the precise definition.

entitled On the Analytic Representation of Direction: An Attempt, did this rep- resentation crystallize. Wessel’s discovery was not adopted immediately but,

by 1830, most mathematicians routinely used the bijection a C bi $ .a; b/ between complex numbers and points of the plane. The complex plane has

gone by other names in its history: for example, Argand Diagram and Gaus-

sian Plane.

Algebraic Operations

In Section 3.1, you saw that mathematicians were forced to add and multiply complex numbers. However, without precise definitions of the operations or of

the complex numbers themselves, they could not trust many of their results.

The complex plane allows us to resolve the many doubts our ancestors had

about the algebra of complex numbers.

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3.2 Complex Numbers 93

In a linear algebra course, R2 is often viewed as a vector space with real

scalars; we continue using these operations in the complex plane.

Definition. Define addition C � C ! C by

.a C bi/C .c C di/ D .a C c/C .b C d/i:

In terms of ordered pairs, .a; b/C .c; d / D .a C c; bC d/. Define scalar multiplication R � C ! C by

r.aC bi/ D raC rb i;

where r 2 R. In terms of ordered pairs, r.a; b/ D .ra; rb/.

As in linear algebra, it is useful to look at each point in the plane (and,

hence, each complex number) as an arrow with tail at the origin (sometimes

we say vector instead of arrow). For example, we’ll think of z D 3C 2i either as the point P D .3; 2/ or as the arrow ��!OP [where O D .0; 0/]. The context will make it clear which interpretation we are using.

Addition is illustrated by the parallelogram law (see Figure 3.2). If P D .a; b/ and Q D .c; d /, then R D .a C c; b C d/. Of course, this needs a geometric proof, especially when points don’t lie in the first quadrant or they

are collinear; see Exercise 3.33 on page 98.

O

P a,b= ( )

R a+c,b+d= ( )

Q c,d= ( )

Figure 3.2. Parallelogram Law.

Scalar multiplication of complex numbers has the same geometric interpre- tation as scalar multiplication of vectors. View a complex number z D a C ib as

��! OP , where P D .a; b/. If r 2 R, then we may view rz as the vector r��!OP ;

that is, if r � 0, then it’s an arrow in the same direction as ��!OP whose length has been stretched by a factor of r (if r > 1) or shrunk by a factor of r (if

r < 1/; if r < 0, then r ��! OP is the arrow in the reverse direction whose length

has been changed by a factor of jr j. The eight properties listed in the next proposition are precisely the defining

properties of a vector space with scalars in R.

Proposition 3.8. Let z D a C bi , w D c C di , and u D e C f i be complex numbers, and suppose that r; s 2 R.

(i) z C w D w C z (ii) z C .w C u/ D .z C w/C u

(iii) z C 0 D z

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94 Chapter 3 Renaissance

(iv) There is a complex number �z such that z C .�z/ D 0In Exercise 3.21 on page 98, you’ll show that

negatives are unique. (v) r.sz/ D .rs/z

(vi) 1z D z (vii) r.z C w/ D rz C rw

(viii) .r C s/z D rz C sz

Proof. The proofs are routine, just reducing each to a familiar statement about

real numbers, and so we’ll only prove the longest such: associativity of addi-

tion. It is clearer if we use ordered pairs.

z C .w C u/ D .a; b/C Œ.c; d /C .e; f / D .a; b/C .c C e; d C f / D .a C .c C e/; b C .d C f // D ..a C c/C e; .bC d/C f / D .a C c; bC d/C .e; f / D Œ.a; b/C .c; d /C .e; f / D .z C w/C u:

In linear algebra, every vector .a; b/ has a decomposition into components

ae1 C be2 with respect to the standard basis e1 D .1; 0/; e2 D .0; 1/:

aC bi D .a; b/ D .a; 0/C .0; b/ D a.1; 0/C b.0; 1/:

It follows that the + in the notation aC bi really does mean add and that bi is the product of b and i ; that is, bi D b.0; 1/ D .0; b/.

The set C of complex numbers has more algebraic structure: any two com-

plex numbers can be multiplied, not just when one of them is real. The defini-

tion arises from pretending that

.a C bi/.c C di/ D ac C adi C bci C bdi2

D .ac � bd/C i.ad C bc/;

where we have set i2 D �1. This is precisely what our ancestors did, which motivates the formal definition. But our definition involves no pretending.

Definition. Define multiplication C � C ! C by

.a C ib/.c C id / D .ac � bd/C i.ad C bc/:

In terms of ordered pairs, .a; b/.c; d / D .ac � bd; ad C bc/.

Notice that

i2 D .0; 1/.0; 1/ D .�1; 0/ D �1;

for ac D 0 D ad D bc. We are now obliged to prove that the familiar properties of multiplication

actually do hold for complex multiplication.

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3.2 Complex Numbers 95

Proposition 3.9. Let z D a C bi , w D c C di , and u D e C f i be complex numbers.

(i) zw D wz (ii) z.wu/ D .zw/u

(iii) 1z D z (iv) z.w C u/ D zw C zu.

Proof. Again, the proofs are routine, for each reduces to a familiar statement

about real numbers. We’ll only prove associativity. As in the proof of Proposi- tion 3.8, it is clearer if we use ordered pairs.

z.wu/ D .a; b/ Œ.c; d /.e; f / D .a; b/.ce � df; deC cf // D .a.ce � df /� b.de C cf /; b.ce � df /C a.de C cf // D .ace � adf � bde � bcf; bce � bdf C ade C acf /:

On the other hand,

.zw/u D Œ.a; b/.c; d / .e; f / D .ac � bd; bc C ad/.e; f / D .ac � bd/e � .bc C ad/f; .bc C ad/e C .ac � bd/f / D .ace � bde � bcf � adf; bceC ade C acf � bdf /:

Hence, z.wu/ D .zw/u.

The operations of addition and multiplication in C extend the definitions

in R; for example, if r and s are real numbers, then their sum rC s is the same, whether you think of doing the addition in R or in C. See Exercise 3.22 on page 98.

In Section 1.4, we displayed nine properties of addition and multiplication

in R, and we have just seen that eight of them also hold in C. We could now

define subtraction, as we did there, and prove results like z.w�u/ D zw�zu, z �0 D 0 for all complex numbers z, and the Binomial Theorem. We don’t have to repeat all of this. As we said then, once this is established, the proofs of other properties of addition and multiplication, such as 0 � z D 0 and .�z/.�w/ D zw, go through verbatim.

The ninth property describes reciprocals: If z D a C bi ¤ 0, there is a number z�1 such that z � z�1 D 1 (Exercise 3.21 on page 98 shows that such a number z�1, if it exists, is unique.) Here is an explicit formula for z�1 D 1=z when z D a C bi ¤ 0. If b D 0, then z D a is a nonzero real number, and we know it has a reciprocal 1=a. If b ¤ 0, we can easily find x C yi so that .x C yi/.a C bi/ D 1. Multiply and equate real and imaginary parts:

xa � yb D 1 and ya C xb D 0:

The second equation gives x D �ay=b; substitute this into the first equation and obtain

y D � b a2 C b2 and x D

a

a2 C b2 :

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96 Chapter 3 Renaissance

In either case (b D 0 or b ¤ 0),

z�1 D a a2 C b2

� i b a2 C b2

:

There is a more elegant derivation of this formula. The denominator a2Cb2 can be factored in C:

Hence, a2 C b2 D zz. a2 C b2 D .a C bi/.a � bi/:

This leads to the useful notion of complex conjugate.

Definition. The complex conjugate z of a complex number z D a C bi is defined to be

z D a � bi:

The function C ! C, given by z 7! z, is called complex conjugation.

If z D .a; b/, then z D .a;�b/, so, geometrically, z is obtained from z by reflection in the real axis.

Complex conjugation interacts well with addition and multiplication.

Proposition 3.10. If z D aC bi and w D c C di are complex numbers, then

(i) z C w D z Cw (ii) zw D z w

(iii) z 2 R if and only if z D z (iv) z D z

Proof. We’ll prove (i), leaving the rest to Exercise 3.25 on page 98.

z C w D .aC c/C .b C d/i D .aC c/ � .b C d/i D .a � bi/C .c � di/ D z C w:

Using induction, the first two statements in the proposition can be general- ized:

z1 C � � � C zn D z1 C � � � C zn z1 � � � zn D z1 � � � zn:

The formula for the multiplicative inverse of a complex number can be writ-

ten in terms of conjugates. Informally, cancel z to see that z=zz D 1=z.

Proposition 3.11. Every nonzero complex number z D aC bi has an inverseW

z�1 D a � bi a2 C b2 D

z

z z :

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3.2 Complex Numbers 97

Proof. It’s enough to see that you get 1 if you multiply z by .a�bi/=.a2Cb2/. And so it is:

z

� 1

z z z

� D � 1

z z

� z z D 1:

It wouldn’t be worth introducing the new term complex conjugation if our

only use of it was to give a neat proof of the formula for reciprocals. The notion has many other uses as well. For example, if f .x/ D ax2 C bx C c has real coefficients, then the quadratic formula implies that whenever z is a complex

root of f , then so is z. In fact, this is true for polynomials of any degree, and Is this true if z is real?

the proof depends only on Proposition 3.10.

Theorem 3.12. If f .x/ is a polynomial with real coefficients and a complex

number z is a root of f , then so is z.

Proof. Suppose that

f .x/ D a0 C a1x C � � � C aixi C � � � C anxn;

where each ai 2 R. Saying that z is a root means that

0 D a0 C a1z C � � � C aizi C � � � C anzn:

Hence,

0 D 0 D a0 C a1z C � � � C aizi C � � � C anzn

D a0 C a1z C � � � C aizi C � � � C anzn

D a0 C a1 z C � � � C ai zi C � � � C an zn

D a0 C a1z C � � � C aizi C � � � C anzn (because all ai are real) D f .z/ :

Therefore, z is a root of f .

Exercises

3.20 In Appendix A.4, we considered the subset P of all (strictly) positive real num-

bers; it satisfies: If we define a < b to mean b � a 2 P , then we can prove all the familiar

properties of inequality.

For example, if a < b and

c < 0, then bc < ac. See

page 441.

� if a; b 2 P , then a C b 2 P and ab 2 P ; � if r 2 R, then exactly one of the following is true:

r 2 P; r D 0; or � r 2 P:

(i) Using only the two properties of P , prove that if a 2 R, then either a D 0 or a2 2 P .

(ii) Prove that there is no subset Q � C, closed under addition and multiplica- tion, such that if z 2 C, then exactly one of the following is true:

z 2 Q; z D 0; or � z 2 Q:

Conclude that it’s impossible to order the complex numbers in a way that

preserves the basic rules for inequality listed in Proposition A.51.

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98 Chapter 3 Renaissance

3.21 * Suppose that z is a complex number. Generalize Propositions 1.33 and 1.32.

(i) Show that �z is unique. (ii) Show that �z D .�1/z.

(iii) If z ¤ 0, show that z�1 is unique.

3.22 *

(i) We may think of real numbers r and s as complex numbers. Show that their

sum r C s and their product rs in C are the same as their sum and product in R.

(ii) If z is a complex number and r is a real number, you can think of the complex

number rz in two ways: as the product of scalar multiplication, or as the

product of two complex numbers r C0i and z. Show that the two calculations give the same result.

3.23 If z 2 C, show that z C z D 2.<z/ and z � z D 2.=z/.

3.24 Find a complex number z such that z C z D 14 and zz D 49.

3.25 * Finish the proof of Proposition 3.10. If z and w are complex numbers, prove

(i) zw D z w (ii) z D z

(iii) z D z if and only if z is a real number

3.26 If z is a complex number and n is a natural number, show that

zn D .z/n :

Is this equation true if z ¤ 0 and n is a negative integer?

3.27 * Let z be a complex number and r a real number. Show how to locate rz in the

complex plane in terms of z.

Hint: r.a; b/ D .ra; rb/.

3.28 Solve the following equation for z.

.3 C 2i /z D �3 C 11i:

3.29 Find real numbers a and b such that

(i) a C bi D .8 C i /=.3 C 2i /

(ii) a C bi D .8 C i /=.3 C i /

(iii) .a C bi /2 D �5 C 12i

(iv) .a C bi /2 D 1 C i .

3.30 What’s wrong with this “proof” that 6 D �6?

6 D p

36 D p

.�9/.�4/ D p

�9 p

�4 D 3i � 2i D 6i2 D �6:

3.31 Establish the identity

.a2 C b2/.c2 C d2/ D .ac � bd/2 C .bc C ad/2

for all complex numbers a; b; c; d .

3.32 Use Theorem 3.12 to prove that every cubic polynomial with real coefficients has

a real root.

3.33 * Let z D a C bi and w D c C di . Show that in the complex plane z C w is the fourth vertex of the (possibly degenerate) parallelogram whose other vertices are

0, z, and w.

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3.2 Complex Numbers 99

Absolute Value and Direction

We’ve already seen that addition can be viewed as the parallelogram law, multi-

plication by a real number can be viewed as scalar multiplication, and conjuga- tion can be viewed as reflection in the real axis. There is a beautiful geometric

interpretation of complex multiplication; it is best understood if, first, we con-

sider a different way to describe an arrow in the complex plane using absolute

value and direction.

Definition. The absolute value (or length or modulus) of z D aC bi is

jzj D p a2 C b2:

The absolute value of a real number is its distance to the origin, and so we

have just extended the notion of absolute value from R to C. Thus, if z D a C bi , then jzj is the distance from the point P D .a; b/ to the origin O ; equivalently, it is the length of the arrow

��! OP . Because z z D a2 C b2, we can

write Does this equation hold if

z is a real number?jzj D p z z:

Proposition 3.13. Let z D aC bi and w D c C di . (i) jzj � 0, and jzj D 0 if and only if z D 0.

(ii) .Triangle Inequality/: jz C wj � jzj C jwj. (iii) jzwj D jzj jwj.

Proof. (i) Both statements follow from the definition jzj D p a2 C b2, be-

cause a2 C b2 D 0 if and only if a D 0 D b. (ii) If P D .a; b/ andQ D .c; d /, then z is the arrow ��!OP and w is the arrow

��! OQ. As in Figure 3.2, zCw D ��!OR. The inequality we want is the usual triangle inequality, which follows from the length of a line segment being

the shortest distance between its endpoints.

Make sure you can justify

each step in the proof.

Would this proof work if

either z or w (or both) is

real?

jzwj D p .zw/ .zw/

D p .zw/ .z w/

D p .z z/ .w w/

D p .z z/

p .w w/

D jzj jwj:

What do we mean by direction? The most natural way to indicate direction in the plane is to point: “He went thataway!”—an arrow shows the way. The

arrow may as well have its tail at the origin and, since the length of the arrow

doesn’t affect the direction, we may as well assume it is a unit vector; that is,

it has length 1. If we denote the tip of the arrow by P D .a; b/, then P lies on the unit circle. There are various geometric ways to describe P . One way Why did our ancestors

divide a circle into 360

parts? We can only guess

why. Perhaps it was related

to calendars, for a year has

about 360 days.

is to consider the angle � between the x-axis and ��! OP ; hence, P D .a; b/ D

.cos �; sin �/. This angle can be described with degrees; the ancients divided

the circle into 360 equal degrees. The angle can also be described with radians;

the circumference of the unit circle is 2� , and � is the length of the arc from

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100 Chapter 3 Renaissance

.1; 0/ to P . When we view the point P D .cos �; sin �/ on the unit circle as a complex number, it is equal to cos � C i sin � .

Figure 3.3 shows z D a C bi as the tip of an arrow of length jzj D r . The direction of this arrow is the same as the direction of the unit vector

��! OP having

the same direction as z. If � is the angle between the x-axis and ��! OP , then the

coordinates of P are jOAj D cos � and jAP j D sin � .

(1, 0)

r |z|=

q O A

P = (cos , sin )q q

z a, b= ( )

Figure 3.3. Absolute value and argument.

Definition. If z is a nonzero complex number, then its argument, denoted by

arg.z/;

is the counterclockwise angle � from the positive real axis to ��! OP .

In Figure 3.3, we see that the coordinates of P are cos � and sin � ; that is,Finding arg.z/ requires some way of comput-

ing values of inverse

trigonometric functions.

Nowadays, we use com-

puters; in earlier times,

tables of values of cosine

and sine were used. Fairly

accurate trigonometric

tables were known over

two thousand years ago.

if P D .a; b/, then a D cos � and b D sin � . Thus, for any nonzero complex number z D a C bi , not necessarily of absolute value 1, the definitions of cosine and sine (in terms of right triangles) give arg.z/ D � , where cos � D a=jzj and sin � D b=jzj. Note that

z D jzj � a

jzj C i b

jzj

� D jzj .cos � C i sin �/ : (3.4)

How to Think About It. Technically, the argument of a complex number is only determined up to a multiple of 360 (if we measure in degrees) or 2�

(if we measure in radians). For example, arg.1 C i/ is 45ı or � 4

radians, and

this is the same direction as 405ı or 9� 4

radians. There is a fussy way to make statements like “arg.1 C i/ D 45ı” precise (introduce a suitable equivalence relation), but we prefer, as do most people, to be a bit sloppy here; the cure is

worse than the disease.

Actually, the trig functions do make angles precise, for cos � and sin � have

the same values when � is replaced by either � C 360n degrees or � C 2�n radians.

The polar form of z is

z D jzj .cos � C i sin �/ ;

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3.2 Complex Numbers 101

where � D arg.z/. Just as a complex number z is determined by its real and imaginary parts, so too is it determined by its polar form: its absolute value

and argument.

Proposition 3.14 (Polar Form). Every complex number z has a polar formW

z D r.cos � C i sin �/;

where r � 0 and 0 � � < 2� . If z ¤ 0, then this expression is unique.

We may paraphrase

uniqueness of polar forms:

two vectors are equal if

and only if they have the

same length and the same

direction.

Proof. Existence is given by Eq. (3.4). Uniqueness is almost obvious. Suppose

that z D r .cos � C i sin �/ for some r > 0 and � . Then

jzj D jr .cos � C i sin �/j D jr j jcos � C i sin � j D jr j D r;

since r � 0 and jcos � C i sin � j D cos2 � C sin2 � D 1; thus, r D jzj. But then

cos � C i sin � D 1 r z D 1jzjz D

a

jzj C i b

jzj ;

so that

cos � D ajzj and sin � D a

jzj and � D arg.z/.

Example 3.15. If z D 3 C 4i , the Pythagorean Theorem gives jzj2 D 32 C 42 D 25, so that jzj D 5; your favorite computer gives arg.z/ D � D cos�1.3

5 / � 53:13ı. Thus, the polar form of z is

z D cos � C i sin � � 5 � cos 53:13ı C i sin 53:13ı

� : N

The Geometry Behind Multiplication

We’re ready to give a geometric interpretation of complex multiplication. Propo-

sition 3.13(iii) tells part of the story: the absolute value of a product is the prod-

uct of the absolute values. To finish the geometric analysis of multiplication, We know that zw sits on a circle of radius jzjjwj, centered at the origin. But

where?

we need to know how arg.zw/ is related to arg.z/ and arg.w/. We may as well

assume that z and w are unit vectors (why?), so that z D cos ˛ C i sin ˛ and w D cosˇ C i sinˇ. Multiply them together and collect real and imaginary parts.

zw D .cos ˛ C i sin˛/.cos ˇ C i sinˇ/ (3.5) D .cos ˛ cosˇ � sin˛ sinˇ/C i.cos ˛ sinˇ C sin ˛ cosˇ/: (3.6)

Do <.zw/ and =.zw/ look familiar? They are the addition formulas for sine and cosine:

cos.˛ C ˇ/ D cos ˛ cosˇ � sin ˛ sinˇ

and

sin.˛ C ˇ/ D cos ˛ sinˇ C sin ˛ cosˇ:

These formulas will give a beautiful characterization of the product of two

complex numbers. We now prove them, beginning with a familiar lemma that

uses Figure 3.4.

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102 Chapter 3 Renaissance

W r, s= ( )

U c, d= ( )

O

Figure 3.4. Orthogonality and dot product.

Lemma 3.16. If W D .r; s/ and U D .c; d /, then the arrows ��!OW and ��!OUWe are proving that per- pendicularity is equivalent

to the dot product being 0. are perpendicular if and only if rc C sd D 0.

Proof. We use the Pythagorean Theorem and its converse, Exercise 1.11 on

page 7: ��! OU ? ��!OW if and only if jUW j2 D jOU j2CjOW j2. Let h D jUW j;

then

h2 D .r � c/2 C .s � d/2

D r2 � 2rc C c2 C s2 � 2sd C d 2

D r2 C s2 C c2 C d 2 � 2.rc C sd/:

But jOW j2 D r2 C s2 and jOU j2 D c2 C d 2. Hence,

jOW j2 C jOU j2 D .r2 C s2/C .c2 C d 2/

and

h2 D r2 C s2 C c2 C d 2 � 2.rc C sd/:

Therefore, h2 D jOU j2 C jOW j2 if and only if rc C sd D 0.

In Figure 3.4, the coor-

dinates of U are labeled

.c; d/. This lemma shows

that .c;d/ D .�s; r/.

Now for the addition formulas.

Theorem 3.17 (Addition Theorem). Let ˛ and ˇ be angles.

(i) cos.˛ C ˇ/ D cos ˛ cosˇ � sin˛ sinˇ: (ii) sin.˛ C ˇ/ D cos˛ sinˇ C sin ˛ cosˇ:

Proof. In Figure 3.5, we have a picture of the unit circle. Let Z D .a; b/ DWe are looking at points here as elements of R2, al-

though we’ll soon interpret

this diagram in the complex

plane.

.cos ˛; sin˛/ and W D .r; s/ D .cos ˇ; sinˇ/. Rotate 4OQZ counterclock- wise through †ˇ to get 4OQ0Z0, so that 4OQZ and 4OQ0Z0 are congru- ent. Thus, Z0 D .cos.˛ C ˇ/; sin.˛ C ˇ// : Our task is to find the coordinates of Z0 in terms of r; s; a, and b.

Define U D .�s; r/. Since W D .r; s/ is on the unit circle, we have r2 C s2 D 1, and so U D .�s; r/ is also on the unit circle. Moreover, since .�s/r C rs D 0, Lemma 3.16 says that ��!OU is orthogonal to ��!OW . Therefore, O Q0 Z0M is a rectangle.

Decompose ��! OZ0 as the sum of two vectors:

��! OZ0 D

��! OQ0 C ��!OM;

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3.2 Complex Numbers 103

a

a b

Q a= ( , 0)

Z¢

O

M

Z a, b= ( )

W r, s= ( )

U –s, r= ( )

Q¢

Figure 3.5. Addition Theorem.

where ��! OM is the projection of

���! Q0Z0 onto

��! OU . We can get explicit expres-

sions for Q0 and M . First, ��! OQ0 is a scalar multiple of

��! OW and, because

jOQj D a, we know the scalar:

Q0 D a.r; s/ D .ar; as/:

Second, ��! OM is a scalar multiple of

��! OU , where U D .�s; r/; as jOM j D

jQZj D b, we know the scalar:

M D b.�s; r/ D .�bs; br/:

Therefore,

��! OZ0 D

��! OQ0 C ��!OM D .ar; as/C .�bs; br/ D .ar � bs; as C br/:

Making the appropriate substitutions for a, b, r , and s, we have the desired

result:

cos.˛ C ˇ/ D cos ˛ cosˇ � sin ˛ sinˇ and

sin.˛ C ˇ/ D cos˛ sinˇ C sin˛ cosˇ:

Here is the result we have been seeking.

Theorem 3.18 (The Geometry of Multiplication.). If z and w are complex In words, the length of a product is the product

of the lengths, and the

argument of a product is

the sum of the arguments.

The equality in Theo-

rem 3.18(ii) holds up to a

multiple of 2� .

numbers, then

(i) jzwj D jzj jwj, and (ii) arg.zw/ D arg.z/C arg.w/.

Proof. The first statement is Proposition 3.13, and the second follows from

Theorem 3.17 and Eq. (3.5) on page 101.

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104 Chapter 3 Renaissance

If we set arg.z/ D ˛ and arg.w/ D ˇ, then Theorem 3.18 has an especially pleasing restatement in polar form.

Corollary 3.19. If z D jzj.cos ˛ C sin˛/ and w D jwj.cosˇ C sinˇ/, then

z � w D jzwj .cos.˛ C ˇ/C i sin.˛ C ˇ// :

Proof. Both sides equal jzj .cos˛ C i sin˛/ � jwj .cosˇ C i sinˇ/.

It follows easily, by induction on k � 1, that if z is a complex number and k 2 Z , then

ˇ̌ zk ˇ̌

D jzjk and arg � zk �

D k arg.z/:

How to Think About It. There’s a way to see, without using trigonometry, that angles add in the product of two complex numbers. Essentially, we recast

the proof of Theorem 3.17 in terms of complex numbers. Given z D a C bi and w D r C si , we want to determine arg.zw/ in terms of arg.z/ D ˛ and arg.w/ D ˇ. We can assume that z and w are unit vectors; this implies that zw is also a unit vector, by Proposition 3.13. The key insight is that

zw D .aC bi/w D aw C .bi/w D aw C b.iw/:

You know the geometric effect of scalar multiplication (Exercise 3.27 on page 98),This approach does not seem to be very well

known. It appears in [22]

and Kerins, B. “Gauss,

Pythagoras, and Heron”

(Mathematics Teacher,

96:5, 2003), but we can’t

find any older sources.

you know how to add geometrically (parallelogram law), and you know

iw D i.r C si/ D �s C ri I

using Lemma 3.16, it follows that iw is obtained from w by counterclock-

wise rotation by 90ı. Figure 3.6 below is almost the same as Figure 3.5; the difference is that points are now labeled as complex numbers.

a

a b

O a

aw

b iw( )

u iw –s ri= = +

z a bi= +

w r si= +zw aw b iw= + ( )

Figure 3.6. Complex multiplication again.

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3.2 Complex Numbers 105

Let’s put this all together. Triangle OZ0Q0 is congruent to triangle OZQ, so that †Z0OW D †ZOQ D ˛; hence, arg.zw/ D ˛ C ˇ. We have shown that Theorem 3.18(ii) follows without any mention of trigonometry.

We have just used plane geometry to derive the geometric interpretation of

complex multiplication, avoiding the trigonometric addition formulas. Aside from proving these ideas for students who haven’t yet seen the addition for-

mulas, complex numbers can now be used to derive these formulas. That’s an

additional bonus, especially for a precalculus class; it allows us to use complex

numbers when trying to establish other trigonometric identities that depend on

the addition formulas. For example, to get a formula for cos � � 4

C � � , calculate

like this:

cos � � 4

C � �

C i sin � � 4

C � �

D � cos �

4 C i sin �

4

� .cos � C i sin �/

D 1p 2 .1C i/ .cos � C i sin �/

D 1p 2

� .cos � � sin �/C i.cos � C sin �/

� :

Hence

cos � � 4

C � �

D 1p 2 .cos � � sin �/

and, as a bonus,

sin � � 4

C � �

D 1p 2 .cos � C sin �/:

Over the next two centuries, people became comfortable with the fact that

polynomial equations with real coefficients can have complex solutions. It was eventually proved that every polynomial f .x/ D xnCcn�1xn�1C� � �Cc1xC c0 with real coefficients has a factorization

f .x/ D .x � ˛1/ � � � .x � ˛n/;

where ˛1; : : : ; ˛n are complex numbers. This amazing result holds for any

nonzero polynomial f with complex coefficients; it is known as the Funda- mental Theorem of Algebra. We won’t prove this result here because, in spite

of its name, it is a theorem of analysis, not of algebra; you can find a readable

account in [4], pp. 142–152.

Exercises

3.34 If z D a C bi , prove that the arrow corresponding to z, namely ��!OP , where P D .a; b/, is perpendicular to the arrow corresponding to iz.

3.35 If z and w are complex numbers with w ¤ 0, show that

arg.z=w/ D arg.z/ � arg.w/:

3.36 (i) Prove that the quadratic formula holds for polynomials with complex coeffi-

cients (use Proposition 3.6).

(ii) Find the roots of x2 C 2ix � 1. Why aren’t these roots conjugate?

3.37 If z and w are complex numbers, find a necessary and sufficient condition that

jz C wj to equal jzj C jwj.

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106 Chapter 3 Renaissance

3.38 * If z D cos ˛ C i sin ˛, show that

z D 1=z D cos.�˛/ C i sin.�˛/:

3.39 Let n � 0 is an integer and � D cos.2�n / C i sin. 2� n /. If z is a complex number,

give a geometric description of how �z is located with respect to z on the complex

plane.

3.40 Preview. Plot the roots in the complex plane for each of the polynomials x2 � 1, x3 � 1, x4 � 1, x6 � 1, and x12 � 1.

3.41 Preview. Let � D cos.45ı/ C i sin.45ı/. (i) Show that �8 D 1

(ii) Show that the distinct roots of x8 � 1 are precisely 1; �; �2; : : : ; �7. (iii) Plot these roots in the complex plane.

(iv) Show that �147 is a root of x8 � 1. (v) Show that �147 is equal to one of the roots in part (ii). Which one?

3.42 Preview. Let n be a nonnegative integer and � D cos.2� n

/ C i sin.2� n

/.

(i) Let � D cos.2�n / C i sin. 2� n /. Show that �

n D 1.

(ii) If k is any nonnegative integer, show that �

�k �n

D 1.

(iii) Give a geometric description of the subset f�k W k � 0g of the complex plane.

3.43 Take It Further. If f .x/ is a polynomial with complex coefficients, define f to

be the polynomial you get by replacing each coefficient in f by its conjugate.

Prove the following statements.

(i) f C g D f C g: (ii) fg D f g:

(iii) f D f if and only if f .x/ has real coefficients. (iv) f f has real coefficients.

(v) f .z/ D f .z/:

3.44 Take It Further. Suppose f .x/ is a polynomial with coefficients in C. If a

complex number z is a root of f , show that z is a root of f .

3.45 Take It Further. Suppose f .x/ is a polynomial with coefficients in C. Then

by Exercise 3.43 on page 106, if we define g.x/ D f .x/f .x/, then g.x/ has coefficients in R. Show that if g.z/ D 0, either f .z/ D 0 or f .z/ D 0. Hence conclude that if every polynomial with real coefficients and degree at least 1 has

a root in C, then every polynomial with complex coefficients and degree at least

1 has a root in C. (The Fundamental Theorem of Algebra says that every polyno-

mial with complex coefficients has all its roots in C. This exercise shows that it’s

enough to prove this for polynomials with real coefficients.)

3.3 Roots and Powers

We saw in the previous section that every point z on the unit circle can be written as z D cos � C i sin � for some angle � . Theorem 3.18 tells us that arguments add when complex numbers are multiplied. In particular,

.cos � C i sin �/2 D .cos � C i sin �/ .cos � C i sin �/ D cos.� C �/C i sin.� C �/ D cos.2�/C i sin.2�/:

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3.3 Roots and Powers 107

On the other hand, complex multiplication gives

.cos � C i sin �/2 D � cos2 � � sin2 �

� C i2 cos � sin �:

Equating real parts and imaginary parts gives the double angle formulas:

cos.2�/ D cos2 � � sin2 � sin.2�/ D 2 cos � sin �:

We now generalize this to any positive integer power.

Theorem 3.20 (De Moivre). For every angle � and all integers n � 0,

.cos � C i sin �/n D cos.n�/C i sin.n�/:

Proof. We prove equality by induction on n � 0. The theorem is true when n D 0, for cos 0 D 1 and sin 0 D 0. Here is the inductive step.

.cos � C i sin �/n D .cos � C i sin �/n�1 .cos � C i sin �/

D �

cos � .n � 1/�

� C i sin

� .n � 1/�

�� .cos � C i sin �/

D cos � .n � 1/� C �

� C i sin

� .n � 1/� C �

�

D cos.n�/ C i sin.n�/:

Example 3.21. (i)

.cos 3ı C i sin 3ı/40 D cos 120ı C i sin 120ı D �1 2

C i p 3 2 :

(ii) Let z D cos.45ı/C i sin.45ı/ D 1p 2 .1C i/. We compute z6 in two ways:

� With the Binomial Theorem:

z6 D � 1p 2 .1C i/

�6

D � 1p 2

�6 .1C i/6

D 1 8

� 1C 6i C 15i2 C 20i3 C 15i4 C 6i5 C i6

�

D 1 8 .1C 6i � 15 � 20i C 15C 6i � 1/

D �i:

� With De Moivre’s Theorem:

z6 D � cos.45ı/C i sin.45ı/

�6 D cos.6 � 45ı/C i sin.6 � 45ı/ D cos.270ı/C i sin.270ı/ D �i: N

Polar Decomposition and De Moivre’s Theorem combine to give a nice

formula for computing powers of any complex number.

Corollary 3.22. If z D r � cos ˛ C i sin˛

� is a complex number, then

zn D rn � cos.n˛/C i sin.n˛/

� :

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108 Chapter 3 Renaissance

We were unable to find a cube root of z D a C ib earlier, but there’s no problem now.

Corollary 3.23. Let r.cos �Ci sin �/ be the polar form of a complex number z. If n � 1 is an integer, then

h n p r �

cos � �=n

� C i sin

� �=n

��in D r.cos � C i sin �/ D z:

Example 3.24. In Example 3.15, we saw that the polar form of z D 3C 4i isOf course, we must find the polar form of z,

which involves finding

� D cos�1.a=jzj/.

approximately 5 .cos.53:13ı/C i sin.53:13ı//. Now .53:13/=3 D 17:71, and so a cube root of z is approximately

3 p 5 .cos.17:71ı/C i sin.17:71ı//. Our

calculator says that

� 3 p 5 � cos 17:71ı C i sin 17:71ı

��3 D 3:000001C 3:99999i: N

We are now going to describe a beautiful formula discovered by Euler. Re- call some power series formulas from calculus. For every real number x,

ex D 1C x C x 2

2Š C � � � C x

n

nŠ C � � � ;

cos x D 1 � x 2

2Š C x

4

4Š � � � � C .�1/

nx2n

.2n/Š C � � � ;

and

sin x D x � x 3

3Š C x

5

5Š � � � � C .�1/

n�1x2nC1

.2nC 1/Š C � � � :

We can define convergence of power series P1 nD0 cnz

n for z and cn com-

plex numbers, and we can then show that the series

1C z C z 2

2Š C � � � C z

n

nŠ C � � �

converges for every complex number z. The complex exponential ez is defined

to be the sum of this series. In particular, the series for eix converges for all

real numbers x, and

eix D 1C ix C .ix/ 2

2Š C � � � C .ix/

n

nŠ C � � � :

Theorem 3.25 (Euler). For all real numbers x,It is said that Euler was delighted by the special

case

ei� C 1 D 0;

for it contains five important

constants in one equation.

eix D cos x C i sinx:

Sketch of proof. We will not discuss necessary arguments involving conver- gence. As n varies over 0; 1; 2; 3; 4; 5; : : :, the powers of i repeat every four

steps: that is, the sequence

1; i; i2; i3; i4; i5; i6; i7; i8; i9; i10; i11; : : :

is actually

1; i; �1; �i; 1; i; �1; �i; 1; i; �1; �i; : : : I

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3.3 Roots and Powers 109

the even powers of i are all real, whereas the odd powers all involve i . It

follows, for every real number x, that .ix/n D inxn takes values

1; ix; �x2; �ix3; x4; ix5; �x6; �ix7; x8; ix9; �x10; �ix11; : : :

Thus, in the definition of the complex exponential,

eix D 1C ix C .ix/ 2

2Š C � � � C .ix/

n

nŠ C � � � ;

the even powers of ix do not involve i , whereas the odd powers do. Collecting

terms, one has eix D even terms C odd terms. But

even terms D 1C .ix/ 2

2Š C .ix/

4

4Š C � � �

D 1 � x 2

2Š C x

4

4Š � � � �

and

odd terms D ix C .ix/ 3

3Š C .ix/

5

5Š C � � � :

D i.x � x 3

3Š C x

5

5Š � � � � /:

Therefore, eix D cos x C i sin x. � As a consequence of Euler’s Theorem, the polar decomposition can be

rewritten in exponential form: every complex number z has a factorization

z D rei� ;

where r � 0 and 0 � � < 2� . We have chosen to denote

P1 nD0

.ix/n

nŠ by eix, but we cannot assert,

merely as a consequence of our notation, that the law of exponents, eixeiy D ei.xCy/, is valid. But this is precisely what Corollary 3.19 says once it is trans- lated into exponential notation.

Theorem 3.26 (Exponential Addition Theorem). For all real numbers x and

y,

eixeiy D ei.xCy/:

Proof. According to Corollary 3.19,

eixeiy D .cos x C i sinx/.cos y C i siny/ D cos.x C y/C i sin.x C y/ D ei.xCy/:

We can also translate De Moivre’s Theorem into exponential notation.

Corollary 3.27 (Exponential De Moivre). For every real number x and all

integers n � 1,

.eix/n D einx :

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110 Chapter 3 Renaissance

Proof. According to De Moivre’s Theorem,

.eix/n D .cos x C i sinx/n D cos.nx/C i sin.nx/ D einx :

It is easier to remember the trigonometric addition formulas in complex

form. For example, let’s find the triple angle formulas. On the one hand, De

Moivre’s Theorem gives

ei3x D cos.3x/C i sin.3x/:

On the other hand,

ei3x D .eix/3

D � cos x C i sinx

�3

D cos3 x C 3i cos2 x sin x C 3i2 cos x sin2 x C i3 sin3 x D cos3 x � 3 cos x sin2 x C i

� 3 cos2 x sin x � sin3 x

� :

Equating real and imaginary parts, we have

cos.3x/ D cos3 x � 3 cos x sin2 x

and

sin.3x/ D 3 cos2 x sinx � sin3 x:

Roots of Unity

De Moivre’s Theorem can be used to find the roots of an important family of

polynomials: those of the form xn � 1.

Theorem 3.28. The distinct roots of xn � 1 are

1; �; �2; � � � ; �n�1;

where � D �n D cos.2�=n/C i sin.2�=n/. These numbers are equally spaced on the unit circle and are the vertices of a regular polygon, called the unit

n-gon.

Proof. By Corollary 3.23, �n D 1, so that � is a root of xn � 1. Furthermore, for any nonnegative integer k, we have .�k/n D .�n/k D 1, so that all �k are also roots of xn � 1. But there are repetitions on the list 1; �; �2; � � � . By the Division Algorithm, for any j , we have j D qn C r , where 0 � r � n � 1. Hence,

�j D �qnCr D �qn �r D �r ;

because �qn D 1. On the other hand, all the �k , for 0 � k � n � 1, are distinct. After all, by

De Moivre’s Theorem,

�k D cos.2�k=n/C i sin.2�k=n/;

and Proposition 3.14, uniqueness of polar forms, applies, for 0 � 2�k=n < 2� are n distinct angles. Therefore, we have displayed n distinct roots of xn � 1.

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3.3 Roots and Powers 111

These are all the roots of xn � 1, for a polynomial of degree n can have at most n distinct roots. We’ll give a proof of this later (see Theorem 6.16) but,

since we haven’t yet proved this result, we now proceed in a different way. If z 2 C is a root of xn � 1, then 1 D jznj D jzjn, so that jzj D 1,

by Exercise 1.73 on page 41, and z D cos � C i sin � for some � . By De Moivre’s Theorem, 1 D zn D cos.n�/ C i sin.n�/, so that n� D 2�k for some integer k; hence, � D 2�k=n. Write k D qnC r , where 0 � r < n, and

z D cos.2�k=n/C i sin.2�k=n/ D cos.2�r=n/C i sin.2�r=n/:

Thus, z is equal to the root �r already displayed. Finally, since arg

� �k �

D k arg .�/ D 2�k=n, the roots �k are equally spaced around the circle and, hence, they are the vertices of a regular n-gon.

(See Figure 3.7 for the case n D 8.)

Definition. The roots of xn � 1 are called the nth roots of unity. An nth root of unity � is a primitive nth root of unity if n is the smallest positive integer

for which �n D 1.

For every n � 1, we see that � D e2�i=n is a primitive nth root of unity, for if 1 � m < n, then �m D cos.2�m=n/ C i sin.2�m=n/ ¤ 1. In particular, i D cos.2�=4/ C i sin.2�=4/ is a primitive fourth root of unity, and ! D 1 2 .�1C

p 3/ D cos.2�=3/C i sin.2�=3/ is a primitive cube root of unity.

Corollary 3.29. Let �k D cos.2�k=n/Ci sin.2�k=n/ be an nth root of unity. (i) �k is a primitive nth root of unity if and only if gcd.k; n/ D 1.

(ii) If �k is a primitive nth root of unity, then every nth root of unity is a power

of �k .

Proof. (i) Suppose that �k is a primitiventh root of unity. If d D gcd.k; n/ > 1, then n=d < n, and

.�k/n=d D .�n/k=d D 1:

This contradicts n being the smallest positive integer with .�k/n D 1. Suppose that �k is not primitive; that is, .�k/m D 1 for some m < n.

Since, by hypothesis, gcd.k; n/ D 1, there are integers s and t with 1 D sk C tn; hence, m D msk Cmtn. But now

�m D �mskCmtn D �msk�mtn D 1;

which contradicts � being a primitive nth root of unity.

(ii) Every nth root of unity is equal to �j for some j . If gcd.k; n/ D 1, then there are integers s and t with 1 D sk C tn. Hence,

�j D �jskCjtn D �jsk�jtn D .�k/js :

Definition. For every integer n � 1, define the Euler �-function �.n/ by

�.n/ D number of k with 1 � k � n and gcd.k; n/ D 1:

For example, �.1/ D 1 and, if p is prime, �.p/ D p � 1.

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112 Chapter 3 Renaissance

Corollary 3.30. For every positive integer n, there are exactly �.n/ primitive

nth roots of unity.

Proof. This follows at once from Corollary 3.29(i).

Example 3.31. The complex number cos.2� n / C i sin.2�

n / is a primitive nth

root of unity, by Theorem 3.28. The 8th roots of unity (shown in Figure 3.7)

are

cos.2� 8 /C i sin.2�

8 /; cos.6�

8 /C i sin.6�

8 /;

cos.10� 8 /C i sin.10�

8 /; cos.14�

8 /C i sin.14�

8 /I

that is, the primitive 8th roots of unity are all those cos.2k� 8 /C i sin.2k�

8 / for

which gcd.k; 8/ D 1. N

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Figure 3.7. 8th roots of unity.

The nth roots of unity enjoy some remarkable properties that we’ll use in

upcoming chapters; here are some of them. (See Exercise 3.51 on page 115

and Proposition 6.63 for some other interesting properties.)

Theorem 3.32. Let � be an nth root of unity.

(i) 1C �2 C �3 C � � � C �n�1 D 0. (ii) �k D 1=�k for every nonnegative integer k.

(iii) If k D qnC r , then �k D �r .

Proof. (i) We have xn � 1 D .x � 1/q.x/, and we find q by long division:

.x � 1/ � 1C x C x2 C � � � C xn�1

� D xn � 1:

Now set x D � to see that

.� � 1/.1 C � C �2 C � � � C �n�1/ D �n � 1 D 0:

But � � 1 ¤ 0, and the result follows. (ii) This follows from Exercise 3.38 on page 106.

(iii) This follows from �qn D 1.

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3.3 Roots and Powers 113

Theorem 3.28 establishes an intimate connection between the nth roots of

unity and the geometry of the unit n-gon. The next examples illustrate this

connection for small n.

Example 3.33. The vertices of unit n-gons for small values of n can be calcu- lated with plane geometry.

See Exercise 3.49 on

page 115 for more exam-

ples.

� The vertices of the unit 3-gon are 1; 1 2 .�1 C i

p 3/; 1

2 .�1 � i

p 3/:

� The vertices of the unit 4-gon are 1; i; �1; �i . � The vertices of the unit 6-gon are

1; 1 2 .1Ci

p 3/; 12 .�1Ci

p 3/; �1; 12 .�1�i

p 3/; 12 .1�i

p 3/: N

Example 3.34 (Regular Pentagon). Since a primitive 5th root of unity is � D cos.2�=5/C i sin.2�=5/, by Theorem 3.28, the vertices of the unit 5-gon are

� D cos.2�=5/C i sin.2�=5/ �2 D cos.4�=5/C i sin.4�=5/ �3 D cos.6�=5/C i sin.6�=5/ �4 D cos.8�=5/C i sin.8�=5/ �5 D cos.10�=5/C i sin.10�=5/ D 1 D �0:

1

Figure 3.8. Unit 5-gon.

Can we find explicit expressions for these vertices that don’t involve trigonom-

etry? We’ll obtain such an expression for cos.2�=5/, but we’ll leave the rest of the details for you (Exercise 3.49(i) on page 115); after all, you can evaluate,

say, cos.8�=5/.

We have � D �4, by Theorem 3.32(ii). Inspired by Lemma 3.2, we define

g D � C �4 and h D �2 C �3:

Now

g D � C �4 D 2 cos.2�=5/;

a real number that is twice the number we are after. Similarly,

h D �2 C �3 D 2 cos.4�=5/;

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114 Chapter 3 Renaissance

another real number. By Theorem 3.32(i),

g C h D � C �4 C �2 C �3

D �1:

Thus, we know that g C h D �1. Do we also know gh? Let’s see.

gh D � � C �4

� .�2 C �3/

D �3 C �4 C �6 C �7

D �3 C �4 C � C �2 by Theorem 3.32(iii) D �1:

Hence, g C h D �1 D gh, and so g and h are the roots of

x2 C x � 1:

Now x2 C x � 1 has a positive root and a negative one. Since g > h (why?), the positive root is g, and so

cos.2�=5/ D 1 2 g

D 1 4 .�1C

p 5/: N

An ancient problem, going back to the Greeks, is to determine which regular

n-gons can be constructed with ruler and compass. As we’ll see in Chapter 7,

the problem comes down to finding an expression for cos.2�=n/ that doesn’t

mention trigonometry, only the operations of arithmetic and iterated square

roots (as in Example 3.34). We’ve essentially shown that the regular pentagon

is so constructible. This argument, grouping the �k into convenient subclusters, was greatly generalized and refined by Gauss (when he was only 17 years old!)

to show that the vertices of the unit 17-gon can be constructed (Euclid did not

know this!). (This and much more is in Gauss’s masterpiece, Disquisitiones

Arithmeticae.) Gauss requested that his tombstone portray a regular 17-gon,

but the stonemason was unable to carve it, saying it would look more like a circle than a polygon.

Exercises

3.46 Is De Moivre’s Theorem true for negative integer exponents? Explain.

3.47 Let z D cos � C i sin � . Show, for all nonnegative integers n, that

zn C .z/n D 2 cos n� and zn � .z/n D 2 sin n�:

3.48 This exercise shows that there’s something special about a 72ı angle: there’s only one isosceles triangle (up to similarity) whose base angle is twice the vertex an-

gle, namely, the “72-72-36 triangle.” Let the equal sides of such a triangle have

length 1, and let q denote the length of the base.

(i) Bisect one of the base angles of the triangle.

(ii) Show that the small triangle is similar to the whole triangle.

(iii) Use (ii) to show that 1 q

D q 1�q ; and solve for q.

(iv) Show that q=2 D cos 72ı.

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3.3 Roots and Powers 115

1 1

q

72°72°

36°

Figure 3.9. 72-72-36 triangle.

1

1 – q

q

q

q

Figure 3.10. Its construction.

3.49 * Find explicit formulas (i.e., without trigonometry) for the vertices of the unit

(i) pentagon. (ii) decagon. (iii) 20-gon.

3.50 * Let n be a positive integer and let � D e2�i=n. (i) Establish the identity

xn � 1 D .x � 1/.x � �/ � x � �2

� � � � �

x � �n�1 �

:

(ii) If x and y are integers, show that

xn � yn D .x � y/.x � �y/ �

x � �2y �

� � � � x � �n�1y

� :

(iii) If x and y are integers and n is odd , show that

xn C yn D .x C y/.x C �y/ �

x C �2y �

� � � � x C �n�1y

� :

3.51 Take It Further. We saw, on page 111, that �.p/ D p � 1, where p is prime and � is the Euler-� function. Note that if n D p1p2,

then .p1 � 1/.p2 � 1/ D n� n

p1 � n

p2 C n

p1p2 :

(i) Suppose n is the product of two primes, n D p1p2 . Show that

�.n/ D .p1 � 1/.p2 � 1/:

(ii) Suppose n is the product of two primes powers, n D pe11 p e2 2 . Show that

�.n/ D n � n p1

� n p2

C n p1p2

D n �

1 � 1 p1

�� 1 � 1

p2

� :

(iii) Generalize to show that, if n D pe11 p e2 2 : : : p

en n , then

�.n/ D n nY

kD1

� 1 �

1

pk

� :

3.52 Prove or disprove and salvage if possible. If a and b are positive integers,

�.ab/ D �.a/�.b/:

3.53 Find explicit formulas (i.e., without trigonometry) for the vertices of the unit n-

gon if

(i) n D 3 (ii) n D 4 (iii) n D 6 (iv) n D 8 (v) n D 12 (vi) n D 16

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116 Chapter 3 Renaissance

3.54 For all integers n between 3 and 9, find all the primitive nth roots of unity.

3.55 Find a primitive 12th root of unity �. Is � unique?

3.56 Suppose � D cos.2� 7

/ C i sin.2� 7

/.

(i) Plot the roots of x7 � 1 in the complex plane. (ii) Show that ˛ D �C�6, ˇ D �2C�5, and D �3C�4 are real numbers.

(iii) Find a cubic equation satisfied by 2 cos.2�7 / by finding the values of ˛CˇC , ˛ˇ C ˛ C ˇ , and ˛ˇ .

3.57 If �n D cos.2�n / C i sin. 2� n

/, evaluate Pn�1 kD0 �

k n :

3.58 Show that cos.2�5 / C cos. 4� 5 / D �

1 2 :

3.59 Take It Further. If n is a nonnegative integer, how many irreducible factors over

Z does xn � 1 have? In other words, we’re looking for a pattern in the outputs of the function n 7! # of factors of xn � 1 over Z: (Use a computer).

n Number of Factors of xn � 1 1

2

3

4

5

6

7

8

9

10

11

12

3.4 Connections: Designing Good Problems

This section will use complex numbers to help create mathematics problems

that “come out nice.” When launching a new topic, you want to start with

examples which focus on the new idea; there shouldn’t be any distractions—

for example, numbers should be simple integers or rationals. Indeed, this is why the Babylonians introduced Pythagorean triples.

Norms

We begin by introducing a function C ! R, called the norm, that is closely related to absolute value. It will be an important tool for our applications; it will also be very useful in Chapter 8 when we do some algebraic number theory.

Definition. The norm of a complex number z D aC bi is

N.z/ D zz D a2 C b2:

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3.4 Connections: Designing Good Problems 117

Here are some basic properties.

Proposition 3.35. Let z D aC ib and w be complex numbers. (i) N.z/ is a nonnegative real number, andN.z/ D 0 if and only if z D 0.

(ii) N.z/ D jzj2. (iii) N.zw/ D N.z/N.w/.

Proof. (i) This follows at once from N.aC bi/ D a2 C b2. (ii) This follows at once from jzj D jaC bi j D

p a2 C b2.

(iii) N.zw/ D zw zw D zw z w D zz ww D N.z/N.w/:

It follows from Proposition 3.35(iii) that

N.zk/ D N.z/k

for all z and all k � 0. Here is an application of the norm.

Example 3.36. Let’s revisit Example 3.5, the “bad example,” in which the

cubic formula gives the roots of x3 � 7x C 6 D .x � 1/.x � 2/.x C 3/ in unrecognizable form.

Imagine again that you have just left the contest in Piazza San Marco, think-

ing about how g C h could possibly equal 1, where

g3 D �3C i 10 9

p 3 and h3 D �3 � i 10

9

p 3:

Had you known about conjugates, you’d have seen that g3 D h3. It would have been natural to guess that the cube roots g and h are also complex conjugates (you’d have guessed right: see Exercise 3.64 on page 127); thus, g D a C ib and h D a� ib. Now if gC h D 1, as your opponent loudly proclaimed, then .a C ib/C .a � ib/ D 2a D 1; that is,

g D 1 2

C ib and h D 1 2

� ib:

You really want to find g and h now—what is b? Using the norm function, you

see that

N.g/3 D N.g3/ D .�3/2 C � 10

p 3

9

�2 D 343

27 :

Since norms are always real numbers, you conclude that

N.g/ D 3 q 343 27

D 7 3

(the other cube roots are complex; they are 7 3 ! and 7

3 !2, where ! is a primitive

cube root of unity). But if g D 1 2

C ib, thenN.g/ D 1 4

Cb2. Hence, 1 4

Cb2 D 7 3

, and b D ˙ 5 2

p 3

. Thus,

g D 1 2

C i 5 2

p 3

and h D 1 2

� i 5 2

p 3 :

Bingo! For these “values” of g and h, we have g C h D 1. You were right! To find the other two roots, see Exercise 3.65 on

page 127. Elated, you run back to the square to show off g and h, but everyone has gone home. N

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118 Chapter 3 Renaissance

Pippins and Cheese

We call this subsection Pippins and Cheese, a phrase borrowed from Shake-

speare’s Merry Wives of Windsor, which describes delicious desserts.

Here are five problems. Solve them now; they are not difficult, but the an- swers may surprise you.

(i) A triangle has vertices .�18; 49/; .15;�7/, and .30;�15/: How long are its sides?

(ii) In Figure 3.11, the side lengths of 4QSU are as marked. How big is †Q?

U

Q S

5 7

8

Figure 3.11. Side lengths.

(iii) An open box is formed by cutting out squares from a 7� 15 rectangle and folding up the sides (see Figure 3.12). What size cut-out x maximizes the

volume of the box?

7

15

Figure 3.12. Making boxes.

(iv) Find the zeros, extrema, and inflection points of the function

f .x/ D 140� 144 x C 3 x2 C x3:

(v) Find the area of the triangle with sides of lengths 13, 14, and 15.

A meta-problem is a problem that asks how to design “nice” exercises of

a particular genre, such as “How do you construct integer-sided scalene trian-

gles having a 60ı angle?” As we mentioned earlier, finding Pythagorean triples was one of the first meta-problems; it was invented by teachers who wanted to

study and apply side-lengths of right triangles. In Section 1.2, we developed

the method of Diophantus for this purpose—rational points on the unit circle

correspond to Pythagorean triples. In this section, we’ll consider two types

of meta-problems: two ways of creating exercises like the five listed above. One meta-problem uses the norm function; the other generalizes Diophantus’sSee Exercises 1.79—1.82

on page 44. chord method of “sweeping lines” by replacing circles with other conic sec-

tions. (There are many other kinds of meta-problems, ranging in topic from

exponential equations to algebra word problems to trigonometry.)

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3.4 Connections: Designing Good Problems 119

Gaussian Integers: Pythagorean Triples Revisited

In Chapter 1, we saw that the Pythagorean equation a2C b2 D c2 corresponds to a rational point

� a c ; b c

� on the unit circle. But, given what you’ve just been

studying, the Pythagorean equation might conjure up another rewrite in your

mind, namely

.a C bi/.a � bi/ D c2;

or even

N.a C bi/ D c2:

So, we’re looking for complex numbers z D a C bi whose norms are perfect squares of integers. The Pythagorean equation now looks like

N.z/ D c2:

For example, N.3C 4i/ D 52, N.5C 12i/ D 132, and N.8C 15i/ D 172. This idea doesn’t work for every complex number. What’s needed are com-

plex numbers whose real and imaginary parts are integers (and, besides, whose norms are perfect squares). We’d like the real and imaginary parts to be posi-

tive integers, but any integers will do, because changing the sign of the real or

imaginary part of a complex number doesn’t change its norm (why?).

Definition. The Gaussian integers is the set ZŒi  of all complex numbers

whose real and imaginary parts are integers. In symbols,

ZŒi  D faC bi 2 C W a 2 Z and b 2 Zg :

Proposition 3.37. (i) The set ZŒi  of Gaussian integers is closed under ad-

dition and multiplicationW If aC bi; c C di 2 ZŒi , then

.a C bi/C .c C di/ D .a C c/C .b C d/i 2 ZŒi  .a C bi/.c C di/ D .ac � bd/C .ad C bc/i 2 ZŒi :

(ii) If z D aC bi , then

N.z/ D a2 C b2:

Proof. The formula for addition is clear; for multiplication, use the fact that

i2 D �1. Of course, part (ii) is just the definition of the norm.

We’ll investigate the Gaussian integers in more detail in Chapter 8. Let’s return to the norm equation N.z/ D c2 arising from Pythagorean

triples, but with z a Gaussian integer. Our question is now “Which Gaussian

integers have perfect squares as norms?” The answer comes from Proposi-

tion 3.35(iii): if z and w are complex numbers, thenN.zw/ D N.z/N.w/: In particular (letting z D w),

N � z2

� D N.z/2:

The left-hand side of this equation is the norm of a Gaussian integer: if z D aC ib, then z2 D .a2�b2/C i2ab; moreover, N.z2/ is a sum of two nonzero

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120 Chapter 3 Renaissance

perfect squares if a > 0, b > 0, and a ¤ b. Now the right-hand side is the square of an integer, namely, N.z/2, which produces a Pythagorean triple. For

example, if z D 3 C 2i , then N.z/ D 13 and z2 D 5 C 12i , and we get the Pythagorean triple .5; 12; 13/, for

52 C 122 D N � .3C 2i/2

� D N.3C 2i/2 D 132:

We now have a quick way to generate Pythagorean triples (by hand or with

a computer; one of our colleagues uses this method to amaze friends at parties). Pick a Gaussian integer r C si (with r > 0, s > 0, and r ¤ s), and square it.

The r; s entry in the following table is Œ.r C is/2; N.r C is/. For exam- ple, the top entry in the first column, arising from r D 2 and s D 1, is Œ.2 C i/2; N.2 C i/ D Œ3 C 4i; 5; the corresponding Pythagorean triple is .3; 4; 5/.

s D 1 s D 2 s D 3 s D 4 r D 2 3C 4i; 5 r D 3 8C 6i; 10 5C 12i; 13 r D 4 15C 8i; 17 12C 16i; 20 7C 24i; 25 r D 5 24C 10i; 26 21C 20i; 29 16C 30i; 34 9C 40i; 41 r D 6 35C 12i; 37 32C 24i; 40 27C 36i; 45 20C 48i; 52

Eisenstein Integers.

Let’s now look at the meta-problem of creating triangles with integer side-

lengths and a 60ı angle. Let †C D 60ı in Figure 3.13, so that cos.†C/ D 12 . By the Law of

Cosines,

c2 D a2 C b2 � 2ab cos †C D a2 C b2 � ab:

What’s important here is that the right-hand side of the equation, a2�abCb2, is the norm of a C b!, where ! D 1

2 .�1 C i

p 3/ is a primitive cube root of

unity (Exercise 3.72 on page 128). This leads to the following definition.

b

a c

C

B

A

Figure 3.13. †C D 60ı.

Definition. The Eisenstein integers is the set ZŒ! of all complex numbers of

Eisenstein did extensive

research on complex

numbers of the form

a C b�, where � is a primitive nth root of unity.

Note that ! is a primitive

cube root of unity.

the form a C b!, where ! D 1 2 .�1 C i

p 3/ is a primitive cube root of unity

and a; b are integers. In symbols,

ZŒ! D faC b! 2 C W a 2 Z and b 2 Zg :

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3.4 Connections: Designing Good Problems 121

Here are some properties of Eisenstein integers.

Proposition 3.38. (i) The set ZŒ! of Eisenstein integers is closed under ad-

dition and multiplication: If .a C b!/; .c C d!/ 2 ZŒ!, then

InZŒi, i2 D �1. In ZŒ!; !2 D �1� !.

.aC b!/C .c C d!/ D .a C c/C .b C d/! 2 ZŒ! .a C b!/.c C d!/ D .ac � bd/C .bc C ad � bd/! 2 ZŒ!:

(ii) If z D aC b!, then In ZŒi, i2 D �1. In ZŒ!; !2 D �1� !.

N.z/ D a2 � ab C b2:

Proof. (i) The formula for addition is clear. For multiplication,

.a C b!/.c C d!/ D ac C .bc C ad/! C bd!2:

Since !2 C ! C 1 D 0; we have !2 D �1 � !, and

.a C b!/.c C d!/ D .ac � bd/C .bc C ad � bd/!:

(ii) As we said above, this is Exercise 3.72 on page 128.

Definition. An Eisenstein triple is a triple of positive integers .a; b; c/ such

that

a2 � ab C b2 D c2:

The same idea that produces Pythagorean triples from norms of squares of

Gaussian integers applies to produce Eisenstein triples from norms of squares

of Eisenstein integers. If z is an Eisenstein integer, then

N.z2/ D N.z/2:

The left-hand side of this equation, being the norm of an Eisenstein integer,

is of the form a2 � ab C b2. And the right-hand side is the square of the integer N.z/. Hence a2 � ab C b2 is a perfect square, and we have produced an Eisenstein triple.

Example 3.39. If z D 3C 2!, thenN.z/ D 32 � 3 � 2C 22 D 7, and we have

z2 D 9C 12! C 4!2

D 9C 12! C 4.�1 � !/ D 5C 8!:

Hence, 52 � 5 � 8C 82 D N � z2 �

D N.z/2 D 72; and .5; 8; 7/ is an Eisenstein triple. N

In Figure 3.14, we have †Q D 60ı. We have found a quick way to generate Eisenstein triples (by hand or with

a computer). Pick an Eisenstein integer r C s! (with r > 0, s > 0, and r ¤ s) and square it.

The r; s entry in the following table is .r C s!/2; N.r C s!/. For exam- ple, the top entry in the first column, which arises from r D 2 and s D 1,

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122 Chapter 3 Renaissance

U

Q S

5 7

8

Figure 3.14. All sides have integer length.

is � .2 C !/2; N.2C !/

� D .3C 3!; 3/; the corresponding Eisenstein triple

gives .3; 3; 3/, which is an equilateral triangle. One of our friends calls this

table a “candy store of patterns.” Which entries give equilateral triangles?

s D 1 s D 2 s D 3 s D 4 r D 2 3C 3!; 3 r D 3 8C 5!; 7 5C 8!; 7 r D 4 15C 7!; 13 12C 12!; 12 7C 15!; 13 r D 5 24C 9!; 21 21C 16!; 19 16C 21!; 19 9C 24!; 21 r D 6 35C 11!; 31 32C 20!; 28 27C 27!; 27 20C 32!; 28 r D 7 48C 13!; 43 45C 24!; 39 40C 33!; 37 33C 40!; 37 r D 8 63C 15!; 57 60C 28!; 52 55C 39!; 49 48C 48!; 48 r D 9 80C 17!; 73 77C 32!; 67 72C 45!; 63 65C 56!; 61 r D 10 99C 19!; 91 96C 36!; 84 91C 51!; 79 84C 64!; 76

Eisenstein Triples and Diophantus

There’s another, geometric, way to generate Eisenstein triples, using the same

idea as the method of Diophantus in Chapter 1. If .a; b; c/ is an Eisenstein

triple, so thatSee Exercise 1.79 on page 44.

a2 � ab C b2 D c2;

then dividing by c2 gives

.a=c/2 � .a=c/ .b=c/C .b=c/2 D 1:

Thus, .a=c; b=c/ is a rational point on the ellipse with equation

x2 � xy C y2 D 1:

(See Figure 3.15.) As with the unit circle, the graph contains .�1; 0/, and we can use the chord method idea of Diophantus.See Exercise 3.66 on

page 128.

Proposition 3.40. Let ` be a line through .�1; 0/ which intersects the ellipse with equation x2 � xy C y2 D 1 in a point P . If ` has rational slope, then P has rational coordinates, P D .a=c; b=c/, and

a2 � ab C b2 D c2:

If P D .a=c; b=c/ is in the first quadrant, then .a; b; c/ is an Eisenstein triple.

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3.4 Connections: Designing Good Problems 123

y

x 0.4 0.8 1.2 1.6

0.4

0.8

1.2

–0.4

–0.8

–1.2

–0.4–0.8–1.2–1.6

Figure 3.15. The graph of x2 � xy C y2 D 1.

Proof. The proof is almost identical to the proof of Proposition 1.2. We leave it to you to fill in the details.

For example, if ` has slope 1 4

and equation y D 1 4 .x C 1/, then ` intersects

the ellipse in � 15 13 ; 7 13

� , and .15; 7; 13/ is an Eisenstein triple. So, the triangle

whose side lengths are 15, 7, and 13 has a 60ı angle. Which angle is it?

Nice Boxes

Our next application is to a “box problem.” In an a � b rectangle, cut out little squares at the corners, and then fold up the sides to form an open-top box (see

Figure 3.16). What size cut-out maximizes the volume of the box? For most rectangles, the best cut-out has irrational side length. The meta-problem:

How can we find a and b to make the optimal cut-out a rational number?

a

b

x

Figure 3.16. Box problem.

As we tell our students, let the size of the cut-out be x. Then the volume of

the box is a function of x:

V.x/ D .a � 2x/.b � 2x/x D 4x3 � 2.aC b/x2 C abx;

and its derivative is

V 0.x/ D 12x2 � 4.a C b/x C ab:

We want V 0.x/ to have rational zeros, and so its discriminant

16.aC b/2 � 48ab

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124 Chapter 3 Renaissance

should be a perfect square. But 16 is a perfect square, and so

.a C b/2 � 3ab D a2 � ab C b2

should be a perfect square. This will be so if a and b are the legs of an Eisen-

stein triple .a; b; c/.

For example, from the Eisenstein triple .7; 15; 13/, we get a 7�15 rectangle that can be used to create a box whose maximum volume occurs at a rational-

length cut-out. The volume of the resulting box is

V.x/ D .7 � 2x/.15 � 2x/x D 4x3 � 443x2 C 105x:

So, V 0.x/ D 12x2 � 88x C 105. The roots of V 0.x/ are 3 2

and 35 6

. Both are

rational, but only 3 2

fits the context and maximizes V . (Why doesn’t 35 6

fit the context? What significance does it have? Also, see Exercise 3.69 on page 128.)

Nice Functions for Calculus Problems

Our next meta-problem is one that has occupied faculty room discussions about

calculus teaching for years.

How do you find cubic polynomials f .x/ with integer coefficients and

rational roots, whose extrema and inflection points have rational coordi-

nates?

No cheating: we want the extrema points and inflection points to be distinct. We’ll actually create cubics in which all these points have integer coordinates.

Using the notation of Theorem 3.3, we can first assume that the cubic f is reduced; that is, it has form

f .x/ D x3 C qx C r:

This immediately guarantees that f 00.x/ D 6x has an integer root, namely 0 (the inflection point of the graph is on the y-axis). Next, if we replace q by

�3p2 for some integer p, then f 0.x/ D 3x2�3p2, and f 0.x/ also has integer roots. So, our cubic now looks like f .x/ D x3 � 3p2 x C r . This will have rational extrema and inflection points (what are they?), so all we have to do is ensure that it has three rational roots.

If f has two rational roots, it has three (why?), and so it’s enough to make

two roots, say �˛ and ˇ, rational (we use �˛ instead of ˛ because we’ve experimented a bit and found that this makes the calculations come out nicer).

But if f .�˛/ D f .ˇ/ D 0, we have

�˛3 C 3p2˛ D ˇ3 � 3p2ˇ

or

ˇ3 C ˛3 D 3p2.˛ C ˇ/:

We can divide both sides by ˛ C ˇ, for ˛ C ˇ ¤ 0 (lest �˛ D ˇ; remember that we want our roots distinct); we obtain

˛2 � ˛ˇ C ˇ2 D 3p2:

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3.4 Connections: Designing Good Problems 125

Eisenstein integers again. This is the same as

N.˛ C ˇ!/ D 3p2:

This time we want an Eisenstein integer whose norm is 3 times a square.

We’re in luck: the equation a2 � abC b2 D 3 has several integer solutions, including .1; 2/. So, 3 D N.1 � !/. Hence we just need to take ˛ C ˇ! to be 1 � ! times the square of an Eisenstein integer. Indeed, if

˛ C ˇ! D .1 � !/.r C s!/2;

then

˛2 � ˛ˇ C ˇ2 D N.˛ C ˇ!/ D N � .1 � !/.r C s!/2

�

D N .1 � !/N � .r C s!/2

�

D 3N.r C s!/2;

which is 3 times the square of an integer.

Example 3.41. Let’s take s D 1 and r D 3. Then we have

˛ C ˇ! D .1 � !/.3 C !/2

D 13C 2!:

This tells us several things:

(i) Since N.13C 2!/ D 147, our cubic is

f .x/ D x3 � 147x C r:

(ii) But because 147 D 3 � 72 (so, p D 7), f 0.x/ D 3x3 � 3 � 49 will have rational roots: ˙7.

(iii) Since two roots of our cubic are �˛ and ˇ, two roots are �13 and 2. This lets us find r . Since

23 � 147 � 2C r D 0;

we have r D 286. Hence our cubic is

f .x/ D x3 � 147x C 286:

You can check that the third root is 11 and that the extrema and inflection

points are rational. N

Creating examples like this is not hard by hand, but a computer algebra

system makes it automatic. The next table was generated by a CAS, and it

shows the results of our algorithm for small values of r and s.

s D 1 s D 2 s D 3 r D 2 54 � 27x C x3 �128 � 48x C x3

r D 3 286 � 147x C x3 286 � 147x C x3 �1458 � 243x C x3

r D 4 �506 � 507x C x3 3456 � 432x C x3 �506 � 507x C x3

r D 5 �7722 � 1323x C x3 10582 � 1083x C x3 10582 � 1083x C x3

r D 6 �35282 � 2883x C x3 18304 � 2352x C x3 39366 � 2187x C x3

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126 Chapter 3 Renaissance

All these cubics have coefficient of x2 equal to 0. If you’d like examples

where this is not the case, just replace x by, say, x C 1 and simplify. Again, aSee Exercise 3.70 on page 128. CAS makes this easy.

Lattice Point Triangles

Our last meta-problem arises when illustrating the distance formula.A lattice point is a point with integer coordinates.

How can you find three lattice points A, B , and C in the plane so that

the distance between any two of them is an integer?

Clearly, solutions are invariant under translation by a lattice point; that is,

if A, B , and C form a lattice point solution and U is any lattice point, then

A� U , B � U , and C � U form another solution: since d.A � U;B � U / D d.A; B/ (where d.P;Q/ is the distance between points P and Q), we have d.A; B/ D jA� Bj. Hence, we can assume that one of the points, say C , is at the origin.

Now view the plane as the complex plane, so that lattice points are Gaussian

integers. Thus, we want Gaussian integers z and w such that jzj, jwj, and jz �wj are integers. But if z D aC bi , then

jzj D p a2 C b2 D

p N.z/:

Hence, to make the length an integer, make the norm a perfect square and, to

make the norm a perfect square, make the Gaussian integer a perfect square in

ZŒi . That is, we want Gaussian integers z and w so that z, w, and z � w are perfect squares in ZŒi . Hence, we choose z and w so that

z D ˛2 for some ˛ 2 ZŒi  w D ˇ2 for some ˇ 2 ZŒi 

z �w D 2 for some 2 ZŒi :

In other words, we want Gaussian integers ˛, ˇ, and so that

˛2 � ˇ2 D 2

or

˛2 D ˇ2 C 2:

The punchline is that one of our favorite identities,

� x2 C y2

�2 D � x2 � y2

�2 C .2xy/2 ;

which holds in any commutative ring, holds, in particular, in ZŒi . So, the trick

is to pick Gaussian integers x and y, set

˛ D x2 C y2

ˇ D x2 � y2;

and then let

See Exercise 1.25 on

page 14.

z D ˛2

w D ˇ2:

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3.4 Connections: Designing Good Problems 127

Example 3.42. Pick x D 2C i and y D 3C 2i . Then

˛ D x2 C y2 D 8C 16i and ˇ D x2 � y2 D �2 � 8i:

Now put

z D ˛2 D �192C 256i and w D ˇ2 D �60C 32i:

Hence, .0; 0/, .�192; 256/, and .�60; 32/ are vertices of an integer-sided tri- angle. Moreover, adding a lattice point to each vertex produces another such triangle with no vertex at the origin. Once again, a CAS can be used to generate

many more. N

This is just the beginning; many research problems are generalizations of

meta-problems. Fermat’s Last Theorem started as a search for integer solutions

to equations like the Pythagorean equation but with larger exponents.

There are many other meta-problems that yield to these two methods: norms from ZŒi  or ZŒ!; rational points on the unit circle or on the graph of x2 � xyCy2 D 1. Still others can be solved with norms from other number systems or from rational points on other curves. In Chapter 9, we will see that congruent

numbers lead to rational points on certain cubic curves.

Exercises

3.60 For each integer n between 3 and 9, find a polynomial of smallest degree with

integer coefficients whose roots are the primitive nth roots of unity.

3.61 * Let a and b be real numbers, and let z be a complex number.

(i) Show that a C bz D a C b z: (ii) Show that N.a C bz/ D a2 C 2<.z/ab C b2N.z/.

3.62 * If z and w are complex numbers, show that N.z/ < N.w/ if and only if

jzj < jwj.

3.63 Let  be an isosceles triangle with side lengths 13, 13, and 10.

(i) Show that the altitude to the base has length 12, and that it divides  into two

5, 12, 13 triangles.

(ii) Show that the altitude to one of the sides of length 13 divides  into two right

triangles whose side lengths are rational.

(iii) Each of the side lengths can thus be scaled to get a Pythagorean triple. Show

that one triple is similar to .5; 12; 13/ and that other comes from .5 C 12i /2. (iv) Generalize this result to any isosceles triangle formed by two copies of a

Pythagorean triple, joined along a leg.

3.64 * Let g and h be complex numbers such that g3 D h3. (i) Show that g is equal to either h, !h, or !2h, where ! D 12 .�1 C i

p 3/.

(ii) If gh is also real, show that g D h.

3.65 * Suppose that g D 12 C 5

2 p 3

i , h D g, and ! D cos.2�3 / C i sin. 2� 3 / (see

Example 3.36). Find the value of

(i) g! C h!2

(ii) g!2 C h!

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128 Chapter 3 Renaissance

3.66 * Sketch the graph of x2 � ax C y2 D 1 for (i) a D �1

(ii) a D 1 (iii) a D 2 (iv) a D 3 (v) a D 1

2

3.67 In Theorem 1.5, we saw that every Pythagorean triple is similar to one of the form

.2xy; x2 � y2; x2 C y2/:

Show how this can be obtained via the “norm from ZŒi ” method.

3.68 Obtain a formula for Eisenstein triples analgous to the one for Gaussian inte-

gers in Theorem 1.5 using norms from ZŒ! and rational points on the graph of

x2 � xy C y2 D 1.

3.69 * Assume that the square of the Eisenstein integer r C s! is used to generate an Eisenstein triple, and that the triple is used to create a “nice box,” as on page 123.

Express the volume of the box in terms of r and s.

3.70 * Replace x by x C 1 in several of the cubics in the table on page 125 to produceReplacing x by x C 1 just translates the graph by one

unit. Which way? nice cubics whose coefficient of x2 is nonzero. Show that your cubics are indeed

nice.

3.71 Describe where Gaussian integers are situated in the complex plane.

3.72 * Suppose that a and b are real numbers and

! D cos.2�3 / C i sin. 2� 3 / D

1 2 .�1 C i

p 3/:

Show that

N.a C b!/ D a2 � ab C b2:

3.73 Describe where Eisenstein integers are situated in the complex plane.

3.74 Find an integer-sided triangle one of whose angles has cosine equal to 3=5.

Hint: Let � D cos.35 / C i sin. 3 5 / and consider norms from ZŒ�. What conic

would help here?

3.75 A Heron triangle is a triangle with integer side lengths and integer area. In Ex-

ercise 1.26 on page 14, you found a Heron triangle by joining two Pythagorean

triangles together along a common leg. Show that the following method also pro-

duces Heron triangles.

Pick a rational point .cos �; sin �/ on the unit circle, where 0 < � < � , and let

˛ D � cos � C i sin � . Then pick any number z of the form r C s˛, where r and s are rational numbers and r > s > 0.

(i) What is the norm of r C s˛? (ii) Show that

˛2 C 2˛ cos � C 1 D 0:

(iii) Show that if z2 D a C b˛, then the triangle with side lengths a and b and included angle � will have a rational number, say c, as its third side length

and a rational number as an area. (This triangle can be then scaled to produce

a Heron triangle.) Use this method to generate a few Heron triangles.

3.76 Show that a triangle with lattice point vertices and integer side-lengths is a Heron

triangle.

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3.4 Connections: Designing Good Problems 129

3.77 Take It Further. Here’s a typical current problem, taken from B. Kerins, Gauss,

Pythagoras, and Heron, Mathematics Teacher, 2003, 350-357:

A boat is making a round trip, 135 miles in each direction. Without a cur-

rent, the boat’s speed would be 32 miles per hour. However, there is a con-

stant current that increases the boat’s speed in one direction and decreases

it in the other. If the round trip takes exactly 9 hours, what is the speed of

the current?

(i) Solve the problem.

(ii) Solve the corresponding meta-problem: find a method for generating current

problems that come out nice.

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4 Modular Arithmetic Theorems about integers can be generalized to other interesting contexts. For

example, an early attack on Fermat’s Last Theorem was to factor xn C yn (n odd) as in Exercise 3.50 on page 115:

xn C yn D .x C y/.x C �y/ � � � .x C �n�1y/;

where � D e2�i=n is an nth root of unity. It turns out that the most fruitful way to understand this factorization is within the system ZŒ� of cyclotomic integers, the collection of all polynomials in � with coefficients in Z (a com- We’ll discuss cyclotomic

integers in Chapter 8.mon generalization of Gaussian integers ZŒi  and Eisenstein integers ZŒ!).

Numbers in these systems can be added and multiplied, and they satisfy all but

one of the nine fundamental properties that ordinary numbers do (reciprocals

of cyclotomic integers need not be such); we will call such systems commuta-

tive rings. But for some roots of unity �, the commutative ring ZŒ� does not enjoy the unique factorization property that Z, ZŒi , and ZŒ! have, and this

caused early “proofs” of Fermat’s Last Theorem to be false. Dealing with the

lack of unique factorization was one important problem that led naturally to

the modern way of studying algebra.

In Section 4.1, we shall see that the distinction between even and odd can be generalized, using congruences: studying remainders in the Division Algo-

rithm. It turns out, as we’ll see in Section 4.3, that, for any fixed positive integer

m, the set of its remainders, 0; 1; :::; m � 1, can be viewed as a commutative ring, as can cyclotomic integers, and they behave in many, but not all, ways It turns out that many of

the “number systems”

studied in high school are

commutative rings.

as do ordinary integers. Finally, in Section 4.5, we’ll apply these results to an analysis of decimal expansions of rational numbers.

4.1 Congruence

It is often useful to know the parity of an integer n; that is, whether n is even

or odd (why else would these words be in the language?). But n being even or

odd is equivalent to whether its remainder after dividing by 2 is 0 or 1. Modular arithmetic, introduced by Euler around 1750, studies the generalization of par-

ity arising from considering remainders after dividing by any positive integer.

At a low level, it will help us answer questions of the following sort:

� London time is 6 hours ahead of Chicago time; if it is now 9:00 AM in Chicago, what time is it in London?

� If April 12 falls on a Thursday this year, on what day of the week is May 26?

131

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132 Chapter 4 Modular Arithmetic

At a more sophisticated level, it will allow us to solve some difficult number

theoretic problems.

Definition. Let m � 0 be an integer. If a; b 2 Z, then a is congruent to b modulo m, denoted by

Note that a � 0 mod m if and only if m j a. a � b mod m;

if m j .a � b/.

Etymology. The number m in the expression a � b mod m is called the modulus, the Latin word meaning a standard unit of measurement. The term

modular unit is used today in architecture: a fixed length m is chosen, say,

m D 1 foot, and plans are drawn so that the dimensions of every window, door, wall, etc., are integral multiples of m.

We claim that integers a and b have the same parity if and only if a � b mod 2. Assume that a and b have the same parity. If both are even, then

a D 2a0 and b D 2b0. Hence, a � b D 2.a0 � b0/, 2 j .a � b/, and a � b mod 2. Similarly, if both are odd, then a D 2a0 C 1 and b D 2b0 C 1. Hence, a � b D .2a0 C 1/ � .2b0 C 1/ D 2.a0 � b0/, 2 j .a � b/, and a � b mod 2 in this case as well. Conversely, suppose that a � b mod 2. If a and b have different parity, then one is even, the other is odd, and so their difference is

odd. Hence, 2 − .a � b/, and a 6� b mod 2. Having proved the contrapositive, we may now assert that a and b have the same parity.

Example 4.1. If a � r mod m, then r is obtained from a by throwing out a multiple of m. For example, let’s compute the time of day using a 12-hour

clock. When adding 6 hours to 9:00, the answer, 3:00, is obtained by takingModular arithmetic is called “clock arithmetic” in some

introductory texts. 9C 6 D 15 � 3 mod 12 (i.e., we throw away 12). In more detail, let 0 denote 12:00, 1 denote 1:00, : : : ; 11 denote 11:00. Three hours after 9:00 is 12:00;

that is, 9 C 3 D 12 � 0 mod 12; 4 hours after 9:00 is 1:00; that is, 9 C 4 D 13 � 1 mod 12, and 6 hours after 9:00 is 3:00; that is, 9C6 D 15 � 3 mod 12.

The same idea applies to calendars. Let 0 denote Sunday, 1 denote Mon-

day, : : :, 6 denote Saturday.

Sun Mon Tues Wed Thurs Fri Sat

0 1 2 3 4 5 6

If today is Tuesday, what day of the week is 90 days from now? Since 2C90 D 92 � 1 mod 7, the answer is Monday.

Let’s now answer the question: if April 12 falls on Thursday this year, on

what day of the week is May 26? There are 18 days to April 30, so there are

18C 26 D 44 days until May 26 (for April has only 30 days). Now Thursday corresponds to 4, so that May 26 corresponds to 4 C 44 D 48 � 6 mod 7; therefore, May 26 falls on Saturday. N

There are at least two ways to state the solutions of Exercises 3.2 and 3.3

on page 89. We expected you to say then that in D im if and only if n and m leave the same remainder when divided by 4 and, if ! is a primitive cube root

of unity, that !n D !m if and only if n and m leave the same remainder when

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4.1 Congruence 133

divided by 3. In light of the next proposition, we can also say that in D im if and only if 4 j .n � m/; that is, n � m mod 4, and !n D !m if and only if 3 j .n �m/; that is, n � m mod 3.

Proposition 4.2. Let m � 2 and a; b 2 Z.

(i) If a D qmC r , then a � r mod m. (ii) a � b mod m if and only if each of a and b have the same remainder

after dividing by m.

Proof. (i) Since a � r D qm, we have m j .a � r/; that is, a � r mod m. (ii) Assume a � b mod m. Let r; r 0 be the remainders after dividing a; b, re-

spectively, bym; that is, a D qmCr and b D q0mCr 0, where 0 � r < m and 0 � r 0 < m. We want to show that r 0 D r . If not, suppose that r 0 < r (the argument is the same if r < r 0). Then a � b D m.q � q0/C .r � r 0/ with 0 < r � r 0 < m. Now Exercise 1.46 on page 29 gives m j .r � r 0/. Hence, m � r � r 0, by Lemma 1.13, contradicting r � r 0 < m.

Conversely, if a D qmC r , b D q0mC r 0, and r D r 0, then a � b D m.q � q0/ and a � b mod m.

Notice that Proposition 4.2 generalizes the fact that integers a and b have

the same parity if and only if a � b mod 2. We are now going to see that congruence modulo m behaves very much

like ordinary equality; more precisely, it is an equivalence relation (see Ap- pendix A.2): it is reflexive, symmetric, and transitive.

Proposition 4.3. Let m � 0. For all integers a; b; c, we have

(i) a � a mod mI (ii) if a � b mod m, then b � a mod mI

(iii) if a � b mod m and b � c mod m, then a � c mod m.

Proof. All are easy to check. We have a � a mod m, because a � a D 0 and m j 0 is always true (even whenm D 0). Since b�a D �.a�b/, ifm j .a�b/, thenm j .b � a/. Finally, .a� b/� .b � c/ D a� c, so that ifm j .a� b/ and m j .b � c/, then m j .a � c/.

How to Think About It. Congruence mod 1 makes sense, but it is not very

interesting, for a � b mod 1 if and only if 1 j .a� b/. But this latter condition is always true, for 1 is a divisor of every integer. Thus, every two integers are congruent mod 1. Similarly, congruence mod 0 makes sense, but it, too, is not

very interesting, for 0 j c if and only if c D 0. Thus, a � b mod 0 if and only If 0 j c, then there is some k with c D 0 � k D 0; that is, c D 0.

if 0 j .a � b/; that is, a � b mod 0 if and only if a D b, and so congruence mod 0 is just ordinary equality. You should not be surprised that we usually

assume that m � 2.

Corollary 4.4. If m � 2, then every integer a is congruent mod m to exactly See Exercise 4.5 on page 140 for a generaliza-

tion. one integer on the list

0; 1; : : : ; m� 1:

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134 Chapter 4 Modular Arithmetic

Proof. By the Division Algorithm, we have a D qmC r , where 0 � r < m; that is, a � r mod m.

If a were congruent to two integers on the list, say, r < r 0, then r � r 0 mod m by transitivity, so that m j .r 0 � r/. Since 0 < r 0 � r < m, this would contradict Lemma 1.13.

Congruence gets along well with addition and multiplication.

Proposition 4.5. Let m � 0. (i) If a � a0 mod m and b � b0 mod m, then

aC b � a0 C b0 mod m:

More generally, if ai � a0i mod m for i D 1; : : : ; k, then

a1 C � � � C ak � a01 C � � � C a0k mod m:

(ii) If a � a0 mod m and b � b0 mod m, then

ab � a0b0 mod m:

More generally, if ai � a0i mod m for i D 1; : : : ; k, then

a1 � � �ak � a01 � � �a0k mod m:

(iii) If a � b mod m, then

ak � bk mod m for all k � 1:

Proof. (i) If m j .a � a0/ and m j .b � b0/, then m j .a C b/ � .a0 C b0/, because .aC b/� .a0 C b0/ D .a� a0/C .b � b0/. The generalization to k summands follows by induction on k � 2.

(ii) We must show that ifm j .a� a0/ and m j .b � b0/, thenm j .ab � a0b0/. This follows from the identity

ab � a0b0 D ab � ab0 C ab0 � a0b0 D a.b � b0/C .a � a0/b0:

The generalization to k factors follows by induction on k � 2. (iii) This is the special case of part (ii) in which all ai D a and all a0i D b.

How to Think About It. The key idea in calculating with congruences

mod m is that every number can be replaced by its remainder after dividing

by m, for this is precisely what Proposition 4.5 permits; it allows you to “re-

duce as you go” in calculations, as the next example shows.

Example 4.6. The last (units) digit of a positive integer is the remainder when

it is divided by 10. What is the last digit of

103243 C 2348 � 5267‹

We could do this by brute force: cube 10324, multiply 2348 and 5267, add, and

look at the last digit. But, as one of our friends says, why should the calculator

have all the fun? You can do this more cleverly using congruence.

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4.1 Congruence 135

� To compute 103243, first look at 10324.

10324 � 4 mod 10; so that 103243 � 43 mod 10:

Now 43 D 64 � 4 mod 10, so that 103243 � 4 mod 10, and the last digit of 103243 is 4.

� To multiply 2348 and 5267, note that 2348 � 8 mod 10 and 5267 � More simply, think of multiplying 2348 and 5267

by hand. What’s the last

digit? This is what most

middle school students

would do. We just want

to illustrate the general

principle here.

7 mod 10. Hence,

2348 � 5267 � 8 � 7 D 56 � 6 mod 10: � Thus,

103243 C 2348 � 5267 � 4C 6 D 10 mod 10;

and 0 is the last digit; 103243 C 2348 � 5267 is divisible by 10.

Now you try one: what is the last digit of 752843 C 10988 � 310767? N

The next example uses congruence to solve more difficult problems.

Example 4.7. (i) If a 2 Z, then a2 � 0, 1, or 4 mod 8. If a is an integer, then a � r mod 8, where 0 � r � 7; moreover,

by Proposition 4.5 (iii), a2 � r2 mod 8, and so it suffices to look at the squares of the remainders. We see in Figure 4.1 that only 0, 1, or 4 can be

a remainder after dividing a perfect square by 8.

r 0 1 2 3 4 5 6 7

r2 0 1 4 9 16 25 36 49

r2 mod 8 0 1 4 1 0 1 4 1

Figure 4.1. Squares mod 8.

(ii) n D 1003456789 is not a perfect square. Since 1000 D 8 � 125, we have 1000 � 0 mod 8, and so

1003456789D 1003456 � 1000C 789 � 789 mod 8:

Dividing 789 by 8 leaves remainder 5; that is, n � 5 mod 8. But if nwere a perfect square, then n � 0; 1, or 4 mod 8.

(iii) There are no perfect squares of the form 3m C 3nC 1, wherem and n are positive integers.

Again, let’s look at remainders mod 8. Now 32 D 9 � 1 mod 8, and so we can evaluate 3m mod 8 as follows: if m D 2k, then 3m D 32k D 9k � 1 mod 8; if m D 2k C 1, then 3m D 32kC1 D 32k � 3 � 3 mod 8. Thus,

3m � ( 1 mod 8 if m is even

3 mod 8 if m is odd:

Replacing numbers by their remainders after dividing by 8, we have the

following possibilities for the remainder of 3m C 3n C 1, depending on

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136 Chapter 4 Modular Arithmetic

the parities of m and n:

3C 1C 1 � 5 mod 8 3C 3C 1 � 7 mod 8 1C 1C 1 � 3 mod 8 1C 3C 1 � 5 mod 8:

In no case is the remainder 0, 1, or 4, and so no number of the form

3m C 3n C 1 can be a perfect square, by part (i). N

Many beginning algebra students wish that .aCb/p D apCbp in Z; if only .a C b/2 D a2 C b2! The next proposition (which paraphrases Exercise 7.27 on page 293) would delight them. If theorems were movies, Proposition 4.8 would be X-rated: only adults would be allowed to see it.

Proposition 4.8. If p is a prime and a; b are integers, then

.a C b/p � ap C bp mod p:

Proof. The Binomial Theorem gives

.a C b/p D pX

rD0

p

r

! ap�rbr :

But � p r

� � 0 mod p for all r with 0 < r < p, by Proposition 2.26. The result

now follows from Proposition 4.5(i).

The next theorem (sometimes called the Little Fermat Theorem to distin-

guish it from Fermat’s Last Theorem) turns out to be very useful.

Theorem 4.9 (Fermat). Let p be a prime and a 2 Z.See Corollary 4.67 for another proof.

(i) ap � a mod p. (ii) ap

n � a mod p for all n � 1. (iii) If p − a, then ap�1 � 1 mod p.

Proof. (i) We first prove the statement when a � 0, by induction on a. The base step a D 0 is obviously true. For the inductive step, the inductive hypothesis is ap � a mod p. Hence, Proposition 4.8 gives

.a C 1/p � ap C 1 � a C 1 mod p:

To complete the proof, consider �a, where a > 0; now

.�a/p D .�1/pap � .�1/pa mod p:

If p is an odd prime (indeed, if p is odd), then .�1/p D �1, and .�1/pa D �a, as desired. If p D 2, then .�a/2 D a2 � a mod 2, and we are fin- ished in this case as well.

(ii) The proof is by induction on n � 1: the base step is part (i), while the inductive step follows from the identity ap

n D � ap

n�1�p .

(iii) By part (i), p j .ap � a/; that is, p j a.ap�1 � 1/. Since p − a, Euclid’s Lemma gives p j .ap�1 � 1/; that is, ap�1 � 1 mod p.

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4.1 Congruence 137

Later in this chapter, we will use the next corollary to construct codes that

are extremely difficult for spies to decode.

Corollary 4.10. If p is a prime andm � 1 mod .p�1/, then am � a mod p for all a 2 Z.

Proof. If a � 0 mod p, then am � 0 mod p, and so am � a mod p. Assume now that a 6� 0 mod p; that is, p − a. By hypothesis,m � 1 D k.p � 1/ for some integer k, and so m D 1C .p � 1/k. Therefore,

am D a1C.p�1/k D aa.p�1/k D a � ap�1

�k � a mod p;

for ap�1 � 1 mod p, by Theorem 4.9(iii)

We can now explain a well-known divisibility test.

Proposition 4.11. A positive integer a is divisible by 3 if and only if the sum

of its (decimal) digits is divisible by 3.

Proof. The decimal notation for a is dk : : : d1d0; that is,

a D dk10k C � � � C d110C d0;

where 0 � di < 10 for all i . Now 10 � 1 mod 3, and Proposition 4.5(iii) gives 10i � 1i D 1 mod 3 for all i ; thus parts (i) and (ii) of Proposition 4.5 give a � dk C � � � C d1 C d0 mod 3. Therefore, a is divisible by 3 if and only if a � 0 mod 3 if and only if dk C � � � C d1 C d0 � 0 mod 3.

How to Think About It. The proof of Proposition 4.11 shows more than its

statement claims: the sum of the (decimal) digits of any positive integer a is

congruent to a mod 3, whether or not a is divisible by 3. For example,

172 � 1C 7C 2 mod 3I

that is, both 172 and 10 (the sum of its digits) are � 1 mod 3.

Since 10 � 1 mod 9, Proposition 4.11 holds if we replace 3 by 9 (it is often called casting out 9s): A positive integer a is divisible by 9 if and only if the Is the sum of the decimal

digits of an integer a

congruent mod 9 to a

itself?

sum of its digits,†.a/, is divisible by 9.

Define two operations on the decimal digits of a positive integer a.

(i) Delete all 9s (if any) and delete any group of digits whose sum is 9

(ii) Add up all the digits.

It is easy to see that repeated applications of these operations to a positive integer a yields a single digit; call it r.a/. For example,

5261934 ! 526134 ! 561 (for 2C 3C 4 D 9) ! 12 ! 3:

(It is now clear why this procedure is called casting out 9s.) In light of a � †.a/ mod 9, we have †.a/ � r.a/ mod 9, so that r.a/, which seems to de- pend on a choice of operations (i) and (ii), depends only on a, for the variation

of Proposition 4.11 for 9 says that†.a/ is the remainder after dividing a by 9.

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