Elementary Number Theory : An Algebraic Approach

Elementary Number Theory An Algebraic Approach

Ethan D. Bolker Bryn Mawr College

W. A.

Benjamin, Inc. 1970

New York

1�/mmmtwy Nwnlu'l' 'rlwo�y

An

AJ.c;c·lu·ct/c' ApproliC'It

Copyright © 1969 by W. A. Benjamin, Inc. All rights reserved

Standard Book Number 8053-1018-5 Library of Congress Catalog Card Number 76-92217 AMS 1 968 Classification 4065 Manufactured in the United States of America 2345M321

The manuscript was put into production on May 28, 1969; this volume was published on January 1, 1970

Benjamin, Inc. New York, New York 10016 W. A.

Preface

Elementary number theory is frequently taught only to those who have studied little mathemati cs and plan to stu dy no more . Ambitious students le arn " abstract algebra " in ste ad, but all too often they find the axiomatic s tudy of groups and rings sterile and irrelevant . To remedy both il ls I have tried to capture in this book the excitement of my d i sco very that the alge bra I had known for years was the perfect setting in which to recreate the tradi­ tional first theorems in numb6r theory we owe to Ferm at, Euler, and Gauss. Th e exposition is tied to the study of three classi cal pro blems : the structure of the group of u n its of zn ' integers representable in the form x2 - my2' and the Fermat equati on xn + yn = zn for n = 2, 3, and 4. I have concen­ trated on the parts of these pr o bl ems in which the number theory and the algebra each serve to deepen the reader' s understa nding of the other. I therefore omitted to pics such as continued fractio ns, elementary analytic number theory, and the beginnings of a general theory of quad rati c forms which, though acc e s si ble to be ginni ng students, did not lend themselves to an elementary algebrai c trea tm e nt . Moreo ver, I have st ressed the algebraic aspects of s om e of the traditional theorems. Wilson's theorem is derived from the unique factorization of po lyno mial s with coefficients in a fi eld, the structure of the group of units of zn from a the orem on products of cyclic groups proved for that purp o se earlier in the book. I have assumed only the algeb ra re qu ired to carry out these aims, less than is found in any s ta nd ard course in "modern algebra." An interested v

•

instructor could probably Sllf)ply the necessary background and teach the course to good students with no experience in alge bra . The prospective reader must know or learn the definitions of group, ring, homomorphism, equivalence relation, and quotient structure, and a few simple theorems, such as Lagrange ' s , which asserts that the order of a finite group is a multiple of the order of any of its subgroups. I state each such theorem in the text the first time it is used. Often an alternative, even more algebraic approach to a topic covered in the text is treated in the problems. Appendix I contains some essential algebra, which is not all rudimentary and may be new to the reader: the traditional definitions of the theory of divisibility in an integral domain and a proof that a Euclidean domain enjoys unique factorization. That theorem is applied to the integers in Chapter I, to the ring of polynomials with coefficients in a field in Chapter 3, and to some rings of algebraic integers defined in Chapter 6. We have all been familiar with the arithmetic of the integers since element­ ary school, so the study of number theory is an ideal place to discover that mathematics is an experimental science. The subject of our experiments is the well-known sequence 1, 2, 3, . . . ; the results of those experiments are theorems which show that observed patterns and regularities are not coin­ cidental. This book, like most, almost always gives only the theorems and suppresses the experimental evidence that would be costly to include and dull to read. The reader is urged to reconstruct it by computing numerical special cases of each definition and theorem. To encourage this habit many of the problems begin " Investigate ... " rather than "Prove.... " The problems are important and often difficult. They consist of applica­ tions and examples of theorems and techniques in the text, numerical ex­ amples which show that arguments have been pushed to their natural boundaries, special cases of topics treated later in the book, and material often included in more traditional books on elementary number theory. The more time the reader spends on them the better. I have marked the harder or longer problems with an asterisk, but that subjective classification is not always reliable. A starred problem may yield to a special trick while one I think easy proves surprisingly stubborn. Sections are numbered consecutively through the book, equations con­ secutively within each chapter. The notation m.n refers to the nth numbered item in Section m; it may be a theorem, definition, lemma, corollary, or example. The bibliography lists only the general works on which I relied most heavily; other references occur where relevant in the text and problems. Finally, I should like to thank my algebra class at Bryn Mawr, which suffered through false starts while I learned the number theory I was teaching; Mary Wolfe, whose lecture notes were an invaluable zeroth draft of the manuscript; Bryn Mawr College, for generous support during a leave of absence; William Adams, who read and commented on an early

Preface version of the manuscript; Russ Fallowes, who wrote produced Appendix 3; and my wife Joan

for

fort, from the routine of reading proof to the

vii

the program that of aid and com­

many kinds sublime.

Ethan D. Bolker Bryn Mawr, Pennsylvania Apri/1969

To my parents

Contents

Preface

Chapter 1

v·

Linear Diophantine Equations

1

2

3 4

5

6

Chapter 2

1

Sums of Squares Divisibility and Unique Factorization The Diophantine Equation ax + b = c The Diophantine Equation a,x, + + a.x. The Infinitude of the Primes Problems · · ·

Congruence

7 8 9

10

11

3

= c

4

5

6·

7

10

Arithmetic in z.: Solving Congruen(n) which contains k. i s the identity element of cl>(n).

Proof. 9.7

p,f' k.

Fermat's (Little) Theorem.

Thus k'P(n) 1 . Thus x = ba'�'-1 i s a solution t o the original congruence.

For example,

5x = 4(12) has X = 4 5'1'(1Z)-1 •

as a solution.

=

4 53= 500 •

Of course, since

500

=

41 . I2 + 8

500= 8

=

-4 (12)

so x = -4 and x = 8 are also solutions. We can also prove the Chinese remainder. theorem (8. 1) in the following new way. Suppose (n;, ni) = I when i =1: j. As in the proof in Section 8, let N= n n, and N; = N/n;. Then (n;, N;) = I and N;= O(ni) when 1 j =1: j. Let • • •

Then

and Theorem 8.1 is proved. This construction of x has the computational advantage of requiring no long division, though x is likely to be large. 10.

MORE A BOUT

rp(n)

Consider the n rational numbers

1 2 '

n n

'

. . .. '

n . n

(9)

18

Congruence

How many of these fractions are written in lowest terms? Since mfn is in lowest terms if and only if (m, n) = I, the answer to this question is qJ(n). Suppose now that we reduce each fraction in (9) to lowest terms. The denominators which occur when we have finished are just the divisors of n. How often does each divisor appear? If d I n, then each of the rational numbers 1/ d, . . . , dfd is mfn for some m and hence appears in (9). However, of these d fractions with denominator d just qJ(d) are in lowest terms. There­ fore the denominator d appears just qJ{d) times when the fractions in (9) are written in lowest terms. But all the fractions have now been accounted for, so n, the number of fractions, is just the sum of the numbers qJ(d) for d dividing n. We have notation summarizing that clumsy sentence:

n

=} c/J(d).

(10)

dr,.

Note that implicit in Eq. (10) are the assumptions that and that d = 1 and d = n are counted as divisors. If we rewrite Eq. (10) as

n

qJ(n) = n - ) ({J(d) dr,.

d 1 if and only if k is a multiple of p. How many multiples of p less than pfl are there ?

Proof (Second version).

so that of the pfl integers 1, 2, . . . , pfl

pfl .

The others are in 4>(pfl ).

- 1 exactly pfl - 1 - 1 share a factor with

Therefore

Notice that we used Greek letters for exponents in Theorem 1 0. 1 . We shall to adhere to this convention in the following pages .

try

10.2

\l'(mn)

Definition. A function '¥ : Z � Z is multiplicative if and only if '¥ (m)'¥ (n) whenever m and n are relatively prime.

=

The identity function is multiplicative ; so is the absolute value function, since for these '¥, 'P(mn) '¥(m)'¥(n) for all m and n. Problems 11.18, 1 1 .25, and 1 1 .29 consider examples of multiplicative functions. Problem 1 1 .30 concerns the theory of such functions. A multiplicative function '¥ is known when its values for prime powers are known, because if n = p�' · · · p'k", then

=

Our

next

task is to show cp is multiplicative.

20

Congruenc

10.3

Lemma.

Suppose (m, n) = 1 . Then (xm + yn, mn) = (x, n)(y, m).

(13

Proof. It suffices to show that the two members of Eq. 1 3 have the sam( prime power divisors. Suppose pa. divides either member. Then pa. l mn Since (m, n) 1 , ei ther pa. l m and (pa., n) = 1 or pa. l n and (pa., m) = 1. Sincf the theorem and the alternatives above are symmetrical in m and n, we shall treat only the first. That is, suppose pa. l m and, necessarily, (pa. , n) = 1 .

=

Then

pa. l (xm

+ yn, mn) pa. l xm + yn Pa. l Y

Pa. l (y, m) Pa. l (x, n)(y, m). 10.4

Theorem.

The Euler qJ-function is multiplicative.

Let X be a set of coset representatives of the cosets in W(n). We may as well take for X the positive integers less than n and relatively prime to n. Similarly define Y using m i nstead of n. Then X ha s qJ(n) e lements ; Y has qJ (m) . We shall show that the set

Proof.

W = {xm

+ nyI x e X, y e

Y}

has qJ{m) qJ (n) elements and is a complete set of representatives of the cosets in w(mn). The theorem will then be proved .

Suppose x E X and implies that

y E Y.

Then (x, n) = (y, m) = 1 so Lemma

10.3

(xm + yn, mn) = 1 . That is, W represents only cosets in (mn). Next, w e show no two elements of W are congruent modulo mn. It wil l follow that the elements of W represent different cosets in (mn) and hence , a fortiori, no two elements of W are equal and W has qJ(m)qJ(n) eleme nts .

Suppose

xm

+ yn =

x'm + y'n (mn)

(14)

10.

21

More about rp(n)

for x, x' E X and y, y' E (14) implies that

We must prove x = x' and y = y'. Congruence

Y.

mn I (x - x')m + (y -y')n. Apply Lemma 10.3 :

mn I (x - x', n)(y - y ', m).

Since m and n are relatively prime, it follows that and

n l x - x'

m I Y - y' ,

but distinct elements of X are incongruent modulo n, so x must equal x'. Similarly, y = y'. Finally, we must show that W represents every coset in cl>(mn). Since (m, n) = 1 , we can always solve the Diophantine equation

w = xm + ny (Theorem 3.1). If w lies in an element of cl>(mn) then (w, mn) = 1. Lemma then implies that

10.3

(x, n) = (y, m) = 1 .

[f we replace x by the element of X to which i t i s congruent modulo n and similarly replace y, we do not change w in Zmn · Therefore W represents every coset in (mn). 10.5

Theorem

cp(n) = n n

primes pl n

(1 - �)P

Proof. First a remark O:Q. the notation. The symbol n is to products as L iN to sums. That is, Theorem 10.5 may be restated as : if P1 > , p, are

d i fferent

primes and

• • •

(15)

then

cp(n)

=

=

( 1 - :J ( 1 - :J ( ;) n ft ( �)· ···

n

1 -

1 -

22

Congruence

For example, cp( 12)

=

cp(22 3 ) •

=

12(1 -!)( 1

- j)

=

4

with what we already know. Now to prove the theorem. Suppose n = p"', a prime power. Then

which checks

(Theorem 10.1) so the theorem is true. Now suppose n any integer. Write n as in Eq. (1 5). Since cp is multiplicative,

. . . cp(p�·) p�· ( 1 - :J . . . p�· ( 1 - :J p�· . . p�· ( - :J . . . ( 1 - :J = n TI ( 1 - .!.) . Pi

cp(n) = cp(p�') =

.

=

1

primes p! l n 11.

is odd, then a 2 6.1 and 6.4.

PROBL EMS =

11.1

Prove : If a

11.2

When do es ax = bx(n) imply a =

facts help in Problems 11.3

11.4

-

···

1,

b(n) ?

n,).

Prove the

of order n

If a is even, then a 2 = 0(4).

a = b(n,) for mutually relatively

Prove: If

then a = b(n,

1 (8).

prime

i ntegers n . ,

These

.. .

, n, ,

following theorem due to Lucas : If y > 0 . Then since x2 - y2 = (x - y)(x y), Eq. (4) implies x

-y

so x

=

=

1

p+l

2-

-

and x+ y

and

y

=

p

p-1

""" -- . 2

Thus the prime p is a difference of integral squares if and only if it is odd. In fact, we have discovered a little more. 14.6

Theorem.

Every odd integer is the difference of two consecutive

squares.

Proof.

If n is odd, then (n ± 1)/2 e Z and

n (n ;· lr - (n ; lr. =

Problem 15. 10 considers the representation of an even integer as a difference of squares.

15.1

Let F be a field.

15.

PRO BLEMS

Show that the units in F[x] are the constant polynomials.

15.2 If h e z.[x], we may regard h as a function from Zp _. zp by substituting for x the elements of z. . Show that for fand g e Z[x] the following are equivalent :

(a) f(n) = g(n)(p) for all n e Z. (b) I = g(x• - x) in z.[x]. That is, xP - X I I - g in Zp[X]. (c) I and ii yield the same function from z. -+ z• .

Show by example that none of these conditions implies I = g.

15.3* Prove that there are infinitely many prime (that is, irreducible) poly­ nomials in Z,[x). Write down the pl'ime polynomials of degree less than or equal to 3 in Z2[x] and Z3[x]. 15.4 Show that n is prime if and only if every linear polynomial in Z.[x] has at most one root in z. .

15.5 Let f e Z[x] be monic, that is, suppose its leading coefficient is 1. that the only rational roots of f are integers.

Show

I '.

l'roblc•nl.l'

1 �.6

()ed u ce

power of ul'

33

an

from Problem 1 5. 5 that m 1 1" is irrational unless m is the nth integer.

1�.7"' Prove Wilson's theorem by counting the number of p-Sylow subgroups t he symmetric group on p symbols.

1 5.8

1 5.9"'

Prove the converse of Wilson's theorem.

Then

The kth elementary symmetric function of n variables, St", is defined by

}J

(X -xJ) =

j� ( - 1)• - •st•(xl , . . . , x.)x•

in F[X] for every field F. The fundamental theorem on symmetric functions that any polynomial in n variables with coefficients in F which is invariant 1111der all permutations of its arguments is a polynomial in the elementary symmetric functions. For example, i� true

snys

x2 +

y 2 + z2

=

(St 3(x, y, z)) 2 - 2S23(x, y, z).

I nvestigate s.•(l , 2, . . . , p - 1) modulo A 1- 2 2 + · · · + (p - 1 ) 2 in Zp .

1

15.10

p

when

p

is an odd prime.

Investigate

Show that an even integer is a difference of squares if and only if it is

doubly even, that is, divisible by 4.

15.11 The argument preceding Theorem 14.6 shows that the representation of a prime as a difference of squares is unique. However

1 5 = 82 - 72 = 42 - P.

We can count the number· of solutions to the Diophantine equation (5)

Let be the factorization of n as a product of powers of distinct primes. Let N = (ex - l)(ex 1

+

1)

• · ·

(ext + 1).

Prove that Eq. (5) kas N/2 positive solutions if N is even and (N + 1)/2 if N is odd. 15.12"'

integers ? 15.13"'

Which integers can be written as a sum of consecutive odd positive Of two or more consecutive odd positive integers ? .

Answer the questions posed in Problem 1 5 . 12 when the word " odd "

is deleted. 15.14"' Reread Section 1 and Problems 6.1 , 6.2, and 6.3. Theorem 1 4.5 and Problem 1 5. 11 may suggest new conjectures on representable integers and the number of ways to represent them.

4 The

Group of Units of Zn

We shall show in this chapter that �. regarded as a group valued function of n, is multiplicative. That fact together with an analysis of the structure of �(n) when n is a power of a prime will allow us to answer classical questions about the congruence X'= m(n). -

16.

DECIMAL EXPANSIONS

In this section we shall investigate the form of the decimal expansion of l/n ; the questions raised by th at investigation motivate the subsequent discussion

of the group �(n). First we shall do some arithmetic to provide ourselves with numerical examples .

7 = 0.142857 . 1

--

The digits under the bar are to be repeated, that is,

7 = 0. 142857 1

142857 142857 . . . . 34

(1)

11'1.

35

f)£'dmal Expansions

ignore all questions about the convergence of the infinite decimals Any question the reader wishes to raise he must answer for himself. l!quation (1) follows from

We Nhall

we u sc.

1 428 57 711.000000 .

-7

30

- 28 20 - 14 60 - 56

40

- 35

50 - 49 1

The sequence of remainders, "

"

which

1 , 3, 2, 6, 4, 5 = 1 , 3, 2, 6, 4,

appear in bold face, is

5,

1 , 3,

2, 6, 4,

5,

. . . .

Similar computations show that

1 = 0.01250, 80 where the remainders are

(2)

1, 10, 20, 40, 0 ; that 1 = 0.076923, 13 --

where the remainders are 1 , 10, 9,

12, 3,

4; and that

1 = 0 .01 136 , 88 -

where the remainders are 1,

(3 )

10, 12, 32, 56.

(4)

The Group of Units of Z

36

Now fix a positive integer n.

S u pp o se that

1

-=

n

where 0

::;;

O.a1a2

• • •

,

a1 � 9, and that

where 0 ::;; r1 < n, is the sequence of remainders which occurs in the Ion� division algorithm. We wish to consider the remainders r1 both as integer� and as elements of Zn . The rule " bring down the next 0 " shows that

( 7)

Therefore

r1 + 1 = 1 0r1 (n).

(8)

r;+ 1 = 101

(9)

Since r0 = 1 , (8) implies (n).

The infinite sequence (6) of remainders lies in the finite set {0, 1 , . . . , s o there must be a first repetition

n

- 1 },

( 1 0) Then the long division algorithm implies that the sequence of remainders is just (1 1) and the corresponding decimal fraction is (12) We call A the period of the expansion in Eq. (12) ; that expansion is p urely if and only if Jl = 0, or, equivalently, the first repetition in the sequence (1 1 ) i s r;. = 1 . The expansion terminates if and only if rP. = 0. In that case all succeeding remainders will be 0. We wish to discover how Jl and A depend on n. To do so we review some elementary group theory. Let G be a finite group with i dentity e, and g an element of G. Then the

periodic

l fl ,

l>t•C'Imal Expansions

l i l'll l

•·cpctition in the sequence

37

e, g, g 2 , . . . iN

of the form g

"

(13)

=

e and A. is the order of g. The map 'P : Z � G given by g" is then a homomorphism with kernel A.Z, so that it may be regarded us an isomorphism between the additive group z, Z/A.Z and the subgroup j c•, g, . . , g A - l } of G spanned by g.

'l'(n)

=

·=

.

I 6.1 Theorem. Each of the following three groups of statements consists ,,r equivalent statements. For any particular positive integer n exactly one uf these groups consists of true assertions. I

(a) (b) (c) II (a) (b) (c) III (a) (b) (c)

The decimal expansion for 1/n is purely periodic. (10, n) 1 ; that is, neither 2 nor 5 divides n. For some A. > 0, r, 1. The decimal expansion for 1/n terminates. For some Jl, n 1 10" ; that is, n has no prime factors other than 2 or 5. For some Jl, r,. = 0. The decimal expansion for 1/n is not purely periodic and does not terminate. 2 or 5 and some third prime divide n. For i > 0, r; is never 0 or 1 .

=

=

Proof Let n be a positive integer. Elementary logic shows just one of I(a), II(a), or III(a) and just one of I(c), II(c), and III(c) is true. A little reflection shows just one of l(b) , II(b) , or III(b) is true. Thus to prove the theorem it suffices to show I(a) l(b) I(c) and II(a) II(b) e- II(c). We treat case I first. Suppose (10, n) = 1 (I(b)). Then 10 e Cl»(n) (Theorem . 9.2). Congruence (9) now implies that sequence (6) of remainders is just the sequence ( 1 3) when g = 10 in the group Cl»(n). Therefore the first repetition is r, = 1 and I(a) is true. Suppose I(a) true. Then r, r0 1 , so I(c) follows. Finally, suppose I(c) true. Then

=

1

=

=

1 0" = 10 10" - 1(n). ·

That is, 10 is invertible in Zn , so (10, n) = 1 (I(b)) (Theorem 9.2). Case II is simpler. We show II(a) => II(b) => II(c) => II(a). If the decimal expansion for n is

11re Group of U11its of z.

38

then 1 0"/n i s an integer, so n 1 10". If n 1 10", then (9) shows r�" = 0. If r�" = 0, then for all k ;;:;: 0, rp, + k and hence ap, + k + l is zero. In our examples above, 7 and 1 3 are covered by Case I, 80 by Case II, and 88 by Case III. For the remainder of this section we shall restrict our attention to Case I. 16.2 Corollary. Suppose (10, n) = 1 . Let A be the subgroup of (n) generated by 10. Then the period .A.(n) of the decimal expansion of 1/n is the order of A and hence divides qJ(n).

The only special significance of 10 in this section is the fact that we have 10 fingers and so write numbers decimally. The methods we used really prove more than we have so far made explicit. The following theorem states a consequence of Case I for expansions to any base. 16.3 Theorem. Suppose (m, n) = 1 . Let Am be the subgroup of (n) generated by m. Then the period .A.m(n) of the expansion of lfn in the base m is the order of Am and hence divides qJ(n).

We shall continue to write .A.(n) for .A.1 0(n). A question commonly asked is : For which n does .A.(n) = qJ(n) ? The integer 7 enjoys this property ; 1 3 does not. The question is equivalent to : For which n is (n) cyclic with 1 0 as a generator ? The bulk of this chapter is devoted to the structure of (n) ; when we are done we shall know when (n) is cyclic. The problem of deciding whether 10 happens to be a generator is unsolved. For example, we shall see that the 1 2 element group (13) is cyclic, though we know that the order of 1 0 in that group is only 6. In general (p) is always cyclic when p is prime. Part of a conjecture due to Artin asserts that 10 generates (p) for infinitely many primes p. We close this section with some remarks on the decimal expansion of kfn. Suppose (10, n) = 1 and that k/n is in lowest terms, so that (k, n) = 1 . If k happens to be in A, then it is just one of the remainders which appeared when we worked out the decimal expansion of 1/n. Then the decimal expansion of k/n is purely periodic ; its block of digits is a cyclic permutation of the block for n. For example,

10/13 = 0.769230

and

9/1 3 = 0.692307.

( 'y('/lr Groups

I I,

I I' k

39

¢ A we must begin again. Then the successive remainders in the

d i v ision algorithm for kfn are just the numbers l O ;k. Thus they exhaust 11 coset of A in �(n). Therefore, k/n too has a purely periodic expansion w i l h period A.(n), and the cyclic permutations of its block of digits determine I he expansions of the other elements of the A coset of k in �(n). 1 7.

In

C YCLIC GRO UPS

this section we shall find criteria for decidi�g when a group is cyclic or a

llruduct of cyclic groups. Using them we shall be able to prove that �(p'") is cyclic when p is an odd prime and that �(n) is a product of cyclic groups in n useful way when n is divisible by several primes.

g

17.1 Definition. Let generate a cyclic group G of order n. For G let the index of a relative to g be the least nonnegative integer m for wh ich a

uE

= gm. Write m = ind11{a).

Then 0 � ind11(a) � n -

g indf;� = a.

1,

and

� �nd� (a� Q. .

{14)

When we regard the index as a map ind11 : G -+ Z,. , + from G to the additive group of the ring Z,. , then it is a group isomorphism. That is, to multiply two elements of G simply add their indices modulo n. The index of e is 0, the index of is 1 . The index map should be thought of as a logarithm to the base for it turns multiplication in G into addition in Z,. :

g

·

g,

indg(a � b) = ind11(a) + indg{b) (n ). Choosing a generator

g as a base for the indices is equivalent to choosing

a particular isomorphism of G with Z,. ,

+.

These introductory remarks show that the study of finite cyclic groups is equivalent to the study of the groups Z,. , + . For the remainder of this section Z,. will mean simply the additive group of the ring Z,. . 17.2

Lemma.

The order of a e Z,. is nf(a, n).

Proof. Remember that we are discussing Z,. as an additive group. Thus the order of a is the least positive k for which ka = O(n). But Theorem 7. 1

40

The Group of Units of z.

tells us how to find all such k ; 0 is a soluti on and the solutions are unique modul o n/(a, n). Therefore nf(a, n) is the least positive s o l uti o n Note that this result is valid eve n for a = 0 since the order of 0 is 1 and (0, n) = n. .

17.3 Corollary. The element a generates Z, if and only if a e 1, then no d c an be ns large as mn, which is the order of Zm x Zn , so that group is not cy­ clic. Conversely, if (m, n) = 1 , then ( 1 , 1) has order mn and so generates Zm X zn . 17.9

Proof Then

Theorem.

If G

x

H is cyclic, then G and H are cyclic.

Let (g, h) generate G

mn

=

x

H.

S uppose G has order m and H order n.

order of (g, h) in G

x

H

= l.c.m. {order of g, order of h}

which is a divisor of mn. 17.10

Theorem.

Thus the order of g is m, and the order of h is n.

Let G be an abelian group.

Supp ose

g1 ,

• • •

,

g, are

Tile

42

elements of G of orders nl >

. • •

, n, respectively.

Group of Units of z.

Then the map

given by

( 1 5) is a group homorphism onto the subgroup G' of G generated by g1 , =

Proof If the order of g is n, gkg1 is performed modulo n. Thus

. . . , g, .

gk+ 1 where the addition in the exponent

-r((kl, . . . ' k,) ( 11> . . . ' 1,)) = g�· . . . g�·g't' . . . g� = -r(( kl + 11,

where addition in the ith place is modulo n1 • morphism. Since -r((O, when the

.

.

.

, 1,

.

.

.

.

. ' k,

+ 1,))

Therefore -r is a group homo­

. , 0)) = g1

1 is in the ith place, the image of -r contains each g1 and thus is G'.

The proof above depends in a subtle way on the notation we used. We implicitly invoked the identification of Zn, with { I , 2, . . . , n;} in order to define -r by Eq. ( 1 5) and then conveniently ignored the identification for the rest of the proof. The argument is however essentially correct. Rather than make it more pedantic by resolving the ambiguities of the notation we shall give another, more abstract version. Write zr for z X X z (r times). The map • • •

T: Z' --. G given by

T( (k1,

• • •

,

k,)) = g�'

···

g:·

is clearly a group homomorphism, and no notational ambiguity besets its definition. The kernel of T contains the subgroup H = n1 Z x · · · x n,Z of zr' so the fundamental theorem of group homomorphisms implies that there is a homomorphism

frlt,

'11tl'

Group (p)

What is the period of the decimal expansion of 1/1 3 ?

A. (1 3) = order of 10 in (13) =

order of indz{10) 12 (10, 12)

= --=

=

10 in Z1 2

(Lemma 1 7 .2)

6,

which we already knew. Solve

X8 =

3 in (13) - 8 ind2 X = ind23 in Z1 2 - 8 ind2 x = 4 (12) - ind2 x

- ind2 x -

x

=

=

=

2 + 3n (12)

(Theorem 7. 1)

2 , 5, 8 or 1 1

4, 6, 9, or 7.

We showed above that 6 and 7 are primitive roots for 1 3. each solve (17), we know

Since 6 and 7

That equality is a coincidence ; ind2 3 = 4 -:1- 8. In general the index of x in (n) depends on the existence and the choice of a primitive root for n. So far we know of their existence only for primes. The index calculus is only useful once a primitive root has been found ; we have given no procedure other than trial and error for finding one. No universal shortcut is known though we shall show that in some special cases it is possible to find a primitive root for n without doing as much arithmetic as we required to find 2 for 1 3. When we locate a primitive root g by trial and error, the computations which prove g a primitive root also serve to build the table of indices to the base g. Appendix 2 contains a short table of primitive roots for primes. The index calculus is of practical value if we have many actual computations to make modulo a fixed prime p. For most theoretical purposes what is important is just the existence of a primitive root. That is, we are often interested in consequences of the fact that (p) is cyclic. For example,

Tlte Gro11p of UJtlts of z.

46

Theorem 17. 5 shows that when d I P - 1 , (l)(p) contains (2'") is isomorphic to Z2

X

Z2• - 2 when

ex :0::

3.

Proof Apply Theorem 1 7 . 1 0 with 91 = - 1, 92 = 5, n1 = 2, and n2 = 2'" - 2 • The map -r : z 2 X z 2 . - 2 � (2'") s o constructed i s surjective because - 1 and 5 generate C1>(2'"). Since both the domain and range of -r have q> (2'") = 2'"- 1 elements, -r is injective as well and hence is an isomorphism. 20.

THE GRO UP !f}(p")

In this section we use techniques similar to those we just developed to study W(p'") when p is an odd prime. The results are nicer than those in Section 1 9. We shall discover that (p") is always cyclic by finding an integer 9 which is a primitive root for p" for all ex ; 9 plays a role for p analogous to that played by 5 for 2. 20. 1

Theorem.

following is true :

(a) (b)

Let

9

be a primitive root for p".

Then just one of the

The integer 9 is a primitive root for p" + 1 ; 9"'(P"') =

1 (p'"+ 1).

Proof Let m be the exponent to which 9 belongs modulo m I q>(p'" + 1 ). Moreover

so 9m = 1 (p'").

Hence q> (p"')I m

«+ 1

p

•

Then

20.

49

The Group G>(p•)

since g is a primitive root for p«.

Therefore (21)

19.2.

which we could have proved directly by applying Lemma or

(b) m

=

({)

Thus either

(p«)

since p is prime.

Suppose g is a primitive root for p. Then gP - 1 = 1(p), so gp- 1 = 1 + mp for some integer m. If p ,.f' m, then Theorem 20. 1 tells us g is a primitive root for p2 as well. In fact , more will be true. We can prove the following analogue of Theorem 19.3. 20.2

Theorem.

Let g be a primitive root for p such that

gP - 1 = 1 + mp

and p ,.f' m. Proof.

Then g is a primitive root for p« for all oc

>

0.

Let the induction hypothesis be g 'P

(p� )

¢. 1 (p"' + 1 )

oc.

In particular, we

n ow

'111

51

Let p be an odd prime.

Then (p«) and (2p«) are

Proof. We have just found a primitive root g for p«, so {p«) is cyclic. Moreover, we may assume g is odd, for if it is even then g + pf¥. is odd and is still a primitive root for p«. Then (g, 2p«) = I . The exponent n to which g belongs modulo 2pf¥. can be no less than that to which it belongs modulo pf¥., which is cp(p«). But cp (2p") so

n =

=

cp (2)cp (p")

=

cp (p«)

cp(2p«), and g is a primitive root for (2p").

In the next section we shall show that we have found all the integers n for which (n) is cyclic, namely, n = 2, 4, p«, or 2pf¥.. 21.

THE GRO UP (f)(n)

We are now ready to complete our study of (n). We shall show that (mn) is isomorphic to (m) X (n) when m and n are relatively prime. Then we can find out all about (n) by factoring n as a product of powers of primes and using our knowledge of the structure of the groups (p«) . The route we follow is straightforward and comput ational . In the next theorem we use the Chinese remainder theorem and Theorem 17.10 to write (n) as a product of cyclic groups. Let n = 2«p�1 • • • p�· be the factorization of n into products of primes. Then (n) is isomorphic to 21.1

Theorem.

which is in turn isomorphic to the product of cyclic groups (24) when rx � 3. When rx = 0 or 1 , omit the first two factors in Eq. (24) ; when rx = 2, omit the second factor. Proof. We shall prove explicity only the case rx � 3. When rx = 0, 1 , or 2, the argument is similar but simpler. Begin by choosing a primitive root b; for p�', i = 1 , . . . , r (Theorem 20.4). Let b00 = - 1 and b0 = 5 ; these are the analogues of primitive roots for 2...

52

TIU! Group of Units ofZn

Let h00 simultaneously solve the X =:

r

+ 1 congruences

- 1 (2")

(25)

X =: 1 (pj')

Let h0 simultaneously solve X =: X =:

i = 1,

.

..

, r.

(26)

·

(27)

5 (2'")

1 (pj')

i

= 1,

.

.

.

, r.

(28)

For each j between 1 and r let h1 simultaneously solve X =:

X =:

X =:

h0 , h1 ,

1 (2")

bJ (pji)

1 (pj')

(29) (30)

1 � i � r, i =F j.

(3 1)

The Chinese remainder theorem (8 . 1) guarantees the existence of h00 , , h, . Finally let • • •

i = 00, 0, 1 , . . . ,

r.

The order of 9; in
for n, that is, if

1 (n)

whenever (a, n) 1 . Show 561 i s an F-number. It happens to be the smallest composite F-number ; The next two are 1 105 and 1 729. For more information on F-numbers see Ore, Chapter 14 (Reference 6 in the Bibliography). =

In the next problems we build a theory for solving the congruence

(m, n)

x ' = m (n),

=

1.

(38)

The method is to establish some easy facts about products of groups and then to apply Theorem 21 . 1 . The results generali ze parts of Problems 22. 10 and 22. 12. Let

22.20

fJ be an integer and

This definition makes sense whenever x belongs to a group. When G is an abelian group u, : G --+ G is a homomorphism (why ?) ; then let G6 be its kernel. Prove (a) (G x H)11 Gp X Hp . (b) Given g e G the solutions (if any) in G to =

xfl = g form a coset Let

22.21

(a)

(b) (c)

of Gp .

k11(n) be the number of solutions in cll (n) to the congruence xll

=

1

(n).

Prove k11 is multiplicative. Sho w kp(p•) (fJ, cp(p")) when p is an odd prime. What is ki2") ? =

(a) Show (38) can be solved for rp(n)fkp(n) elements m of il>(n). When (38) has a solution, it has exactly k6(n) of them in «P(n). If f3 = fJ' ( v(n)), then (38) is equivalent to

22.22

(b) (c)

x11 ' (d)

=

m (n).

Congruence (38) has a solution for every

m if and only if (fJ, v(n)) = 1 .

22. Problems 22.23

59

Solve x

8

=

256 (3 1 5)

X5 =

256 (3 1 5)

8 x =

x5 =

263 (3 1 5) 263 (3 1 5)

using either the techniques of the problems above or those developed in Appendix 3.

22.24*

Show that x4

=

- 1 (p)

has a solution for the odd prime p if and only if p infinitely many primes of this kind. 22.25*

When

have a solution ?

=

1 (8).

Deduce that there are

does

(Compare Problem 22.4 and Theorem 1 3 .2.)

5 Q,uadratic Reciprocity

In this chapter we shall discuss the congruence

x2 = a (p)

(1)

where p is a prime which does not divide a. We solved this problem in principle in Section 1 8 ; choose a primitive root g for p and solve

2 ind9 (x) = ind9 (a) (p

-

1 ).

(2)

Unfortunately, the computation of indices is nontrivial. Our objective now is Gauss's Quadratic Reciprocity Law, with which we can decide quite easily whether or not (I) has a solution. Finding one still requires the index calculus or some equivalent. We have already seen and solved one special case of (I), namely, x2

=

- 1 (p)

has a solution if and only if p = 1(4) (Theorem 1 3 .2). This fact allowed us to prove that there are infinitely many primes congruent to 1 modulo 4.

60

23.

lk�·idues

61

The Quadratic Reciprocity Law will help us show that some other arithmetic progressions contain infinitely many primes. We shall also use it to study the Di ophanti n e equation

which we looked at in Chapter 3 for m 23.

=

1 and m =

-

1.

R ESID UES

We say that a is an mth power residue of congruence

n

x"' = a (n)

when (a, n)

=

1 and the (3)

has a solution. Problems 22.9, 22. 10, 22. 12, and 22.20 through 22.25 take up various aspects of the theory of mth power residues. We are interested now in the special case m = 2, n prime, when (3) coincides with (1). We say then that a is a quadratic residue, or residue, of p. Numbers which are not residues are nonresidues. Si nce every odd integer a is an mth power residue of 2 for every m, we shall restrict our attention to odd prime s p. The intro­ ductory remarks ab ove show that - 1 is a residue of p if and only if p is congruent to 1 modulo 4. We shall follow our customary useful but ambiguous procedure and think of the residues of p b oth as integers and as elements of ZP . We can find the residues of p by squaring each of the elements of (p). For example,

( ± 1) 2 = 1 (7)

( ± 2) 2 = 4 (7) and ( ± 3) 2

=

2 (7)

so 1 , 2, and 4 are the residues of 7. 23.1 Lemma. Let g be a primitive root for p. of p if and only if indg(a) is even.

Then a e -p J

,, �

2

is

larger

than p/2,

Quadratic Reciprocity

66 if and only if

Since there are [p/4] positive integers less than p/4, there are

n=

p

; l - [�]

remainders ri greater than p/2. How can we discover the parity of n in terms of p ? The key is to write p modulo 8. Since p is odd

p = 8k + v where v = 1 , 3, 5, or 7. v

Next compile the table (8k + v) - 1

[8k: ]

n

4k

2k

2k

2

1

4k +

5

4k + 2

7

4k + 3

Thus n is even if and only if v = 24.3

1

3

Theorem.

1 or 7.

24.4

Corollary.

Proof

2k + 1 2k + 1

2k

2k + 1 2k + 1

2k + 2.

We have proved the next theorem.

(�) 1 if and only if =

( 1�) = (�) = 1 modulo 8 .

v

but

p ==

± 1(8).

(1�)

=

For example,

-

1

.

There are infinitely many primes congruent to ± I

There are infinitely many primes modulo which the polynomial for which 2 is a residue. (Compare Problem 22.24.)

x2

-

2 has a root (Theorem 12. 1) ; these are the primes

The Lemma of Gauss

24.

67

Now we return to the general stituation and prove the lemma of Gauss. s1 , , sn be the remainders ri > p/2. Let m ((p - l)/2) - n and t1 , tm be the remainders ri < p/2. Let

Let

• • •

=

• • • ,

Ut

=

p

St ,

-

• • •

, Un

p - Sn

=

•

Then

p 1 < - u - < J 2

j

=

1, . . . ,

n

and p

2

•

p-1

.

z

1 ::;; t. < ­

=

- - n. 1, . . . , m = 2

(1 1)

Let us show that the (p - 1)/2 integers

(12)

are incongruent m o dulo p. numbers

Observe

that for any choice of signs the (p - 1)/2

p-1 ± 1 , ± 2 , . . . , ± -2 Then since (p) is a group, the sequence

are mutually inco ngruent modulo p.

± a , ± 2a, . . . , ± also

p- 1

2- a

-

( 1 3) .

represents (p - 1 )/2 distinct elements of (p) for any ch oice of signs. some i

If ri < p/2, then for

t; while if ri > p/2,

=

ri

=. ja (p)

then for some i u;

=p -

s;

=

p - ri = -ja (p).

Therefore the sequence (12) of t;'s and u/s is congruent modulo p to a of the sequence (1 3) for a particular choice of n minus signs.

rearran gement

Quadratic Reciprocity

. 68

Hence the t /s and u/s are mutually incongruent modulo p. This fact and the inequalities in (11) together imply that the sequence (12) is just a re­ arrangement of the sequence I , 2, . , (p - 1)/2. Therefore

.. (E..:_!) ' = t . . . . . . u 2 P 1 = ( - 1t ( ; )! a