A Simple Proof of the Four-Squares TheoremAuthor(s): Charles SmallSource: The American Mathematical Monthly, Vol. 89, No. 1 (Jan., 1982), pp. 59-61Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2320997 .

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NOTES

EDITED BY J. ARTHUR SEEBACH, JR.

Material for this department should be sent to Professor J. Arthur Seebach, Jr., Department of Mathematics, St. Olaf College, Northfield, MN 55057.

A SIMPLE PROOF OF THE FOUR-SQUARES THEOREM

CHARLES SMALL

Department of Mathematics and Statistics, Queen's University, Kingston, Ontario, Canada K7L 3N6

The fact that every positive integer is a sum of four squares was known long before it was first proved (in 1770, by Lagrange) and it has been given many different proofs since. The purpose of this note is to give a short proof that makes a lovely and convincing application in undergraduate algebra courses.

In this approach the four-squares theorem follows from a certain factorization theorem for 2 X 2 matrices over the ring Z[i] of Gaussian integers. The proof is due to M. Newman (see [1, p. 813]). The version given here is a simplification of that given by G. M. Bergman [2].

The starting point is a well-known result:

LEMMA. Let p be a prime and let Zp denote the ring of integers mod p. Then every element r E Zp is a sum of two squares.

Proof. We may assume p # 2. Put S1 = {x2 1 xCE Zp} and S2 ={r-y2 1 yZE Zp}. Then S1 and S2 each has (p + 1)/2 elements. Therefore S1 n S2 is -nonempty; so r is a sum of two squares.

REMARK. The same proof shows more generally that in any finite field F any equation ax2 + cy2= r (a, c # O) has solutions: let S1 = {ax2 x E F} and S2 = {r-cy2 2Y cF}. The reader is invited to prove, more generally still, that ax2 + bxy + cy2 r has solutions if b2- 4ac =# 0. Thus "every nondegenerate binary quadratic form over a finite field is universal."

COROLLARY. Let n be squarefree and let Z,, denote the ring of integers mod n. Then every element r E Z,, is a sum of two squares.

Proof. This follows from the lemma with the aid of the Chinese Remainder Theorem: n, being squarefree, is a product of distinct primes, say n = Pi I * Pk. Then Zn- Zp, X * * * X Zp, and therefore writing r as a sum of two squares in each Zp writes r as a sum of two squares in Z,.

REMARK. The converse of the corollary is false: every element of Z25 is a sum of two squares. The problem this suggests-given n, determine the smallest m such that every element of Z, is a sum of m squares-is solved in [3, ?3], and the generalization from squares to kth powers is handled in [3] and [4].

Now, to see that a positive integer n is a sum of four squares, we may assume that n is squarefree; for if n = a2n' (n' squarefree) and if n' = w2 + X2 + y2 + Z2 then n -(aw)2 +

(ax)2 + (ay)2 + (az)2. Once n is squarefree, we use the corollary above (with r =-1) to find integers c, d, and m with - = c2 + d 2 - mn. Now consider the matrix

A= ( n. c+di)

(where of course i = -/ 1). Observe that det A = nm - C2-d 2 = 1. The factorization theorem for such matrices is as follows:

THEOREM. Let

A =(n. c +di)

59

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60 CHARLES SMALL [January

where c, d, m, n are integers and n > 0, and assume det A = nm-c2- d2 = 1. Then A = BB* for some 2 X 2 matrix B over Z[i], where B* denotes the conjugate transpose of B.

The four squares theorem is an immediate consequence:

( : * x (w*i Y Zi)w- :)

gives n = w2 + x2 + y2 + Z2. To prove the factorization theorem we do induction on c2 + d2. If c2 + d2 = 0 the hypotheses

force A = I, in which case B = I will do. For the rest of the proof we assume, therefore, that c2 + d2 > 0; i.e., c and d are not both 0.

Since m is also positive (nm = 1 + c2 + d2), there are two cases to consider: 0 < n s m or 0 K m < n .

Case 1. 0 K n < m. Let A' = MAM* where

M= (I 0) x -yi I

and x and y are integers to be specified momentarily. Then

with c' = c + nx, d' = d + ny, and det A' = I because det M - det M* = 1. If we can choose x and y to ensure c'2 + d'2 K c2 + d2, we are done: for then A' = CC* by induction, and putting B = M-'C gives A M-IA'(M*)-y = M-l(CC*)(M-)* = (M-IC)(M-IC)* BB*.

\\1 c - n 0 n1 * ; \ IF - _ _* I I

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ I

-n/2 c-n 0 n/2 c

FIG. 1

Now as long as c > n/2, we can simply choose x =-l,y = 0 (see Fig. 1); then c'2 = (c - n)2 < c2 and d'2 = d2, so that c'2 + d'2 c C2 + d2. Similarly, if c K -n/2 we can take x, y = 1, 0; if d> n/2 take x, y =0,- 1; and if d K -n/2 take x, y 0,1.

Hence, to complete the proof (in Case ) we need only show that we are necessarily in one of these four situations; that is, I c I > n/2 or I d I > n/2. If n I I this is clear since c and d are not both 0. If n > 1, suppose the contrary: I c I n/2 and; d I s n/2. Then, since O < n < m, we have n2 nm - c2 + d2 + 1 < (n/2)2 + (n/2)2 + 1 = n2/2 + 1 K n2, a contradiction. This com- pletes the proof in Case 1.

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1982] THE TEACHING OF MATHEMATICS 61

For Case 2 (0 < m < n) the argument is similar. Let

A' = MAM* where M ( Yi)

then

A'l (,idi d'i) with c'=c+ mx, d'=d+ my.

Again it suffices to find x and y so that c'2 + d'2 <c2 + d2, and again we can take x, y 1, 0 or 0, ? 1, since I c I > m/2 or I d I > m/2. Done!

Query. What kind of representation theorems for integers can be obtained by studying matrices over Z[c] (c a primitive cube root of 1) instead of Z[i]? Newman's book [5] has some remarks bearing on this question; see Chapter 1 1, especially page 215.

References

1. 0. Taussky, Sums of squares, this MONTHLY, 77 (1970) 805-830. 2. Response to Query #26, Notices of A.M.S., 21 (1974) 220-221. 3. C. Small, Waring's Problem mod n, this MONTHLY, 84 (1977) 12-25. 4. , Solution of Waring's Problem mod n, this MONTHLY, 84 (1977) 356-359. 5. M. Newman, Integral Matrices, Academic Press, New York-London, 1972.

THE TEACHING OF MATHEMATICS

EDITED BY MARY R. WARDROP AND ROBERT F. WARDROP

Material for this department should be sent to Professor Robert F. Wardrop, Department of Mathematics, Central Michigan Universitv, Mount Pleasant, MI 44859.

TEACHING AIDS

Moss E. SWEEDLER Department of Mathematics, Cornell University, Ithaca, NY 14853

Here are two devices that have proved useful in teaching first-year calculus. The first helps students remember the popular values of the trigonometric functions. The key is the pattern

2 2 2 2 2 The student still has to remember that these are the values of sine at 0?, 300, 450, 600, 900 and how to get cosine and the other values of trigonometric functions.

The second is how to convince a class that .999 ... equals 1 and is not the number just before 1. Write

3 .333 ...

on the board, and everyone will agree. Then tell the class to multiply both sides by three. This usually convinces 99.999 ... percent of the students.

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