Download On Representing a Square as the Sum of Three Squares Owen

On Representing a Square as the Sum of Three Squares
Owen Fraser; Basil Gordon
The American Mathematical Monthly, Vol. 76, No. 8. (Oct., 1969), pp. 922-923.
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922
MATHEMATICAL NOTES
[October
THEOREM
1. Let S be a closed and bounded convex subset of Rn whose centroid i s
If f ( x ) i s a convex function and h ( x , () i s a support hyperplane at t , then the
function
$.
has a m i n i m u m at ,$ = 3.
Proof. Two properties of a support hyperplane to f a t ,$ are
h(E,
= f(E)
and
h(x, t ) 4 f (x)
If a is the positive volume of S, integration shows t h a t
By the first of the two properties,
and by the second, equations (1) and (2) reveal that +(E) 2 4 ( 3 ) ; t h a t is, the
minimum value of 4 is a t 3.
An inspection of the proof yields the following generalization:
THEOREM
2. Let S be a closed and bounded subset of Rn that contains its centroid
3. Let f ( x ) be a real-valued function that i s integrable on S and possesses a lower support hyperplane, h ( x , (), at least at [ = 3. T h e n +(() has n m i n i m u m at $.
ON REPRESENTING A SQUARE AS THE SUM OF THREE SQUARES
OWENFRASER,
LOSAngeles Valley College, and BASILGORDON,
University of California, Los Angeles In his book on number theory [3; 1941, Nagell says "it follows from
Lebesgue's identity
that every integral square may be written as the sum of three integral squares."
Since there is always the trivial representation n2=n2+02+02, he presumably
means to imply that there is also a representation n2=x2+y2+z2 with x y z Z O .
This, however, is not always the case; the complete answer is as follows:
THEOREM.
If n i s a positive integer, then the equation n2=x2+y2+z2 has a
solution i n positive integers x , y , z if and only if n i s not of the form 2k or 2"5.
This theorem was stated without proof by Hurwitz [2; 7511, who alsogave a
19691
MATHEMATICAL NOTES
923
formula for the number of solutions. I t appears that his proof was probably
based on the theory of elliptic functions. Here we shall give a completely
elementary proof.
Proof. I t is trivial to verify the unsolvability of our equation for n = 1 and
n = 5. Suppose that k 2 1, and t h a t unsolvability has already been established
for n = 2"l and 2"l.S. Let n be either 2k or 2k-5,and suppose that n 2=x2+y2+z2,
where xyz#O. Since n is even, either two or none of the numbers x, y, z are odd.
In the first case x2+y2+z2=2 (mod 4), contradicting the fact that n 2 = 0 (mod
4). I n the second case, the equation can be divided by 4, giving ( 7 ~ / 2 ) ~ =
( ~ / 2 ) ~ + ( y / 2 ) ~ + ( 2 / 2 )This
~ . contradicts the induction hypothesis, completing
the proof of unsolvability for n = 2k and n = 2" 5.
Now suppose n has neither of these forms. Then n is divisible either by 25, or
by some prime p #2,5. Clearly if n = ab, and if a 2= x2+y2+z2 with xyz #O, then
( b ~with
) ~ (bx) (by) (bz) $0. Hence our theorem will be proved
n" ( b ~ ) ~(by)2+
+
if we can show that each of the numbers 2S2 and p 2 (p#2, 5) is the sum of three
positive squares. We have 2S2= 122+ 152+ 162, so we can suppose from now on
that n = p , where p is a primeZ2, 5. By Lagrange's Theorem we can write
p=a2+b2+c2+d2, and Lebesgue's identity then gives
Consider first the case where p = 3 (mod 4). Here i t is known [1; 2991 that there
is no representation of p 2 as the sum of two positive squares. Hence the number
of positive terms on the right hand side of (1) is either 1 or 3. T h e first term is
clearly odd, so it suffices to show that the second and third terms cannot both
vanish. If ac+bd = a d -bc = 0, then
(ac 4- bdl2
+ (ad
-
b ~ =) (a2
~
+ b2)(~2+ d2) = 0,
Hence either a = b = 0 or c = d = 0. In either case, the formula p = a2+b2+c2+d2
leads to a representation of p as a sum of two squares, which is well known to
be impossible, since p = 3 (mod 4).
If p = 1 (mod 4), we can write p = a2+c2, and Lebesgue's identity reduces to
p2= ( U ~ - C ~ () 2~ a+~ ) Clearly
~.
both terms on the right are positive. Since p #5,
we have p2= 1 or 4 (mod 5). Every square is congruent to O,1, or 4 (mod 5) ; from
this i t is easily seen by enumeration of cases that whenever fi2= u2+v2, we must
have either u = 0 (mod 5) or v = 0 (mod 5). Suppose for definiteness that v = 5w.
~ . we found a representation fi2=
Then p2=u2+25w2 = u 2 + ( 3 ~ ) ~( 4+ ~ ) Since
u2+v2 with UVZO, this leads to a representation p2=x2+y2+z2 with xyz
= u (3v/5) (4v/5) $0. T h e proof of our theorem is now complete.
The second author was sponsored in part by NSF Grant GP-5497.
References
1. G. H. Hardy and
E. M. Wright, An Introduction to the Theory of Numbers, Oxford,
New York, 1960.
2. A. Hurwitz, Mathematische Werke, vol. 2, Basel, 1933.
3. T. Nagell, Introduction to Number Theory, Wiley, New York, 1951.