IMO 2003 SL N5

Origin: KOR | Category: Number Theory

Problem

An integer n is said to be good if |n| is not the square of an integer. Determine all integers m with the following property: m can be represented in infinitely many ways as a sum of three distinct good integers whose product is the square of an odd integer.

Solution

Suppose that m = u+v+w where u, v, w are good integers whose product is a perfect square of an odd integer. Since uvw is an odd perfect square, we have that uvw \equiv1 (mod 4). Thus either two or none of the numbers

u, v, w are congruent to 3 modulo 4. In both cases u + v + w \equiv3 (mod 4). Hence m \equiv3 (mod 4). Now we shall prove the converse: every m \equiv3 (mod 4) has infinitely many representations of the desired type. Let m = 4k + 3. We shall represent m in the form 4k + 3 = xy + yz + zx, for x, y, z odd. (1) The product of the summands is an odd square. Set x = 1 + 2l and y = 1 −2l. In order to satisfy (1), z must satisfy z = 2l2 + 2k + 1. The summands xy, yz, zx are distinct except for finitely many l, so it remains only to prove that for infinitely many integers l, |xy|, |yz|, and |zx| are not perfect squares. First, observe that |xy| = 4l2 −1 is not a perfect square for any l ̸= 0. Let p, q > m be fixed different prime numbers. The system of congruences 1 + 2l \equivp (mod p2) and 1 −2l \equivq (mod q2) has infinitely many solutions l by the Chinese remainder theorem. For any such l, the number z = 2l2 + 2k + 1 is divisible by neither p nor q, and hence |xz| (respectively |yz|) is divisible by p, but not by p2 (respectively by q, but not by q2). Thus xz and yz are also good numbers.