IMO 1985 SL 1

Origin: MON

Problem

Given a set M of 1985 positive integers, none of which has a prime divisor larger than 26, prove that the set has four distinct elements whose geometric mean is an integer.

Solution

Since there are 9 primes (p1 = 2 < p2 = 3 < \cdot \cdot \cdot < p9 = 23) less than 26, each number xj \inM is of the form $9 i=1 paij i , where 0 \leqaij. Now, xjxk is a square if aij + aik \equiv0 (mod 2) for i = 1, . . . , 9. Since the number of distinct ninetuples modulo 2 is 29, any subset of M with at least 513 elements contains two elements with square product. Starting from M and eliminating such pairs, one obtains (1985 −513)/2 = 736 > 513 distinct two-element subsets of M each having a square as the product of elements. Reasoning as above, we find at least one (in fact many) pair of such squares whose product is a fourth power.