1985 IMO Problems/Problem 4

Problem

Given a set $M$ of $1985$ distinct positive integers, none of which has a prime divisor greater than $23$, prove that $M$ contains a subset of $4$ elements whose product is the $4$th power of an integer.

Solution

We have that $x\in M\Rightarrow x=2^{e_1}3^{e_2}\cdots 19^{e_8}23^{e_9}$. We need only consider the exponents. First, we consider the number of subsets of two elements, such that their product is a perfect square. There are $2^9=512$ different parity cases for the exponents $e_1,e_2,...,e_9$. Thus, we have at least one pair of elements out of $1985$ elements. Removing these two elements yields $1983$ elements. By applying the Pigeon Hole Principle again, we find that there exists another such subset. Continuing on like this yields at least $734$ pairs of elements of $M$ whose product is a perfect square. Let $S$ be the set of the square roots of the products of each pair. Then, by the Pigeon Hole Principle again, there exist at least two elements whose product is a perfect square. Let the elements be $x,y$ and let $x=\sqrt{ab},y=\sqrt{cd}$ where $a,b,c,d\in M$. Then, we have $xy=z^2$ for some $z$ which implies $abcd=z^4$ and the claim is proved.

See also

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