Consider a set of $1985$ positive integers, not necessarily distinct, and none with prime factors bigger than $23$. Prove that there must exist four integers in this set whose product is equal to the fourth power of an integer.
For an integer $n$ in this given set we can express $n$ as $$n=2^{\alpha_1} 3^{\alpha_2}\dots 23^{\alpha_9}. $$
Now it seems that for another element $m$ in the set the product $n\cdot m$ is a perfect square if and only if the exponents of the prime factors of $n$ and $m$ are of the same parity? I don't really see where this is coming from. If they have the same parity then I know that the exponents of the primes of the product $n\cdot m$ will be even so my question can be reframed as
Why does the exponents of the prime factors of a perfect square have to be even?