TITLE: Can the product of 4 distinct numbers be equal to a fourth power?
QUESTION [7 upvotes]: The set $M$ consists of 2001 distinct positive integers, none of which is divisible by any prime $p > 23$. Prove that there are distinct $x, y, z, t \in M$ such that $xyzt = u^4$ for some integer $u$.
I guess some kinda Pigeonhole Principle argument should be useful, but I can't figure it out.

REPLY [5 votes]: Just think of even/odd. 
There are 512 classes of integers, depending on the parities of the nine exponents.  So two must belong to the same class.  Pair them off, you have 1999 numbers left.
Another two belong to the same class as each other because 1999>512.  And so on until only 511 numbers are left.  
You now have $(2001-511)/2$ pairs of numbers, the product of each pair is a square.  
Now, how to show that two of the pairs have a product which is a fourth power...