Can one partition the set of positive integers into finitely many pythagorean-triple-free subsets? If so, what is the smallest number of such subsets? Taking a wild guess, I would be least surprised if the answer were 3.
Notice that the 2 subsets of integers such that highest power of 5 that divides them is a) even b) odd manage to split most primitive triples, plus all the multiples of those.
Notice also that Schur proved the positive integers cannot be split into any finite number of sum-free subsets (i.e. no finite partition can split all power-of-1 Fermat triples), while Fermat's theorem proves that all power-of-n (n>2) triples can be split by the trivial partition into 1 set.
Edit: Since this turns out to be a known open problem, we're adding the tag [open-problem] and converting this question to community wiki. The idea is to have a separate answer for each possible approach to solving this problem. If you have some additional insight or a reference to contribute to an answer, you only need 100 rep to do so. We're still figuring out exactly how to handle open problems on MO. The discussion is happening on this tea.mathoverflow.net thread.