The following fun problem was posed in one of the issues of the American Mathematical Monthly (if I am not wrong). I don’t remember the exact issue or the author, but here is the problem anyway.

Prove that \sqrt[n]{2} is irrational for all n > 2 and n \in \mathbb{N}.

Slick solution: We could either use Euclid’s arguments or invoke the rational root theorem to prove the above statement. However, there is a slicker proof!

Assume, for the sake of contradiction, that \sqrt[n]{2} = p/q, where p, q \in \mathbb{N} and p \ne 0. Then, we have 2 = (p/q)^n which implies q^n + q^n = p^n. But this contradicts Fermat’s Last Theorem! And, we are done.