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 is irrational for all and .
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 , where and . Then, we have which implies . But this contradicts Fermat’s Last Theorem! And, we are done.