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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.

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