# Killing a butterfly with a bazooka [duplicate]

Let $$n\ge3$$. Prove that $$\sqrt[n]2\notin\Bbb Q$$.

Let us suppose that $$\sqrt[n]2=p/q$$, that is $$2q^n=p^n$$, so $$q^n+q^n=p^n$$, against FLT.

Do you know similar examples, in which simple problems are solved using huge weapons (maybe in a elegant way)?

## marked as duplicate by Servaes, MJD, José Carlos Santos, Raskolnikov, Morgan RodgersJan 22 at 23:45

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• That's the first example that pops into my head! – José Carlos Santos Jan 21 at 19:11
• Whoever downvoted this has no... I'm not sure what exactly, but they have none of it. – David C. Ullrich Jan 21 at 19:21
• @HagenvonEitzen Here's more: math.stackexchange.com/questions/555316/… – Metric Jan 21 at 19:24
• And here's some more! mathoverflow.net/questions/42512/… – Eevee Trainer Jan 21 at 19:29
• @DavidC.Ullrich We have such a proof for $\sqrt{2}$. If it would be rational, then the right angled triangle with rational sides $\sqrt{2},\sqrt{2},2$ would have area $1$, so $1$ would be a congruent number, which contradicts Tunnel's theorem. – Dietrich Burde Jan 21 at 19:36

## 1 Answer

Every bounded entire complex valued function on the complex plane misses three values in the range and, therefore, is constant by Picard's theorem.