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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)?

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    $\begingroup$ That's the first example that pops into my head! $\endgroup$ Jan 21, 2019 at 19:11
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    $\begingroup$ Whoever downvoted this has no... I'm not sure what exactly, but they have none of it. $\endgroup$ Jan 21, 2019 at 19:21
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    $\begingroup$ @HagenvonEitzen Here's more: math.stackexchange.com/questions/555316/… $\endgroup$
    – Metric
    Jan 21, 2019 at 19:24
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    $\begingroup$ And here's some more! mathoverflow.net/questions/42512/… $\endgroup$ Jan 21, 2019 at 19:29
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    $\begingroup$ @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. $\endgroup$ Jan 21, 2019 at 19:36

1 Answer 1

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

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