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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)?

Joe
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    That's the first example that pops into my head! – José Carlos Santos Jan 21 '19 at 19:11
  • I like the title lol – Cloud JR K Jan 21 '19 at 19:14
  • Fabulous. Now we need to find a similar proof for $\sqrt 2$. – David C. Ullrich Jan 21 '19 at 19:19
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    Whoever downvoted this has no... I'm not sure what exactly, but they have none of it. – David C. Ullrich Jan 21 '19 at 19:21
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    Possible bazookas are the four-colour-theorem or the classification of finite simple groups - now I still have to find a suitable butterfly – Hagen von Eitzen Jan 21 '19 at 19:22
  • The common saying is to kill a fly with a bazooka, but since problems in mathematics are prettier, we have the butterfly. Is this right, OP? – Metric Jan 21 '19 at 19:23
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    @HagenvonEitzen Here's more: https://math.stackexchange.com/questions/555316/what-are-the-most-overpowered-theorems-in-mathematics – Metric Jan 21 '19 at 19:24
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    And here's some more! https://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts – PrincessEev Jan 21 '19 at 19:29
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    @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 '19 at 19:36
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    I think should not kill butterfly ..let them live – Cloud JR K Jan 21 '19 at 19:42
  • @Metric: that's right! – Joe Jan 21 '19 at 22:30
  • @DavidC.Ullrich Killer of joy, reporting for duty. But Hagen von Eitzen's comment captures the essence of my objection; this question is an invitation to showcase our favourite bazooka's by shooting them at arbitrary innocent butterflies. Too broad. – Servaes Jan 21 '19 at 22:30
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    There's Direchlet's proof that there are an infinite number of primes via complex integration: https://dms.umontreal.ca/~andrew/Courses/Chapter3.pdf – fleablood Jan 21 '19 at 22:45

1 Answers1

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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.

DVD
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