Prompted by a recent exchange with Gerry Myerson, I was wondering if anyone has a favorite example of a relatively simple problem with a rather elementary (though perhaps complicated) answer for which there's another answer that relies on an elegant use of a powerful result that's almost certainly beyond the background of the poser of the question.
Swatting flies with a sledgehammer
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6I don't see any reason for the downvote. That's just mean. Downvoter should explain himself. – 2012-07-18
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3How about *[Pascal's mystic hexagon](http://en.wikipedia.org/wiki/Pascal's_theorem)*? It can be proved by elementary planar geometry, but also be attacked in many ways, including [Bézout's theorem](http://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem) in algebraic geometry. – 2012-07-18
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0There are a number of such proofs for the existence of infinitely many prime numbers (and probably stackexchange and/or mathoverflow have threads devoted to this), so I suggest that proofs of infinitely many primes be excluded in order to leave room for examples that are not so well known. – 2012-07-18
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0(continued) In re-reading the question (often a useful thing to do!), I see that the wording of the question (i.e. *though perhaps complicated* followed by *elegant use*) seems to eliminate the prime number examples I was thinking of. – 2012-07-18
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0I'm pretty sure there is a question like this on MathOverflow already. (And I'm pretty sure I left an answer...) – 2012-07-18
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11We have http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts a while ago on MO – 2012-07-18
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0Rick Decker, i think you mean my problem right? – 2012-07-18
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0@MohammedAl-mubark Yes. – 2012-07-18
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0@Dave. Actually, I'm partial to Furstenberg's proof of the infinitude of primes. Who would have expected that one could come up with a topological proof? – 2012-07-18
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0@Mariano. Duh! This is such a natural question for MO, I'm chagrined that it never occurred to me to look there. – 2012-07-18
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0@Rick Decker: You might be interested in [BIG HAMMER for primes](http://mathforum.org/kb/message.jspa?messageID=206856), a sci.math thread I started back on 20 October 1999. I began with a quote from pages 176-177 of L. A. Lyusternik, *The early years of the Moscow mathematics school*, **Russian Mathematical Surveys** 22 #1 (1967), 133-211. – 2012-07-19
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0@Rick Decker: Here's the beginning (but not all) of what I quoted from that paper. (I think it took place around 1920.) *Nina Bari, a student who had just been elected Vice-President of the circle, gave a lecture on "Various proofs of the unboundedness of the set of prime numbers". A couple of months before the lecture, Nina with her customary energy succeeded in organizing an original competition to devise "exotic" proofs of the existence of an infinite set of primes. Luzin condemned as defective exotic proofs by Khinchin and myself and offered one even more exotic. She produced all three.* – 2012-07-19
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0This should be community wiki. If you have objections let me know. – 2014-01-08
3 Answers
If $2^{1/n}$ were a rational $a/b$, with $n>2$, then $a^n=b^n+b^n$, which would contradict Wiles proof of Fermats Last Theorem.
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2I don't think it gets any more OP than this ^^ – 2012-07-18
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16I remembered that this example had been mentioned in [this old MathOverflow question](http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/42519#42519) and in a comment to the answer, BCnrd explains that the argument is basically circular. – 2012-07-18
OK, here's my proof that $49 < 50$.
There's one point in the first quadrant where the circle $x^2+y^2=1$ intersects the line $x=y$, and the tangent line to the circle at that point has slope $-1$ and $x$-intercept $\sqrt{2}$. Every line with the same slope and a slightly smaller $x$-intercept intersects the circle twice, and those with a larger intercept do not intersect the circle.
Now suppose we use $7/5$ as an approximation to $\sqrt{2}$, and draw the line with slope $-1$ and that $x$-intercept. Lo and behold, it passes through the points $(4/5,3/5)$ and $(3/5,4/5)$, which are on the circle. So it's a secant line, not a tangent line. Therefore we conclude that $$ \frac75<\sqrt{2}. $$ Therefore $$ 49 = 7^2 < 5^2\cdot2=50. $$
Last semester our grad Analysis II class proved the Isoperimetric Inequality using Fourier analysis.
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5This is pretty standard... – 2012-07-18
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3Sure, but it was new to me! – 2012-07-18
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1What would be the elementary way to prove the isoperimetric inequality in this case? I've always been under the impression that this already is the elementary way (the use of Fourier analysis certainly is elegant and I like it, too, but this is not what's being asked here). – 2012-07-19
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0I was under the impression that this was "known to the Greeks" whatever that may mean. Doing a quick search it definitely seems that they may have discovered a rigorous proof, e.g.: www.cs.nyu.edu/faculty/siegel/SCIAM.pdf – 2012-07-19