# Proving Fermat's last theorem [closed]

Fermat's last theorem states that $X^n+Y^n=Z^n$ when $n$ is greater than or equal to $3$ has no solutions in integers other than the obvious ones. How do you go about proving this theorem? I realize it took the entirety of a guy's life to fully prove this, but why was it so complex?

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## closed as too broad by Will Jagy, Potato, N. S., Dominic Michaelis, iostream007Nov 16 '13 at 7:50

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Wikipedia has a decent summary. –  user61527 Nov 16 '13 at 4:09
I might add: It took the entirety of the lives of many mathematicians to fully prove the theorem, not just Andrew Wiles! –  Mike Miller Nov 16 '13 at 4:10
You may also enjoy a nice fluffy hour long BBC documentary. –  J. W. Perry Nov 16 '13 at 4:26
A better question might be why did people consider it such an interesting question. It is completely unclear to me what fascinates people about it - to me it looks like just any other Diophantine equation. So perspectives of this sort might be interesting. –  Timotej Nov 16 '13 at 4:56
Does anyone seriously think this should be tagged as "elementary-number-theory"? –  Paramanand Singh Nov 16 '13 at 17:52

I will not answer the question of "how one can go about to prove this theorem", for obvious reasons. However, I can say something about the question: "why is it so difficult"? The answer is simple: for a single exponent, the problem is already fantastically difficult.

Have you ever tried showing that $x^3+y^3 =z^3$ has no nontrivial rational solutions? If you are very clever, you might discover a proof on your own.

If you are as clever as Kummer, you might be able to prove it for $x^{17} + x^{17} = z^{17}$. However, you'll need to reinvent the deepest and most beautiful number theory since quadratic reciprocity (class groups of cyclotomic fields, special values of $L$-functions...)

However, even Kummer could not do it for $x^{37} + y^{37} = z^{37}$. Unless I am mistaken, this case resisted until the general proof by Wiles. Even by itself, the non-existence of nontrivial rational solutions to $x^{37} + y^{37} = z^{37}$ is an extremely difficult problem. It is a problem that elementary number theory does not even come close to grazing. With this in mind, Wile's result is truly spectacular.

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That doesn't actually answer the question, of course (although I upvoted you anyway). The deeper question is, why is one equation like $x^{37}+2=y^{37}$ so easy to solve (that example being particularly trivial) but another equation of almost identical shape like $x^{37} + y^{37} = z^{37}$ so difficult to solve? There are fourth, fifth, sixth, and much higher degree equations — with many more variables — that are easier to solve than $x^3+y^3=z^3$. So the question really is, what makes that true? –  Kieren MacMillan Sep 10 '14 at 14:00

Wiles proof uses - group theory, algebraic geometry, commutative algebra, and Galois theory.
There have been numerous attempts to prove this theorem before Wiles by simpler means as many partial proofs of exponents - $3 , 5 , 7 , 10$ etc where given .Look here.
Inspite of many attempts there was no simple way of proving it for a general case .So , Andrew Wiles opted to attempt to "count" and match elliptic curves to counted modular forms.
Thus this being very difficult to realise took the full educational career of Andrew Wiles

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