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I wonder is there an elementary proof for Fermat's last theorem. Why it's so difficult to prove this theorem by elementary method?

Thanks,

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To prove Fermat's last theorem is difficult. To explain why it's difficult, it's perhaps even more difficult :-) –  leonbloy Mar 3 '11 at 16:36
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No elementary proof is known. As Gauss put it, though, it is a trivial thing to write down simple statements that have long, complicated, difficult proofs. There is no reason to expect there to be a connection between the simplicity of the statement and the simplicity of its proof. –  Arturo Magidin Mar 3 '11 at 16:37
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Mathematicians have gotten headaches over FLT for some 300 years. Thus, I'd say that an elementary proofs is either not there or must rely on some extremely clever and hidden idea (maybe that contradicts elementarity?!). –  Andrea Mori Mar 3 '11 at 16:48
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Answer: Not yet.... –  Aryabhata Mar 3 '11 at 17:34
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@Chan: Yes there is but this comment box is to small to explain. –  Eelvex Mar 3 '11 at 18:02
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3 Answers 3

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It is a common false hunch that shortly-stated theorems should have short proofs. This hunch is easily falsified by employing basic results in logic. For any formal system that has nontrivial power (e.g. Peano arithmetic) there is no recursive algorithm that decides theoremhood. Now if there existed a recursive bound $\rm\ L(n)\ $ on the length of proofs of a statement of length $\rm\:n\:,\:$ then we could test for theoremhood simply be enumerating and testing all possible proofs of length $\rm\le L(n)\ $. Hence there can be no such recursive bound on the length of proofs. It follows that there exist short stated theorems with proofs so long that they are probably not amenable to human comprehension (these results date back to Goedel's 1936 paper on speedup theorems).

It remains to be seen whether or not there exists mathematically interesting theorems like this. There may be examples in Collatz-like congruential iterations (similar to the difficult open $\rm\: 3\ x + 1\: $ problem) that were discovered in the wild while analyzing Busy-beaver holdout machines (while attempting to find the smallest universal Turing machines). John Conway has shown that there exists such congruential iterations with undecidable halting problem. That such undecidable problems may be encoded so succinctly in programs for tiny Turing machines should not come as a surprise to anyone familiar with the above simple results from logic. They are a testament to the power of ingenuity - whether it be human (in powerful mathematical theories) or nature (the DNA-based programs designed by evolution).

For a chess-theoretic analog see my post here, which discusses some massive brute-force computated chess endgame databases revealing optimal move sequences that are completely incomprehensible to human experts.

Returning to the specific topic at hand, it is known that Fermat's Last Theorem cannot be proved by certain types of descent proofs similar to the classical simple proofs known for small exponents. References to such work can probably be located by Googling "Tate Shafarevich obstruction".

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thanks for your chess link. I had similar thoughts, but your exposition is very nice. I guess it made me feel more at ease with the fact that i was never able to go past candidate master level. :-) –  Andrea Mori Mar 3 '11 at 19:44
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@Andrea: Thanks. No doubt many mathematicians have similar sentiments. As enticing as chess may be - it places competing demands on the mind so, ultimately, one is forced to choose between the two (unless one is blessed with extreme neural reusability capabilities such as in synesthesia - imagine being able to smell or taste a proof!). –  Bill Dubuque Mar 3 '11 at 20:50
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Have you heard of Harvey Friedman's conjecture? <en.wikipedia.org/wiki/…; –  Samuel Tan Oct 25 '11 at 7:53
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Fermat's Last Theorem, although elementary to state, is a very subtle problem.

The general $ABC$ conjecture (still unproved) states (roughly) that if $C = A + B$ (with $A, B, C$ coprime integers), then it is not possible for $A$, $B$, and $C$ to be simultaneously divisible by high powers of integers.

Fermat's equation considers the special case when $A$, $B$, and $C$ are all taken to be perfect $n$th powers.

The $ABC$ conjecture in general, and Fermat in particular, are then subtle problems relating the additive and multiplicative nature of the integers, and so there it is perhaps not too surprising that they are difficult to prove (or, in the case of $ABC$, that it remains unproved!).

Another famous conjecture relating the additive and multiplicative nature of the integers is the Goldbach conjecture. This is, if you like, the "opposite" of $ABC$; it states that for an even integer $N$, we may write $N = p_1 + p_2,$ where $p_1$ and $p_2$ are prime (which is a kind of "opposite" to being divisible by a high perfect power). It also resists proof.

At a technical level, the tools that have been brought to bear on Fermat, and on $ABC$, are quite different from the tools that have been brought to bear on Goldbach, so perhaps one shouldn't take a comparison between them too seriously. But they do share the common element of getting at something quite deep about the interrelationship between the additive and mutliplicative structure of the integers, and this is what makes them difficult (or so it seems to me).

[Added September 2012:] Shinichi Mochizuki has very recently claimed a proof of the ABC conjecture.

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As far as I know, the only proof there is, is Wiles's proof. And that is not an elementary proof. There is an article about his proof on wikipedia, if you're interested. There is also this video documentary about it that I would highly recommend.

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Love the video ;) Many thanks. –  Chan Mar 3 '11 at 17:48
    
@Elvind: I think you mean "Wiles's"; unless Wiley publishers came up with their own, that is... –  Arturo Magidin Mar 3 '11 at 18:27
    
@Chan: Good to hear. I watched it a couple of weeks ago. It is so inspiring! –  please delete me Mar 3 '11 at 18:27
    
@Arturo: Of course. Thank you. –  please delete me Mar 3 '11 at 18:29
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protected by t.b. Aug 23 '11 at 11:49

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