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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,

roxrook
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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 03 '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 03 '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 03 '11 at 16:48
  • @all: Thank you. Probably because I'm a novice, FLT looked simple to me. By the way, 300 years for this proof was too much, I'd rather think about something else :P! – roxrook Mar 03 '11 at 17:05
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    Depends on what you mean by elementary. If by that you mean simple and small, then probably the answer is no. That problem stayed open for 300 years for a reason. If by elementary you mean that it can be proved in a simple system of arithmetics then some say that it's possible. You can check out the grand conjecture: http://en.wikipedia.org/wiki/Grand_conjecture#Friedman.27s_grand_conjecture – Apostolos Mar 03 '11 at 17:30
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    Answer: Not yet.... – Aryabhata Mar 03 '11 at 17:34
  • @Moron: I really like your answer! Personally, I think nothing is impossible if we keep trying. Hopefully, one day, there will be an elementary proof for this notorious theorem. – roxrook Mar 03 '11 at 17:50
  • @Apostolos: Thank you. – roxrook Mar 03 '11 at 17:50
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    @Chan: Yes there is but this comment box is to small to explain. – Eelvex Mar 03 '11 at 18:02
  • There is an ongoing project to see that Wiles's proof is at least formalizable in a fragment of Peano Arithmetic (PA). Not only this is not obvious but, taken at face value, one could even think (naively) that the argument uses more than the usual system of axioms of set theory (ZFC). The preprint http://www.cwru.edu/artsci/phil/Groth%20found26.pdf by Colin McLarty explains how the background cohomology theory can be formalized within ZFC. Macintyre has been working in showing that PA can prove FLT, and Kisin has provided a significant simplification of Wiles's argument. – Andrés E. Caicedo Mar 04 '11 at 06:54
  • @Andres Caicedo: Dear Andres, It is not quite correct to say that Kisin has provided a significant simplification of Wiles's arguments. There have been some simplicifications of the argument since it was first written (perhaps the most important being due to Fred Diamond and Fujiwara, who saw how to remove the Gorenstein hypotheses on Hecke algebras that Wiles required), and there have been many generalizations of the argument, by Taylor, Kisin, and others. Kisin's paper on Barsotti--Tate groups simplifies and conceptualizes the proof of the full modularity theorem for ... – Matt E Mar 04 '11 at 13:18
  • ... elliptic curves over $\mathbb Q$ by Breuil, Conrad, Diamond, and Taylor, but that more general result is not needed for the applications to FLT (which just requires modularity for semi-stable curves, a result that is technically somewhat easier). Also, Khare's work on Serre's conjecture (much of it joint with Wintenberger, and which was completed by Kisin) leads to a somewhat different proof of FLT, but it is not so different from that of Wiles, and at least as technically demanding, if not more so. Regards, – Matt E Mar 04 '11 at 13:21
  • @Matt: Yes, you are right, thanks! I was looking at the Inventiones paper and I agree that I oversimplified the situation in my comment above. – Andrés E. Caicedo Mar 04 '11 at 15:53
  • Related: see answers here https://math.stackexchange.com/questions/1139251/ – Watson Nov 28 '18 at 19:07
  • https://math.stackexchange.com/questions/3428002/ – a boy Nov 09 '19 at 04:04
  • 'Elementary' does not necessarily mean 'easier'. For example, the elementary proof of the Prime Number Theorem by Erdos and Selberg is more complicated than the complex-variables proof by Hadamard. –  Apr 27 '20 at 22:14

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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".

Bill Dubuque
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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 03 '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 03 '11 at 20:50
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    Have you heard of Harvey Friedman's conjecture? http://en.wikipedia.org/wiki/Grand_conjecture#Friedman.27s_grand_conjecture – Samuel Tan Oct 25 '11 at 07:53
  • I enjoyed your post on chess and if it's beauty was only skin deep. It reminded me of this comic I read recently. I am afraid we might be making the idea of human intelligence too abstract. – Aditya P Apr 10 '19 at 15:41
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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.

Matt E
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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.

Jonas Meyer
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Eivind
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