Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There is a long tradition of mathematicians remarking that FLT in itself is a rather isolated claim, attractive only because of its simplicity. And people often note a great thing about current proofs of FLT is their use of the modularity thesis which is just the opposite: arcane, and richly connected to a lot of results.

But have there been uses of FLT itself? Beyond implying simple variants of itself, are there any more serious uses yet?

I notice the discussion in Fermat's Last Theorem and Computability Theory concludes that one purported use is not seriously using FLT.

share|improve this question
9  
2  
I read somewhere that it was attempting to prove Fermat's Last Theorem which brought about most of algebraic number theory. –  Koundinya Vajjha Aug 26 '13 at 15:11
6  
@Koundinya Vajjha: so, FLT would be just as useful if Wiles had finally disproved it! –  Laurent Moret-Bailly Aug 26 '13 at 16:00
11  
It has been used to good effect to prove the irrationality of $2^{1/n}$ here: mathoverflow.net/a/42519/8430 (sorry, couldn't resist this joke) –  Suvrit Aug 26 '13 at 16:19
4  
Alex B.'s answer is certainly a great application of the $p=7$ case, proved by Lamé in 1839. I would be interested to know if there is a substantial application for which pre-Wiles work did not suffice. –  S. Carnahan Aug 26 '13 at 23:16

4 Answers 4

up vote 30 down vote accepted

Corollary 3.17 in this paper of Stefan Keil uses FLT for exponent 7 to show that if $E/\mathbb{Q}$ is an elliptic curve with a rational 7-torsion point $P$, and $E\rightarrow E'$ is the 7-isogeny with kernel $\langle P\rangle$, then $E'(\mathbb{Q})[7]=0$. There are of course lots of ways of proving this, but the paper does it by writing down a parametrisation of all elliptic curves over $\mathbb{Q}$ with 7-torsion and of their rational 7-isogenies, and then playing with parameters to get a contradiction to FLT.

share|improve this answer
2  
In that corollary he needs to show certain rational numbers $q$ and $q-1$ cannot both be 7th powers, and he gets that immediately from degree 7 FLT. Looks like a serious use to me. –  Colin McLarty Aug 26 '13 at 17:33
2  
@ColinMcLarty: it is an absolutely serious use. There are other ways of doing this (rational points on the corresponding modular curves), but possibly not easier or more elementary than this one. –  Alex B. Aug 26 '13 at 17:40
3  
On the one hand this is, of course, awesome. On the other, this only uses FLT for n=7 so it is possible to argue it doesn't count as a real application. Still very satisfying though! –  Oliver Nash Aug 27 '13 at 10:52
7  
FLT for exponent 7 can be proved using a standard 2-descent. It is arguably simpler than FLT for exponent 3, which requires 3-descent or a 2-descent over number fields. –  Franz Lemmermeyer Aug 27 '13 at 11:58
    
This does not refute the centuries old consensus that FLT is a rather isolated fact in number theory. But it is a serious application. Keil wanted to complete a calculation and the easiest way he could see to do it was FLT. –  Colin McLarty Aug 27 '13 at 22:14

It is perhaps an indication of the average age of today's MOers that nobody remembers the work of Hellegouarch who introduced around 1970 what is nowadays called the Frey curve precisely in order to deduce information about elliptic curves from (the then) known results about Fermat's Last Theorem.

share|improve this answer
1  
See for example math.unicaen.fr/~nitaj/hellegouarch.html. –  Chandan Singh Dalawat Aug 27 '13 at 10:23
    
I Have been reading Hellegouarch but he says much less about the Fermat equations than I expected. He cites and uses many results on elliptic curves but can you point me to some place where he uses a then known case of FLT? Thanks. –  Colin McLarty Aug 27 '13 at 11:59
1  
In what follows $A$ is an elliptic curve defined over the field of fractions $K$ of a Dedekind ring $R$, and all the points of order $2$ on $A$ are defined over $K$. A rational prime $p$ is said to be "idoine'' if $(−2)^m\equiv−1\mod p$ for some integer $m$. When $R=\mathbf{Z}$, the rational integers, it is shown that there is no point of order $p^2$ defined over $K$, where $p$ is an idoine prime, provided that Fermat's equation $x^p+y^p=z^p$ is insoluble. (ams.org/mathscinet-getitem?mr=274451) –  Chandan Singh Dalawat Aug 27 '13 at 12:20
    
Thanks. This led me to a cleaner more general paper by Hellegouarch, Points d’ordre 2p sur les courbes elliptiques. Acta Arith. 26 (1974/75), no. 3, 253–263. MR0379507 (52 #412). –  Colin McLarty Aug 27 '13 at 12:56
1  
It is mentioned in the link in my first comment. –  Chandan Singh Dalawat Aug 27 '13 at 12:58

Do applications to physics count?

Supersymmetry Breakings and Fermat's Last Theorem, Hitoshi Nishino, Mod.Phys.Lett. A 10 (1995) 149-158.

In this paper, we give the first application of Fermat's last theorem (FLT) to physical models, in particular to supersymmetric models in two or four dimensions. It is shown that FLT implies that supersymmetry is exact at some irrational number points in parameter space, while it is broken at all rational number points except for the origin. This mechanism presents a peculiar link between the FLT in number theory and the vacuum structure of supersymmetry. Previously, the only well-known connection between number theory and supersymmetry has been via topological effects, such as instantons and monopoles in supersymmetric models.

share|improve this answer
    
FLT comes into this via their specification of a specific superpotential $W$ which includes a term $\Phi_4[\Phi_1^p+\Phi_2^p+ (t\Phi_3)^p]$ (p.~3). Is there something in the theory that constrains them to that, or do they simply have the freedom to pick a superpotential that will relate to FLT this way? –  Colin McLarty Aug 27 '13 at 12:26
1  
@Colin --- more general non-polynomial superpotentials are considered in section 4. –  Carlo Beenakker Aug 27 '13 at 13:11
1  
I do not see that Section 4 uses FLT. It looks like in this paper FLT is just a nice source of integer polynomials without rational solutions. –  Colin McLarty Aug 27 '13 at 13:34

There is of course the proof of the irrationality of the $n$-th root of $2$, for $n\geq 3$, that was posted on MO some while back....

share|improve this answer
10  
This very popular joke is made in the comments to the question by Carlo Beenakker and again by suvrit who added the precise MO link, and it was made for the case of $n=17$ in another comment since deleted. –  Colin McLarty Aug 27 '13 at 13:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.