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I am investigating solutions to Fermat's equation $$x^n+y^n=z^n$$ with $x,y,z$ in the Gaussian integers, excluding solutions in excluding $\mathbb{Z}$ or $i\mathbb{Z}$ .

I have found out that there are only trivial solutions for the n=3 and n=4 cases, e.g. here.

I would be grateful if you let me know of the current status or if it is already a theorem.

P.S.: This same question was asked on Math.SE but it has now drowned under the fold and I thought I will have better chances of getting answers here.

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It would follow from Conjecture 1.2 in the paper afjarvis.staff.shef.ac.uk/maths/… - from what I know it's an open problem even for Gaussian integers, but maybe I am unaware of some recent developments, so let's wait for what number theorists have to say! –  Vladimir Dotsenko Mar 12 '12 at 9:55
@Vladimir, thank you for bringing up the interesting paper. –  TZE Mar 12 '12 at 10:09
If only Fermat had offered a MO bounty, think how much faster his question would have been settled! –  Tom Leinster Mar 14 '12 at 16:58
(Seriously: nice question, +1.) –  Tom Leinster Mar 14 '12 at 16:58
Tom, I think you meant "had been offered" rather than "had offered" (Fermat had no reason to offer a bounty for a question he believed he had solved himself but giving him an incentive to write a full solution and to present it publicly might have changed the history a bit, indeed). –  fedja Mar 16 '12 at 0:04
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up vote 38 down vote accepted

This is still way open, I should think. "Elementary" methods won't even solve the analogous problem over $\mathbf{Z}$, so you need to use "modular form" methods. The problem is that even if the result were to follow from a Frey curve argument and a potential theorem of the form "all sufficiently nice Galois representations come from automorphic forms", we're a long way from establishing such a theorem.

A few more details: a key problem is that there are two natural candidates for where the automorphic forms will come from, and neither is good enough. The first is the group $GL(2)/\mathbf{Q}(i)$. This is a very natural place to look, but the problem is that this group does not admit Shimura varieties, so it's very hard to even go the "easy" way and to attach a Galois representation to an algebraic automorphic representation (i.e. do what Deligne did), let alone to do what Wiles did. There are some theorems of this nature, but they all have a self-duality hypothesis built into them, which will not hold for the Tate module of the Frey curve in general.

The second place to look is rank 2 unitary groups for the extension $\mathbf{Q}(i)/\mathbf{Q}$. These do have Shimura varieties and their geometry is understood fairly well nowadays, and there are are very strong theorems attaching Galois representations to these automorphic forms (there are even very strong theorems for rank $n$ unitary groups nowadays -- see for example the book by Harris and Taylor, but things have moved on even further since then, e.g. because of recent work of Sug-Woo Shin). However we then run into the same problem -- the unitary groups have some extra symmetry and this means that the associated Galois representations have some extra symmetry (they need to be essentially conjugate self-dual), and this extra symmetry will not in general be true for the Galois representations attached to the Tate module of the Frey curve.

So one needs a good new idea before we can push forward what one might call "Wiles' strategy" in this situation.

This is in marked contrast to the case of totally real fields, where a lot of the machinery works fine and it would not surprise me nowadays if FLT could be proved for several totally real fields. As has been implicitly mentioned in the comments, Jarvis and Meekin did this for $\mathbf{Q}(\sqrt{2})$, and this was a few years ago now, and modularity lifting theorems have moved on tremendously since then, so it would not surprise me if the experts could prove these results for other totally real number fields now. However somehow, after the Jarvis-Meekin work, which is a proof that the machine can be made to work in other cases, perhaps the interesting question now is not something like "is FLT true for $\mathbf{Q}(\sqrt{5})$?" (which one could perhaps hope to answer, perhaps with a lot of work, but with basically existing methods and a lot of hard graft to deal with the small $n$ cases) but more like "is some slightly weakened version of FLT true for all totally real fields?" or some such thing. You need to weaken it a bit because the case $n=3$ is an elliptic curve and it has positive rank for lots of totally real $F$, so now you need to deal with $n=4$ and $n=9$ and $n=6$ by hand, or just declare that you're only interested in $x^p+y^p=z^p$ with $p\geq5$ prime; and then you'll sometimes get reducible mod $p$ Galois representations -- so perhaps the correct weakening is something like "$x^p+y^p=z^p$ has no solutions for $p$ sufficiently large (depending on $F$)". Even then there may be problems in "case 2".

My impression is that the machinery being developed now is not really being developed with generalisations of FLT in mind, but it's generalising this "Wiles machine" in different directions, for example to prove things like the Fontaine-Mazur conjecture and the Sato-Tate conjecture. This is the direction deemed "trendy" -- and in a sense I can see why, because is FLT over a specific number field other than the rationals really a "natural" question? And FLT over a general number field is obviously false, so now you have to weaken things etc etc. On the other hand hard conjectures like Sato-Tate do look to me like natural questions.

IMPORTANT EDIT: I wrote this answer a long time ago -- what is it, 9 days now? -- but life moves on, and I hear from my spies in Toronto that Richard Taylor yesterday announced some results joint with Harris, Lan and Thorne, where they claim that they can attach Galois representations to (not necessarily self-dual) cohomological cuspidal automorphic representations on $GL(n)$ over totally real and CM fields. In particular apparently the theory for $GL(2)/\mathbf{Q}(i)$ is now up to about the state that the theory for $GL(2)/\mathbf{Q}$ was in the late 60s. So give it another 23 or so years and we should have FLT for $\mathbf{Q}(i)$!

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Great answer, Kevin ! –  Joël Mar 14 '12 at 13:36
@Kevin Buzzard, a marvellous answer! –  TZE Mar 14 '12 at 15:20
@Kevin Buzzard, thank you for the news! –  TZE Mar 23 '12 at 11:06
Incredible news! Is there any more information available? –  Joël Mar 26 '12 at 18:17
Yes, incredible. Taylor was reluctant to talk about the result because in general he only speaks about things when there is a manuscript available, but one of the organizers persuaded him that a big conference on the cohomology of Shimura varieties was a very natural place to announce an important result about the cohomology of Shimura varieties...so I guess now we wait for a while... –  Kevin Buzzard Mar 26 '12 at 20:19
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I suppose this is an open problem.

Exponents 5 and 7 might follow from these papers:

Algebraic points of low degree on the Fermat quintic describes all algebraic points of degree at most 2 for exponent 5.

The non-trivial points are $(\eta,\bar{\eta},1)$ and $(\bar{\eta},\eta,1)$ where $\eta$ is primitive 6th root of unity.

For exponent 7 Algebraic points on some Fermat curves and some quotients of Fermat curves: Progress gives all non-trivial points of degree at most 3: $(\eta,\bar{\eta},-1)$ and $(\bar{\eta},\eta,-1)$.

Regarding Vladimir Dotsenko's comment that this will folow from Conjecture 1.2

Conjecture 1.2 (Debarre-Klassen) Let $K$ be a number field of degree $d$ over Q. Then the equation $x^n + y^n = z^n$ has only trivial solutions over K when $n \geq d + 2$. Here, Debarre and Klassen define trivial solutions to mean points $(a, b, c)$ on $x^n + y^ n = z^ n$ where $a + b = c$.

If one considers $x+y=z$ a valid solution the conjecture won't apply and in addition this appears a counterexample to the conjecture for $n=4$ $$ (\sqrt{-7} + 3)^4 + 4^4 = (\sqrt{-7} - 1)^4 $$

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Thanks for pointing that out. I meant to mention something about that but the comment field was too small! (Yes I know I am not Fermat to get away with it.) –  Vladimir Dotsenko Mar 14 '12 at 12:53
@Vladimir saw similar version of the conjecture, but it has $n$ odd prime. –  joro Mar 14 '12 at 13:09
@Joro, thank you very much. –  TZE Mar 14 '12 at 15:24
There's something slightly wrong here, at the time of writing this comment, because $1+0\not=-1$ and yet $1^{100}+0^{100}=(-1)^{100}$. –  Kevin Buzzard Mar 14 '12 at 19:47
@Kevin indeed. Btw, the sqrt(-7) example is not of the form $(x,y,-x-y)$ it is $(x,y,x-y)$ –  joro Mar 15 '12 at 7:37
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