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The first irregular prime is 37. Does FLT(37)

$$x^{37} + y^{37} = z^{37}$$

have any solutions in the ring of integers of $\mathbb Q(\zeta_{37})$, where $\zeta_{37}$ is a primitive 37th root of unity?

Maybe it's not true, but how could I go about finding a counter-example? (for any cyclotomic ring, not necessarily 37)

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    $\begingroup$ The question of for which $n \in \mathbb{Z}^+$ the analogue of FLT holds in the cyclotomic ring $\mathbb{Z}[\zeta_n]$ is surely unsolved in general: this is really hard! Just as surely there must be some work on special cases (possibly including $n = 37$; my blind guess is that's close to the boundary of what can be done with ad hoc techniques). You might want to ask this question on Math Overflow instead, the better to get a sense of exactly how much the experts know. $\endgroup$
    – Pete L. Clark
    Feb 13, 2011 at 21:23
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    $\begingroup$ Prior to Wiles' proof of FLT, one statement that I often heard was that it was known that for each $n > 2$, there can only be finitely many coprime solutions to FLT(n). This is actually due to Falting's theorem, which says that there are only finitely many rational points on an algebraic curve of genus greater than 1. This statement holds in any number field. So, we do know that there can only be finitey many solutions (up to rescaling) to FLT(37) in any number field and, in particular, in $\mathbb{Q}(\zeta_{37})$. Unfortunately, Falting's theorem is not effective, meaning that it does not... $\endgroup$
    – George Lowther
    Feb 13, 2011 at 21:32
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    $\begingroup$ ...provide you with a bound on the largest possible solution so, as far as I know, it doesn't reduce the problem to a finite search. $\endgroup$
    – George Lowther
    Feb 13, 2011 at 21:33
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    $\begingroup$ There's no solutions to the first case of FLT(37) in $\mathbb{Q}(\zeta_{37})$, because $2^{37}=76\not=2$ mod $37^2$. See here: dx.doi.org/10.1070/IM1999v063n05ABEH000262 $\endgroup$
    – George Lowther
    Feb 13, 2011 at 21:55
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    $\begingroup$ I have posted an updated version of this question to mathoverflow. $\endgroup$
    – quanta
    Feb 20, 2011 at 21:13

1 Answer 1

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This question was answered in mathoverflow. I am writing this to close up this question and making this answer as a community wiki according to MSE's guidelines. The answer is due to Tauno Metsänkylä.

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