# Fermat's Last Theorem in the cyclotomic integers.

Kummer proved that there are no non-trivial solutions to the Fermat equation FLT(n): $x^n + y^n = z^n$ with $n > 2$ natural and $x,y,z$ elements of a regular cyclotomic ring of integers $K$.

I am looking for non-trivial solutions to the Fermat equation FLT(p) in the cyclotomic integer ring $\mathbb{Z}[\zeta_{p}]$ for irregular primes p or any information about how the solutions must be (as a step toward constructing them).

George Lowther pointed out in an earlier discussion that by Kolyvagin's criterion any solution in $\mathbb{Z}[\zeta_{37}]$ must be in the second case.

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Kummer's proof apparently had a gap: he "reduced" to the case when a hypothetical solution (x,y,z) in a regular cyclotomic ring of integers was pairwise relatively prime, but you can't reduce to that case if the ring has class number greater than 1. The result was proved by Hilbert. See Chapter 11 of Grosswald's "Topics from the Theory of Numbers" or section V.3 of Ribenboim's "13 Lectures on Fermat's Last Theorem". –  KConrad Feb 20 '11 at 21:41
Actually, there is a solution to $x^5+y^5=z^5$ in $\mathbb{Z}[\zeta_3]$. Consider $\zeta_3^5+(\zeta_3^2)^5=(-1)^5$. I think "a ring of cyclotomic integers" should be replaced by "$\mathbb{Z}[\zeta_n]$" in the question. –  George Lowther Feb 20 '11 at 23:25
Thank you George, that is what I meant to ask! –  Quanta Feb 21 '11 at 0:08
@Quanta: I made some minor edits. I was also thinking of making more significant edits to the first sentence, but don't quite understand your intention. What $n$ did Kummer prove this for? Shouldn't it say "with n > 2 a regular prime" (maybe replace n by p)? And does "a regular ring of integers K" mean "the cyclotomic integers $\mathbb{Z}[\zeta_p]$)? –  George Lowther Feb 21 '11 at 0:39
The solution with cube roots of unity noted by G.Lowther works for any exponent that is not a multiple of 3. Also noteworthy, albeit not directly relevant to the specific question at hand, is the solution $(1 + \sqrt{-7}, 1 - \sqrt{-7}, 2)$ of $x^4+y^4=z^4$. While ${\bf Q}(\sqrt{-7})$ is contained in a cyclotomic extension of ${\bf Q}$ (this is true of all quadratic number fields), the exponent $4$ is not prime. –  Noam D. Elkies Jul 1 '11 at 15:36

Kummer's proof of the nonsolvability of $x^p + y^p = z^p$ for regular primes $p$ used “ideal numbers” (in present-day language: ideals) and was intact, at least basically. Hilbert in his Zahlbericht gave a modified proof. Both proofs cover not only rational integers but also numbers in $\mathbb{Z}[\zeta_p]$. On the other hand, Kummer’s second result concerning irregular primes that satisfy certain additional conditions covers just the rational integers (although Hilbert, in the very last section of Zahlbericht, erroneously says that Kummer had proven this result for $\mathbb{Z}[\zeta_p]$ as well). Thus one cannot exclude the possibility that there is indeed a solution $(x,y,z)$ for $p=37$. And because of "Kolyvagin's criterion" about $(2^{37}-2)/37$, this solution must belong to the second case, that is, at least one of these three numbers $x,y,z$ in $\mathbb{Z}[\zeta_{37}]$ must have a common factor with $37$ (as mentioned by George Lowther).
I don’t know how to find such a solution $(x,y,z)$.
Thank you, Gerry. I have still to learn how to operate here. An addition to my answer: in the possible solution $(x,y,z)$ all the numbers $x,y,z$ cannot be real. This was proved by K. Inkeri (my teacher, by the way) in 1949. –  Tauno Metsänkylä Feb 24 '11 at 8:21