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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    $\begingroup$ 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". $\endgroup$
    – KConrad
    Feb 20, 2011 at 21:41
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    $\begingroup$ 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. $\endgroup$ Feb 20, 2011 at 23:25
  • $\begingroup$ @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 $\mathbb{Z}[\zeta_p]$)? $\endgroup$ Feb 21, 2011 at 0:39
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    $\begingroup$ 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. $\endgroup$ Jul 1, 2011 at 15:36

2 Answers 2


This answer is a bit late; sorry for that.

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).

By the way, this criterion was also proven by Taro Morishima in 1935 (Japan. J. Math. 11, 241-252, Satz 1; but warning: Satz 2 or at least its proof is incorrect since it is based on some incorrect result of Vandiver).

I don’t know how to find such a solution $(x,y,z)$.

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    $\begingroup$ Welcome to MathOverflow, Professor Metsankyla! $\endgroup$ Feb 23, 2011 at 23:31
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    $\begingroup$ 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. $\endgroup$ Feb 24, 2011 at 8:21
  • $\begingroup$ If I read Kummer's article correctly, then he proves FLT for irregular primes not only in rational integes, but in the maximal real subfield of the p-th roots of unity. $\endgroup$ Apr 17, 2017 at 11:53

Very late response but since it is still unresolved, I will answer your question. By Tauno's answer, any solution must belong to the second case, which in $\mathbb{Z}[\zeta]$ looks like $x^p+y^p=z^p$ with $1-\zeta \mid z$ and $x,y$ both coprime with $1-\zeta$. The second case also has no solutions in this ring by another criteria of Kolyvagin (from a different paper: " Fermat Equations over Cyclotomic Fields").
The general criteria is a bit involved to write up here but the prime $p=37$ satisfies a simpler criteria (which applies to both the first and second case): 1) If the index of irregularity $=1$ with $p \mid B_i$ and 2) there is a prime $l \equiv 1 (\text{mod} \ p)$ for which $x^p+y^p=z^p$ has only trivial solutions modulo $l$ and $(1-\zeta(l))^{(l-3)}$, $(1-\zeta(l))^{(i)}$ are not $p$-th powers modulo $l$, then the Fermat equation has no solutions in $\mathbb{Q}(\zeta_p)$. Here $\zeta(l)$ is a primitive $p$-th root of unity modulo $l$ (which exists since $p \mid (l-1)$) and $a^{(j)}$ is the $j$-th component of $a \in \mathbb{F}_l^{\times}/(\mathbb{F}_{l}^{\times})^p$ relative to the Eigenspace decomposition of this group for the operators $\varepsilon_j = -\sum_{a=1}^{p-1}a^j\sigma_{a}^{-1}$.
Since the index of irregularity for $37$ is $1$ with $37 \mid B_{32}$ and $x^{37}+y^{37}=z^{37}$ has only trivial solutions modulo $l=149=1+4\times 37$, then by taking $\zeta(149)=16$, we compute that $(1-16)^{(34)}$ and $(1-16)^{(32)}$ are not $37$-th powers modulo $149$. Therefore $p=37$ satisfies the criteria.


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