In Report on the Theory of Numbers, H.J.S. Smith writes:

"The impossibility of solving [Fermat's] equation has been demonstrated by M. Kummer, first, for all values of $\lambda$ not included among the exceptional primes; and secondly, for all exceptional primes which satisfy the three following conditions:

  1. That the first factor of H, though divisible by $\lambda$, is not divisible by $\lambda^2$.
  2. That a complex modulus can be assigned, for which a certain definite complex unit is not congruous to a perfect $\lambda$-th power.
  3. That $B_{\kappa \lambda}$ is not divisible by $\lambda^3$, $B_{\kappa}$ representing that Bernoullian number $[\kappa \leq \mu-1]$ which is divisible by $\lambda$.

Three numbers below 100, viz. 37, 59, 67, are, as we have seen, exceptional primes. But it has been ascertained by M. Kummer that the three conditions just given are satisfied in the case of each of these three numbers; so that the impossibility of Fermat's equation has been demonstrated for all values of the exponent up to 100. Indeed, it would probably be difficult to find an exceptional prime not satisfying the three conditions, and consequently excluded from M. Kummer's demonstration."

Can anyone cite an exceptional prime that does not satisfy the three conditions?

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    $\begingroup$ I cleaned up the formatting of your question a little. Any idea where one can get a bit more detail on precisely what Kummer's conditions were? (In particular, what is $H$, and what was the "definite complex unit"?) $\endgroup$ May 24, 2011 at 22:09
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    $\begingroup$ Maybe one of the difficulties is that most of the notation isn't defined ;-) $\endgroup$ May 24, 2011 at 22:19
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    $\begingroup$ H seems to be the class number, property 2 some form of generalization of Kummer's lemma (which in the regular case states that if a unit is congruent to an integer modulo $\lambda$, then it is a $\lambda$-th power). If I remember correctly, Kummer's proofs in the irregular case contained quite a few gaps closed much later by Vandiver, so be sure to check out Edwards' and Ribenboim's books on Fermat's last theorem. $\endgroup$ May 25, 2011 at 9:58

1 Answer 1


Washington's book has a table of irregular primes together with the power of p dividing the first factor of the class number. There are plenty for which the first factor is divisible by $p^2$; the first is p=157. So for these Kummer's first condition fails.


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