Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Sophie Germain proved Fermat's Last Theorem $x^p+y^p \neq z^p$ for the special case where p is a Sophie Germain prime and $p\not|xyz$. Does any one know of a proof for the other case, where $p|xyz$? Note: I am looking for a proof restricted to the Sophie Germain primes, as of course, Wiles proved this generally.

share|cite|improve this question
This is a natural question to ask which does not seem to have an answer. – awllower Feb 18 '11 at 14:47
up vote 11 down vote accepted

These two cases are traditionally called ${\bf Case \,\, 1}$ and ${\bf Case \,\, 2}$, and you are after a proof of Case 2.

The essence of Dirichlet's proof of Case 2 when $p=5$ can be found in Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory, by Harold M. Edwards. It's in chapter three, at about page 70.

share|cite|improve this answer
Thank you , Dr Jennings. Do you know of a proof for Case 2 for all Sophie Germain primes? – James Feb 18 '11 at 10:43
@James: There's always Wiles' proof but, of course, that encompasses more than Sophie Germain primes. Sorry, I don't know of a proof restricted only to Sophie Germain primes in Case 2 but I'm fairly sure that the answer to your question is that one does not exist. – Derek Jennings Feb 18 '11 at 13:04
Hmm, I was afraid of that. Thank you anyway, I will check out the Dirichlet proof. – James Feb 18 '11 at 14:21

protected by t.b. Apr 17 '12 at 21:15

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.