In one of his letters to Frenicle, Fermat stated the proposition that no prime of the form $q^2+2$ can divide any number of the form $x^2-2$.

Is there a known proof of this statement? If not, how would one go about proving it?

Many thanks!

  • 4
    $\begingroup$ This is discussed in Weil's book on the number theory of fermat and euler. $\endgroup$ – Franz Lemmermeyer Dec 9 '11 at 15:30

Following up on Noam's comment, here are two possibilities.

  1. (Weil) If $p = a^2 + 2$ divides $x^2-2$, then $p \mid (a^2+2) + (x^2-2) = a^2 + x^2$. Since $p = 4n-1$, this implies $p \mid a$ and $p \mid b$ by a result known to Fermat (all odd prime divisors of a primitive sum of two squares have the form $4n+1$). But this is impossible since $p > a$.

  2. Assume that $p = a^2 + 2$ divides $x^2 - 2$; then $x^2 \equiv 2 \bmod p$. Using Fermat's Theorem this implies $1 \equiv x^{p-1} \equiv 2^{(p-1)/2} \mod p$, and similarly $a^2 \equiv -2 \bmod p$ implies $1 \equiv a^{p-1} \equiv (-2)^{(p-1)/2} \bmod p$. Thus $(-1)^{(p-1)/2} \equiv 1 \bmod p$, which contradicts the fact that $p = a^2 +2 \equiv 3 \bmod 4$.

Of course Fermat did not have congruences, but the statements above are easily translated into simple divisibility results. Both proofs have the advantage of only using the first supplementary law, which was accessible to Fermat early on.

  • $\begingroup$ Lovely! Thanks for the illumination. $\endgroup$ – Kieren MacMillan Dec 10 '11 at 14:27

It's not true: 2 is of the form $q^2+2$, and it divides $2^2-2 = 2$.

Assuming you mean an odd prime, it's an easy exercise in quadratic reciprocity: if $p$ divides $x^2 - 2$, then $x^2 \equiv 2 \pmod{p}$, so (by quadratic reciprocity) $p \equiv \pm 1 \pmod{8}$. But, if $p = q^2+2$, then $q$ must be odd, in which case $p \equiv q^2+2 \equiv 3 \pmod{8}$, a contradiction.

I have no idea what Fermat's proof would have been.

  • 7
    $\begingroup$ PS This probably isn't a suitable level for MO, but it's Friday evening... $\endgroup$ – Martin Bright Dec 9 '11 at 15:18
  • 9
    $\begingroup$ The question of how Fermat might have proved such a result might still be suitable for MO. $\endgroup$ – Noam D. Elkies Dec 9 '11 at 17:08
  • $\begingroup$ Thank you; that makes sense. I concur that it would be interesting to see how Fermat might have proved it: He certainly seemed to have QR-related techniques that pre-dated QR. $\endgroup$ – Kieren MacMillan Dec 9 '11 at 17:31

Not an answer to the question, but to Noam's comment. A survey of the subject is given in:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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