prime factors of x^2 - 2 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!
 A: Not an answer to the question, but to Noam's comment. A survey of the subject is given in:
http://seanelvidge.com/downloads/HistoryQR.pdf
A: Following up on Noam's comment, here are two possibilities.


*

*(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 $p > a$.

*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. 
A: 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.
