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In his letter to Frenicle, dated 18th October, 1640, Fermat states the following (Point 8, translated) :

If you subtract $2$ from a square, the remaining value cannot be divided by a prime which is greater than a square by $2$

For example, take for a square $1,000,000$, from which, subtracted by two, remains $999,998$. I say that the given remainder cannot be divided by $11$ or by $83$, by $227$, and so on.

You can prove the same rule for odd squares and, if I wanted, I would give you the lovely and general rule; but I'm content with having only indicated it to you.

In other words, numbers of the form $x^2 -2$ are not divisible by primes of the form $a^2 + 2$, where $x$ and $a$ are integers.

Questions :

$1)$ What is the general rule Fermat is talking about?

$2)$ Are there any modern references to this problem?

$3)$ How would Fermat have proved it?

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    $\begingroup$ I believe he considered 3 cases : 1) $a^2+2$ is prime, 2) $a^2+2$ is odd, 3) $a^2+2$ is any integer (general case) $\endgroup$ – rsadhvika Dec 7 '18 at 4:19
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    $\begingroup$ $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule $\endgroup$ – rsadhvika Dec 7 '18 at 4:21
  • $\begingroup$ Well, $a^2+2$ primes are always $2$ or $4k+3$ primes (the converse is of course not true). Not sure how that helps though $\endgroup$ – YiFan Dec 7 '18 at 6:04
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    $\begingroup$ Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$? $\endgroup$ – tyobrien Dec 7 '18 at 6:20
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Ad 3): If $p = a^2 + 2$ divides $x^2 - 2$, then $p$ divides $x^2 - 2 + p = x^2 + a^2$. But primes $p = 4n+3$ cannot divide sums of two squares without dividing the squares themselves. This observation is due to Weil.

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Let $p=a^2+2$ be a prime. Suppose $a>0$ as for $a=0$, the statement clearly does not hold. Therefore, $p\equiv 3 \pmod 8$. Thus, the congruence $x^2\equiv 2\pmod p$ has no solutions$($$2$ is a quadratic residue modulo $p$ if and only if $p\equiv ±1\pmod 8$$)$.

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