The general proposition of Fermat

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?

• 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) – rsadhvika Dec 7 '18 at 4:19
• $x^2-2 = m*(a^2+2)$ has no solutions over positive integers, general case/rule – rsadhvika Dec 7 '18 at 4:21
• 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 – YiFan Dec 7 '18 at 6:04
• Could the general rule be that numbers of the form $x^2-n$ are not divisible by primes of the form $a^2+n$? – tyobrien Dec 7 '18 at 6:20

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.
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 $$p\equiv ±1\pmod 8)$$.