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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $x^2-2 = m*(a^2+2)$ 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
  • 1
    $\begingroup$ 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

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$($$2$ is a quadratic residue modulo $p$ if and only if $p\equiv ±1\pmod 8$$)$.


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.