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$s^2 + 1$ is the squared length of a vector in a square lattice for every integer $s$. The number of integer solutions in $a, b$ for $a^2 + b^2 = s^2 + 1$ is of course $> 0$, e.g. $a = s, b = 1$ is a solution. A lemma to Jacobis theorem about the number of solutions of $a^2 + b^2 = n$ with $n \equiv 1 \mod 4$ states, that in the factorization of $n$ every prime factor with $p \equiv 3 \mod 4$ of $n = s^2 + 1$ has to appear with an even exponent. A search up to $s = 10^6$ shows: no prime factors $p \equiv 3 \mod 4$. Is this true for all $s$ and why?

Bill Dubuque
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Helmut Ruhland
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You are indeed correct that $s^2+1$ cannot have a prime factor $p\equiv 3\pmod 4$. Suppose the contrary, that $s^2+1$ does have such a prime factor $p=4k+3$. Then $$s^2+1\equiv 0\pmod p.$$ However this means that $-1\in QR_p$ (i.e. $-1$ a quadratic residue modulo $p$) and this is easily seen to be false. For example, using quadratic reciprocity we know that $$\left(\frac{-1}{p}\right)=-1.$$

timmeh
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    That is not usually called quadratic reciprocity, but is often called one of the "auxiliary laws" coming with quadratic reciprocity. It is also much easier to prove than QR itself: if there existed $x \in \mathbb F_p$ with $x^2=-1$, then $x$ would have order $4$ in the group $\mathbb F_{p}^{\times}$, but this group has order $p-1$, which is not divisible by $4$, contradiction. – R.P. Nov 29 '24 at 15:38
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    It can be called a supplement to quadratic reciprocity – J. W. Tanner Nov 29 '24 at 15:52
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Nov 29 '24 at 22:33
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The given statement is equivalent to proving that all odd primes dividing $s^2+1$ are $1$ mod $4$. Now, consider an odd prime $p\mid s^2+1$. We have $s^2\equiv -1\pmod p$, which means $s^4\equiv 1\pmod p$, so $\mathrm{ord}_p(s)=4$ (where $\mathrm{ord}$ refers to multiplicative order). But since $s^{p-1}\equiv 1\pmod p$ by Fermat's Little Theorem, we must have $4=\mathrm{ord}_p(s)\mid p-1$, or $p\equiv 1\pmod 4$.

We can actually prove a stronger statement:

Claim: If a prime $p\equiv 3\pmod 4$ satisfies $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$.

Proof. Assume not for the sake of contradiction. WLOG let $p\nmid b$. This means $b$ is invertible, so we can rewrite $a^2+b^2\equiv 0\pmod p$ as $(ab^{-1})^2+1\equiv 0\pmod p$. The same proof follows as before with $s=ab^{-1}$.

Note: This is actually a weaker converse of Fermat's Christmas Theorem.

masky
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