Prime divisors of M admitting a square root of -1

Let $M$ be a natural number and let $x$ be an element of $\mathbb{Z}/M\mathbb{Z}$ with $x^2 + 1 = 0$. Let $\ell$ be a prime number dividing $M$. Then $\ell = 2$ or $\ell \equiv 1 \pmod 4$, i.e. $\ell \bmod 4 = 1$ as natural numbers. Equivalently: if the congruence $x^2 \equiv -1 \pmod M$ is solvable, then no prime divisor of $M$ is congruent to $3$ modulo $4$. Note that the case $M = 0$ is permitted by the statement, where $\mathbb{Z}/M\mathbb{Z}$ is $\mathbb{Z}$ and every prime divides $M$; in that case the hypothesis $x^2 + 1 = 0$ is vacuously unsatisfiable, so the assertion holds. The conclusion is a disjunction of natural-number statements about $\ell$ alone; nothing is asserted about the residue of $M$ itself (indeed $4 \mid M$ is excluded only indirectly, via the absence of a prime $\ell \equiv 3 \pmod 4$ being insufficient, so the statement as given genuinely allows $\ell = 2$).

This is the first supplement to quadratic reciprocity in the form used for the solvability of $x^2 \equiv -1 \pmod M$: $-1$ is a square modulo an odd prime exactly when that prime is $1$ modulo $4$. It feeds the derivation that no square root of $-1$ exists modulo an odd $M$ having a prime divisor congruent to $3$ modulo $4$, recorded in ZMod.not_exists_sq_add_one_eq_zero_of_not_two_dvd_of_exists_prime_dvd_mod_four_ne_one.