Prime divisors of M when x²+x+1≡ 0 (mod M)

Let $M$ be a natural number and let $x$ be an element of $\mathbb{Z}/M\mathbb{Z}$ satisfying $x^2 + x + 1 = 0$. Let $\ell$ be a prime number dividing $M$. Then either $\ell = 3$ or $\ell \equiv 1 \pmod 3$, in the sense that the natural-number remainder $\ell \bmod 3$ equals $1$. Note that $M$ is an arbitrary natural number, so the degenerate cases $M = 0$ and $M = 1$ are included; the divisibility hypothesis $\ell \mid M$ is what supplies the reduction map $\mathbb{Z}/M\mathbb{Z} \to \mathbb{Z}/\ell\mathbb{Z}$, and the conclusion is a statement about the residue of $\ell$ modulo $3$ only.

This is the elementary determination of the primes modulo which the cyclotomic polynomial $X^2 + X + 1$ has a root: the splitting primes are exactly $3$ and those congruent to $1$ modulo $3$. It is used to rule out roots of $X^2+X+1$ modulo $M$ when $M$ has a prime divisor that is neither $3$ nor $1$ mod $3$, as recorded by ZMod.not_exists_sq_add_self_add_one_eq_zero_of_not_three_dvd_of_exists_prime_dvd_mod_three_ne_one.