Why does this polynomial equation have exactly $q-1$ solutions modulo $p$?

Question

suppose $p$ is a prime number and $q\mid p-1$. prove the congruence $1+x+...+x^{q-1}$ is congruent to $0$ mod $p$ has exactly $q-1$ solutions. I'm mainly confused on how to prove for exactly $q-1$ solutions because proving there are at most $q-1$ solutions is the proof of lagranges theorem on quadratic residues.

Answer 1

Since $x^q-1=(x-1)(x^{q-1}+\dots+x+1)$, solutions of $x^{q-1}+\dots+x+1 \equiv 0 \pmod p$ are precisely elements of $\mathbb{Z}/p$ of multiplicative order dividing $q$ and different from $1$. Also, since $(\mathbb{Z}/p)^{\times}$ is the cyclic group of order $p-1$, there are exactly $q$ elements of orders dividing $q$ (including $1$).