If $p,q$ are prime, $q$ odd $p \not\equiv 1 \pmod q$, is there an integer $x$ such that $p\mid 1+x+\ldots+x^{q-1}$

Question

Suppose $p,q$ are two distinct prime numbers, $q \geq 3$ and $p \not\equiv 1 \pmod q$. Then I have the following problem: Prove that there is no integer $x \in \mathbb{Z}$ such that $1+x+x^2+...+x^{q-1} \equiv 0 \pmod p$.

It is obvious that $x$ cannot be $0 \pmod p$, and I also found that when $p$ is even, i.e. $p=2$, that this isn't too hard. However, for the rest I only found that $x^q-1 = (1+x+...+x^{q-1})(x-1) \equiv 0 \pmod p$. Where do I go next?

Thanks in advance

Answer 1

Doing arithmetic modulo $\;p\;$ all the time:

$$0=1+x+x^2+\ldots+x^{q-1}=\frac{x^q-1}{x-1}\implies x^q=1$$

But we also have $\;x^{p-1}=1\;$ since clearly $\;x\neq 0\;$ .

Get now your contradiction.

Answer 2

You are looking for an element $x$ of the finite field with $p$ elements such that $x^q=1$ and $x\neq 1$. But for any finite field, the multiplicative group is cyclic, so $q$ (as prime) must be a divisor of $p-1$, so $p=1 (\mod q)$