Let $p$ be an odd prime. Prove that if $p \not \mid r$ and $r \not \equiv 1 \pmod p$ then
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv 0 \pmod p $$
If $r \equiv 1 \pmod p$, then determine the least non-negative residue $x$ in
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv x \pmod p $$
We use the formula for the sum of a geometric series
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv \frac{1-r^{p-1}}{1-r} \pmod p $$
Using Fermat's Little Theorem, $r^{p-1} \equiv 1 \pmod p$, and noting that the denominator $1 - r \not \equiv 0 \pmod p$,
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv \frac{1-1}{1-r} \equiv 0 \pmod p $$
This is the desired conclusion.
If $r \equiv 1 \pmod p$, then
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv \overbrace{1 + 1 + 1 \ldots 1}^{(p-1)} \equiv p-1 \pmod p $$
So the least non-negative residue is $p-1$.