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 start by noting the geometric series formula
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} = \frac{1-r^{p-1}}{1-r} $$
We want to avoid division, so we start by multiplying the series by $1-r \pmod p$, and noting Fermat's Little Theorem $r^{p-1} \equiv 1 \pmod p$ is applicable,
$$ (1 + r + r^2 + \ldots + r^{p-3} + r^{p-2})(1-r) \equiv 1 - r^{p-1} \equiv 0 \pmod p $$
Since we know $1-r \not \equiv 0 \pmod p$, it must be that the series is equivalent to $0$ modulo $p$, that is
$$ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv 0 \pmod p $$
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 $p-1$.