I know that every group of units $\bmod p$ has a generator, in fact $\varphi(p-1)$ of them.
I came across a problem that asked to prove that for such a generator, let's call it $a$ (but see below), and $p$ an odd prime:
$a^{p-1} = 1 + kp$
where $\gcd(k,p) = 1$
It's the last part that is killing me. I can see that $p$ divides $a^{(p-1)/2} + 1$, since it cannot divide $a^{(p-1)/2} - 1$ (this would contradict that $a$ is a generator), but I have no idea how to show that $p^2$ does not divide $a^{(p-1)/2} + 1$.
Any ideas?
EDIT: The answer turned out to be simple (I thought it might be). If it turns out that k and p are not co-prime, replace a with a+p. My apologies for stating the problem incorrectly, as the original problem asked to find an a such that [a] was a generator.