A primitive root modulo p is a primitive root modulo $p^2$ if and only if $g^{p-1} \not\equiv 1 \mod{p^2}$ $p$ is an odd prime. I'm starting with number theory and I'm completly stuck with this question. In general, I don't really know how to approach the proves. Then I'm also supposed to prove that either $g$ or $g+p$ is a primitive root modulo $p^2$ using that first result. I also know about quadratic residues and I think the problem is designed to use them as well.
EDIT: I realised one of the implications is straight-forward, as if  $g^{p-1} \equiv 1 \mod{p^2}$, then the order of $g$ would be, at most, $p-1$, so it can't be a primitve root modulo $p^2$.
 A: $(\Rightarrow)$ As you said in the edit, if $g^{p-1} \equiv 1 \mod p^2$ then the order of $g$ in $\mod p^2$ is at most $p-1$. But, $\phi(p^2)=p(p-1)>p-1$. Hence, $g$ can't be a primitive root. Contradiction.
$(\Leftarrow)$ We have that $g^{p-1} \not \equiv 1 \mod p^2$. We denote $ord(g)$ the order of $g$ $\mod p^2$.
We know that $g^{ord(g)} \equiv 1 \mod p^2 \Rightarrow p^2|g^{ord(g)} - 1 \Rightarrow p|g^{ord(g)}-1 \Rightarrow g^{ord(g)} \equiv 1 \mod p$. Hence, $p-1|ord(g)$ because $g$ primitive root $\mod p$. Also, we know that $ord(g)|\phi(p^2) = p(p-1)$.
Since $p$ is prime, either $ord(g)|p-1$ or $ord(g)=p(p-1)$ (because $ord(g)$ won't divide neither $p$ nor $p-1$). If $ord(g)|p-1$ then, because $p-1|ord(g)$ we have that $ord(g)=p-1$, a contradiction as $g^{p-1} \not \equiv 1 \mod p^2$.
Therefore, $ord(g)=p(p-1)=\phi(p^2)$ and thus $g$ is a primitive root $\mod p^2$
For what you are finally supposed to show, you now know that if $g$ is a primitive root $\mod p^2$, $g^{p-1} \not \equiv 1 \mod p^2$ and $\textbf{vice versa}$. Now try to see what you can do if $g^{p-1} \equiv 1 \mod p^2$. What is $ord(g)$ now? Afterwards, as a hint, you can use the binomial theorem for $(g+p)^{p-1} \mod p^2$. What you observe now? Have we proved something similar for the value you found?
