Let $p$ be a prime number. I'd like to prove that there are primitive roots modulo $p^2$. Could someone check this argument?
Note that if $r\in\mathbb Z$ is a primitive root modulo $p^2$, it must be a primitive root modulo $p$ as well, since it must have powers congruent to $1, 2, 3, ... p$ modulo $p^2$ and therefore those powers will also be congruent to $1, 2, 3, ... p$ modulo $p$.
So, we restrict our search to primitive roots modulo $p$. Let $r$ be such a root, and let's look for one of order $\phi(p^2)=p(p-1)$. Let $n$ be the order of $r$ modulo $p^2$. $n$ must divide $\phi(p^2)=p(p-1)$. But it must also be a multiple of $\phi(p)=p-1$, since $r^n\equiv 1$ modulo $p^2$ and therefore also modulo $p$. This implies that $n/(p-1)$ divides $p$ and thus either $n=p(p-1)$ or $n=(p-1)$.
So we just need to find out if all values of $r$ are of order $p-1$. Well, if $n$ is the order of $r$, then $r+p$ is another primitive root modulo $p$ and thus another candidate in our search. But $(r+p)^n\equiv r^n + pr^{n-1}\equiv1 + pr^{n-1}$ modulo $p^2$. But the $pr^{n-1}$ can't be zero because $r^{n-1}$ is a unit modulo $p^2$. So $(r+p)^n\not\equiv 1$, and in particular $r$ and $r+n$ have different orders.
So, if there is even one value of $r$ of order $p-1$, then there is another that isn't (namely $r+p$), and is therefore of order $p(p-1)$, and therefore is a primitive root. Of course, if there aren't any values of $r$ of order $p-1$, then they're all of order $p(p-1)$, so we still reach the desired result.