This is the last of the stream of number theory problems I have been looking at that I would like to discuss.
Let $p$ be an odd prime number and let $m$ be a positive integer. Prove that any primitive root modulo $p^m$ is also a primitive root modulo $p$.
My idea is as follows: Suppose $r$ is a primitive root modulo $p^m$. Then $r^{p^{m-1}(p-1)} \equiv 1 \mod{p^m}$, and therefore $$\left( r^{p-1} \right)^{p^{m-1}} \equiv 1 \mod p.$$ Can we conclude from this that $r^{p-1}\equiv \pm 1\mod p$, and since $p^{m-1}$ is odd, that $r^{p-1}\equiv 1 \mod p$?
How do I finish?
EDIT: Already answered at Prove that , any primitive root $r$ of $p^n$ is also a primitive root of $p$ . The chosen solution is very simple: If $r$ is a primitive root modulo $p^m$, then there exists an integer $k$ such that $r^k \equiv a \mod{p^m}$ for any $a$ relatively prime to $p^m$. Therefore $r^k \equiv a \mod p$. This can only be true if $r$ is a primitive root modulo $p$.
