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Let $r$ be a primitive root of $p^n$. If $r$ is odd then we can show that $r$ is a primitive root of $2p^n$. If $r$ ain't odd then it can be shown that $r+p^n$ is a primitive root of $2p^n$.let p be an odd prime. Show that if there is a primitive root of p^n, then there is a primitive root of 2p^n. Strategy?
Ok lets first assume r is odd. Then if d divides g(p^n), we have r^d=1 (mod p^n) iffLet $r$ be a primitive root of $p^n$. If $r$ is odd then we can show that $r$ is a primitive root of $2p^n$. If $r$ ain't odd then it can be shown that $r+p^n$ is a primitive root of $2p^n$.
Under certain conditions yes.Can we somehow combine moduli?