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- Mar 10, 2012

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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?

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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 $2p^n$.

d = g(p^n). But g(p^n)=g(2p^n). r is odd so r^d=1 (mod 2). Can we somehow combine moduli?

- Mar 10, 2012

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Under certain conditions yes.Can we somehow combine moduli?

Say $x\equiv a\pmod{m}, x\equiv a\pmod{n}$ and gcd$(m,n)=1$. Then $x\equiv a\pmod{mn}$