Is There a Strategy for Finding Primitive Roots of 2p^n If There is One for p^n?

[Poirot1]

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?

 

[caffeinemachine]

Re: primitive root

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?

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 $2p^n$.

 

[Poirot1]

Re: primitive root

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 $2p^n$.

Ok let's first assume r is odd. Then if d divides g(p^n), we have r^d=1 (mod p^n) iff
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?

 

[caffeinemachine]

Re: primitive root

Can we somehow combine moduli?

Under certain conditions yes.

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