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

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This discussion establishes that if there exists a primitive root for an odd prime power \( p^n \), then a primitive root for \( 2p^n \) also exists. Specifically, if \( r \) is an odd primitive root of \( p^n \), it directly serves as a primitive root for \( 2p^n \). If \( r \) is not odd, then \( r + p^n \) can be shown to be a primitive root of \( 2p^n \). The discussion also touches on the conditions for combining moduli in relation to primitive roots.

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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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Re: primitive root

Poirot said:
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 $2p^n$.
 
Re: primitive root

caffeinemachine said:
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?
 
Re: primitive root

Poirot said:
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 $x\equiv a\pmod{mn}$
 

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