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

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If there is a primitive root for p^n, then there exists a primitive root for 2p^n, particularly if the primitive root r is odd. If r is not odd, then r + p^n can be shown to be a primitive root of 2p^n. The discussion explores the conditions under which moduli can be combined to demonstrate this relationship. Specifically, it highlights that if certain congruences hold true, they can be combined when the moduli are coprime. The overall strategy revolves around leveraging properties of primitive roots and modular arithmetic.
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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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