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Let $a,m,k$ be integers. If $p$ is a prime such that $p\nmid a$. How to prove that $x^m\equiv a\pmod{p^k}$ has either $0$ solutions or exactly $\gcd (m,\varphi (p^k))$ solutions in $\mathbb{Z}_{p^k}$? I know that $ax\equiv c\pmod n$ has either $0$ solution or exactly $\gcd (n,a)$ solutions in $\mathbb{Z}_n$, but this doesn't seem useful here, any help would be appreciated.

commie trivial
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Ethan
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    Here is little simplification which might help. If $x$ is any solution and $b$ is any number such that $b^m \equiv 1$ then $xb$ is also a solution. So the problem reduces to showing the number of such elements $b$ equals gcd$(m, \phi(p^k))$. – Rishi Jul 24 '22 at 16:54
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    One certainly needs to use the fact that there exist primitive roots modulo $p^k$. In particular, the assertion is false for $p=2$ and $k\ge3$ (consider $x^2\equiv1\pmod8$ for example). – Greg Martin Jul 24 '22 at 17:03

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(I will prove the equivalent claim that Rishi proposes in the comments.)

We will use the fact that there exists a primitive root $\color{blue}{g} \bmod p^k$ as long as $p\neq 2$ (as this latter case easily yields counterexamples). We then know that ${\color{blue}{g}}\,^{\color{red}a}\equiv 1\bmod p^k$ and ${\color{blue}{g}}\,^{\color{red}b}=x\bmod p^k$ for some positive integers ${\color{red}a},{\color{red}b}$ since $x$ and $1$ are relatively prime to $p$ by assumption. Hence ${\color{blue}{g}}\,^{\color{red}a}\equiv {\color{blue}{g}}\,^{{\color{red}b}m}\bmod p^k$, which implies that ${\color{red}a}\equiv {\color{red}b}m\bmod \varphi(p^k)$ as $\varphi(p^k)$ is the order of ${\color{blue}{g}} \bmod p^k$. And you know how to deal with this case.

(Have a look at this post by Bill Dubuque on modular order reduction for more information on the last step.)

Dr. Mathva
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