Congruence of a power with 1, $a^x \equiv 1\ mod \ (p^\alpha)$

Question

We have that $a^{\varphi(m)} \equiv 1\ mod(m)$ if $(a,m)$. Then for a prime $p$ that does not divide $a$, $a^{p-1}\equiv 1\ mod\ (p)$. It is true that $a^{\varphi(p^\alpha)}\equiv 1\ mod\ (p^\alpha)$ with $1 <\alpha$.

I've looked at some cases where it's true but haven't able to give a proof or disproved it.

Or, Is there any integer number $x \neq 0 $ such that $a^x \equiv 1\ mod \ (p^\alpha)$ where $p$ is a prime number and $p \nmid a$?.

Answer 1

If $p\not\mid a$, then $(a,p^\alpha) =1$, so by Euler's theorem $a^{\phi(p^\alpha)}\cong1 \pmod {p^{\alpha}}$.

So yes, since $\phi(p^{\alpha})=p^{\alpha}-p^{\alpha-1}\not =0$.

Answer 2

I’m not sure if you are satisfied with this: $$8^2\equiv 1 (\text{mod } 3^2)$$.