Is it true that $x^{au} \equiv 1 \pmod p$ $\implies x^{u} \equiv 1 \pmod p$

Question

$a,u$ are arbitrary integers.

I am trying to see if someone can use whether CRT/Hensel's lemma. We can rewrite $x^{au}$ as $(x^{a})^{u}$ and maybe somehow expand this, but I don't see how to continue any further (if this is indeed a true result).

EDIT: It should also be noted that $a$ and $p$ are relatively prime to each other.

Answer 1

That is false: for instance, $7^2\equiv 4\mod 15$, and $7^4\equiv 4^2\equiv 1\mod 15$.

Answer 2

According to Fermats Little theorem we have (excluding $0$) for any $x$: $x^{p-1}=1$. From some standard group theory, it follows that $a u| (p-1)$. Thus this imply that $a|(p-1)$? Not really, but does this imply that $x^a=1$. I don't think so.

Since we know now a bit about the theory, it is not that hard to find a counterexample. Take $p=7$, $u=4$, $a=3$. Then for any $x$ we have $x^{au}=x^{12}=(x^{7-1})^2=1$ but $x^{u}=x^{4}\neq 1$ for all $x$. Take for example $x=2$, then $x^{4}=16=2\neq1$.