If $\gcd(a,x)=1$ and $a|x^n-1$, then $n|\varphi(a)$?

Question

Does $\gcd(a,x)=1$ and $a|x^n-1$ imply that $n|\varphi(a)$? Where $\varphi$ is the Euler's totient function. Suppose that $n<\varphi(a)$.

Answer 1

This has no chance of even being remotely true. Let $a=2$ and let $x$ be an odd integer.

Then $\gcd(a,x)=1$ and $\varphi(a)=1$ and $a\mid x^n-1$ for any $n>0$, so certainly $n\nmid\varphi(a)$.

More generally, if $n\mid m$ then $x^n-1\mid x^m-1$, so if $a\mid x^n-1$ then there exist arbitrarily large $m$ such that $a\mid x^m-1$. So we certainly can't have $m\mid\varphi(a)$ in general.

For an example with $n<\varphi(a)$ let $a=7$ so that $\varphi(a)=6$. Then for $x=6$ and $n=4$ we get $$6^4-1=1295=5\times7\times37,$$ so $7\mid6^4-1$ but $4\nmid 6$.