I'm trying to figure out the number of solutions to the congruence equation $x^d \equiv1 \pmod{p^2}$ where $p$ is prime and $d\mid p-1$.
For the congruence equation ${x^d}\equiv1 \pmod p$ where $p$ is prime and $d\mid p-1$ I've shown that there are exactly $d$ solutions modulo $p$.
I'm trying to use the above result to extend it to the higher power. I'm aware of something called Hensel's Lemma which says if a polynomial has a simple root modulo a prime $p$, then this root corresponds to a unique root of the same equation modulo any higher power of $p$. We can 'lift' the root iteratively up to higher powers.
I'm unsure exactly how this process works. All I have to start with is that I assume a solution of the form $m+np$ solves the 'base' congruence and I'm trying to somehow extend it to $p^2$.
Any help would be appreciated.
Thanks.
Best Answer
Set $f(x)=x^d-1$. Each solution $m$ mod $p^k$ lifts to a unique solution mod $p^{k+1}$ exactly when $f'(m)\neq 0\pmod{p}$. Fortunately, $f'(m)=dm^{d-1}$. Since $d\mid p-1$, $p\nmid d$. Since $m^d\equiv 1\pmod{p}$, $p\nmid m$. Hence each solution mod $p$ lifts to a unique solution mod $p^2, p^3, \ldots$. There are plenty of resources to read more about Hensel's lemma, such as Wikipedia, or here on mse.