How to determine number of solutions modular

Question

Is there a formula for determine the number of solutions (#S) to $x^{3}=1$ mod $(mn)$ where $n \neq m$ are both prime ( ie in the ring $\mathbb{Z}/mn\mathbb{Z}$)

I think from the Chinese remainder theorem it would tell us that the number of solutions would be the product of the number of solutions in $m$ and in $n$. So would there be a limit of the number of solutions? I guess the most solutions one could have for n or m individually would be 3, so would the most solutions in the whole ring just be 9?

Answer 1

A ring $\mathbb Z_p$ is a field, if $p$ is prime. A polynomial of degree $n$ has at most $n$ roots in such a field. So, the equation $x^3=1$ has at most $3$ solutions in $\mathbb Z_m$ and at most $3$ solutions in $\mathbb Z_n$. So, you correctly assumed that the maximum number of solutions modulo $mn$ is $9$ due to the chinese remainder theorem.

Answer 2

Your reasoning about the Chinese remainder theorem is not only correct, it also implies that one should investigate the number of cubic roots of unity in the finite field $\mathbb Z_p$. According to this article

https://mathoverflow.net/questions/58357/number-of-n-th-roots-of-unity-over-finite-fields

the answer is 3 if $p-1$ is divisible by 3 and 1 otherwise. The smallest prime with 3 solutions is $p=7$: $\{\overline 1, \overline 2, \overline 4\}.$