Q:

Why there are more than 2 solutions for $x^{14} = 1$ in $\mathbb{Z}_{113}$?

How to show that the equation $x^{14}=1$ has more than 2 solutions in the field $\mathbb{Z}_{113}$?

This is the same as solving $x^{14} \equiv 1 \pmod {113}$. I know that $x^{112} \equiv 1 \pmod {113}$ as long as $113$ does not divide $x$. But I am not sure what else can I do.

A:

Hint:
$113$ is prime, so $\phi(113)=112$, which is a multiple of $14$. Remember that groups $\Bbb Z_p^\times$ are cyclic when $p$ is prime.

