 Solution

If $p\mid n$, the field property forces $x=0$, giving exactly one solution. Otherwise, if $x_0$ is any solution, multiplication by $x_0$ is a bijection from the solutions of $z^k=1$ to those of $x^k=n$. Part (i) therefore givesThe two possibilities in the second line distinguish an empty fiber from a coset of the kernel of the power group homomorphism on ${\mathbb{F}}_p^{\times }$.