Problem 1581

Necessity: If $x_{0}$ is a solution of (22), then from $p \nmid a$ we know $p \nmid x_{0}$. This, together with Fermat's Little Theorem, leads to
$$a^{(p-1) / n} \equiv\left(x_{0}^{n}\right)^{(p-1) / n} \equiv x_{0}^{p-1} \equiv 1(\bmod p)$$

Sufficiency: If equation (23) holds, then we have
$$\begin{aligned} x^{p-1}-1 & =\left(x^{n}\right)^{(p-1) / n}-a^{(p-1) / n}+a^{(p-1) / n}-1 \\ & =\left(x^{n}\right)^{(p-1) / n}-a^{(p-1) / n}+p \cdot c, \end{aligned}$$
where $c$ is an integer. From this, we know there must be an integer-coefficient polynomial $q(x)$ such that
$$x^{p}-x=\left(x^{n}-a\right) q(x)+p \cdot c x.$$

From this and Theorem 5, it follows that (22) has a solution and the number of solutions is $n$ (in fact, from the first equation of (24) and Theorem 5, the necessity is also derived).

In the example given above for modulo 11, $n=2,5$ are cases of Theorem 7, and the actual calculations match the conclusions. For the case $n=2$, Theorem 7 is Theorem 2 in $\S 5$.