28. Let $a$ be an integer, and $n, r$ be integers greater than 1. $p$ is an odd prime, and $(n, p-1)=1$. Find the number of solutions to the following congruence equation:
$$x_{1}^{n}+x_{2}^{n}+\cdots+x_{r}^{n} \equiv a(\bmod p)$$

Here, the solutions $\left(x_{1}, x_{2}, \cdots, x_{r}\right)$ and $\left(x_{1}^{\prime}, x_{2}^{\prime}, \cdots, x_{r}^{\prime}\right)$ are considered the same if and only if: for $1 \leqslant j \leqslant r$, we have
$$x_{j} \equiv x_{j}^{\prime}(\bmod p)$$
28. First, prove: For any $b \in \mathbf{Z}$, the congruence equation $x^{n} \equiv b(\bmod p)$ has a unique solution.

In fact, by $(n, p-1)=1$ and Bézout's theorem, there exist $u, v \in \mathbf{N}^{*}$ such that $n u-(p-1) v=1$. If $b \equiv 0(\bmod p)$, then the equation $x^{n} \equiv b(\bmod p)$ has only the unique solution $x \equiv 0(\bmod p)$; if $b \not\equiv 0(\bmod p)$, then the solution to the equation $x^{n} \equiv b(\bmod p)$ satisfies $(x, p)=1$, and $x^{m m} \equiv b^{u}(\bmod p)$. By Fermat's little theorem, we know that $x^{p-1} \equiv 1(\bmod p)$, thus
$$x^{m}=x^{(p-1) v+1} \equiv x(\bmod p),$$

which means $x \equiv b^{u}(\bmod p)$. This shows that the solution to $x^{n} \equiv b(\bmod p)$ is unique.
Using the above conclusion and the structure of the congruence equation, we can get: For any $a \in \mathbf{Z}$, the original congruence equation has exactly $p^{r-1}$ distinct solutions (arbitrarily choose the remainders of $x_{1}^{n}, x_{2}^{n}, \cdots, x_{-1}^{n}$ modulo $p$, and adjust $x_{r}^{n}$ to make $x_{1}^{n}+x_{2}^{n}+\cdots+x_{r}^{n} \equiv a(\bmod p)$, thus obtaining all $p^{r-1}$ solutions).