Four. (This question is worth 50 points)
Given a prime number $p$ that satisfies the following conditions: there exist positive integers $n, u, v$, such that the number of positive divisors of $n$ is equal to $p^{\mu}$, and the sum of these $p^{2}$ positive divisors is equal to $p^{*}$. Find all possible values of $p$.

保留源文本的换行和格式，直接输出翻译结果如下：

Four. (This question is worth 50 points)
Given a prime number $p$ that satisfies the following conditions: there exist positive integers $n, u, v$, such that the number of positive divisors of $n$ is equal to $p^{\mu}$, and the sum of these $p^{2}$ positive divisors is equal to $p^{*}$. Find all possible values of $p$.
Solution: Clearly, $n>1$. Let $n=p_{1}^{a_{1}} \cdot p_{2}^{\alpha_{2}} \cdots \cdots \cdot p_{i}^{a_{n}-}$ be the prime factorization of $n$. $m, a_{1}, a_{2}, \cdots, a_{m} \in \mathbf{N}^{*}$, then
$$
\begin{array}{l}
\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots \cdots \cdot\left(a_{m}+1\right)=p^{a}, \\
\left(1+p_{1}+\cdots+p_{1}^{a_{1}}\right)\left(1+p_{2}+\cdots+p_{2}^{a_{2}}\right) \cdots \cdot\left(1+p_{m}+\cdots+p_{m}^{a_{m}}\right)=p^{2} .
\end{array}
$$

Since $p$ is a prime, it follows from (1) that there exists $s \in \mathbf{N}^{*}$ such that $a_{1}+1=p^{*}$, thus
$$
\begin{array}{l}
1+p_{1}+p_{1}^{2}+\cdots+p_{1}^{a_{1}}=1+p_{1}+p_{1}^{2}+\cdots+p_{1}^{p^{2}-1} \\
=\frac{p_{1}^{p}-1}{p_{1}^{p}-1}=\frac{p_{1}^{p}-1}{p_{1}-1} \cdot \frac{p_{1}^{p_{1}^{p}}-1}{p_{1}^{p}-1} \\
\end{array}
$$

Since the values of the last two fractions in the above expression are integers, and $p$ is a prime, it follows from (2) that there exists $t \in \mathbf{N}^{*}$ such that $p^{t}=\frac{p_{1}^{p}-1}{p_{1}-1}$, hence $p_{1}^{p}-1$.
By Fermat's Little Theorem, we get $p_{1} \equiv p_{1}^{p} \equiv 1(\bmod p)$ 30 points. Let $p_{1}=k p+1, k \in \mathbf{N}^{*}$, then
$$
\begin{aligned}
p^{2} & =\frac{p_{1}^{p}-1}{p_{1}-1}=\frac{(k p+1)^{p}-1}{k p} \\
& =C_{p}^{0}(k p)^{p-1}+C_{p}^{1}(k p)^{p-2}+\cdots+C_{p}^{p-2} k p+C_{p}^{p-1} .
\end{aligned}
$$

When $p \geqslant 3$, $(k p)^{p-1}, C_{P}^{1}(k p)^{p-2}, \cdots, C_{p}^{p-2} k p$ are all multiples of $p^{2}$, but $C_{p}^{p-1}=p$ is not a multiple of $p^{2}$, which makes the right side of the equation impossible to be a power of $p$, a contradiction 40 points.
Therefore, when $p \leqslant 2$. Since $p$ is a prime, hence $p=2$.
In fact, when $p=2$, there exist such positive integers $n$, as long as $n$ is the product of distinct primes of the form $2^{r}-1\left(r \in \mathbf{N}^{*}\right)$, the conditions are satisfied. For example, $n=21=3 \times 7$.
In summary, the only prime $p$ that satisfies the conditions is $p=2$ 50 points.