4. 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 $p^{u}$, and the sum of these $p^{u}$ positive divisors is equal to $p^{v}$. Find all possible values of $p$.


Let $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{m}^{a_{m}}$ be the prime factorization of $n$, where $m, a_{1}, a_{2}, \cdots, a_{m} \in \mathbf{N}^{*}$.
Then $\left\{\begin{array}{l}\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots\left(a_{m}+1\right)=p^{u}, \\ \left(1+p_{1}+\cdots+p_{1}^{a_{1}}\right)\left(1+p_{2}+\cdots+p_{2}^{a_{2}}\right) \cdots\left(1+p_{m}+\cdots+p_{m}^{a_{m}}\right)=p^{v},\end{array}\right.$
Since $p$ is a prime, it follows from (1) that there exists $s \in \mathbf{N}^{*}$ such that $p^{s}=a_{1}+1$,
thus $1+p_{1}+\cdots+p_{1}^{a_{1}}=1+p_{1}+\cdots+p_{1}^{p^{s}}=\frac{p_{1}^{p^{s}}-1}{p_{1}-1}=\frac{p_{1}^{p}-1}{p_{1}-1} \cdot \frac{p_{1}^{p^{s}}-1}{p_{1}^{p}-1}$,
the values of the last two fractions are integers, and since $p$ is a prime, it follows from (2) that there exists $t \in \mathbf{N}^{*}$ such that $\left.p^{t}=\frac{p_{1}^{p}-1}{p_{1}-1} \Rightarrow p \right\rvert\, p_{1}^{p}-1$. Clearly $\left(p_{1}, p\right)=1$,
by Fermat's Little Theorem, we have $p_{1}^{p} \equiv p_{1} \equiv 1(\bmod p)$. Let $p_{1}=k p+1, k \in \mathbf{N}^{*}$, then $p^{t}=\frac{p_{1}^{p}-1}{p_{1}-1}=\frac{(k p+1)^{p}-1}{k p}=C_{p}^{0}(k p)^{p-1}+C_{p}^{1}(k p)^{p-2}+\cdots+C_{p}^{p-2} k p+C_{p}^{p-1}$, noting that $C_{p}^{0}(k p)^{p-1}, C_{p}^{1}(k p)^{p-2}, \cdots, C_{p}^{p-2} k p$ are all multiples of $p^{2}$, while $C_{p}^{p-1}=p$ is not a multiple of $p^{2}$, when $p \geqslant 3$, the right side of the above equation cannot be a power of $p$, which is a contradiction.
Thus $p \leqslant 2$, and since $p$ is a prime, we have $p=2$. In fact, when $p=2$, 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, $21=3 \times 7$. In summary, the only prime number $p$ that satisfies the conditions is $p=2$.