4. $p$ is a prime number, and the sum of all positive divisors of $P^{4}$ is exactly a perfect square. Then the number of prime numbers $p$ that satisfy the above condition is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
4. C Since $p$ is a prime number, $p^{+}$ has 5 positive divisors $1 \cdot p$, $p^{2}, p^{3}, p^{4}$. According to the problem, we have
$1+p+p^{2}+p^{3}+p^{4}=n^{2}$ ( $n$ is a positive integer).

Then $(2 n)^{2}>4 p^{4}+4 p^{3}+p^{2}=\left(2 p^{2}+p\right)^{2}$.
Also, $(2 n)^{2}<4 p^{2}+p^{2}+4+4 p^{3}+8 p^{2}+4 p$
$$
=\left(2 p^{2}+p+2\right)^{2} \text {, }
$$

i.e., $\left(2 p^{2}+p\right)^{2}<(2 n)^{2}<\left(2 p^{2}+p+2\right)^{2}$,
thus
$$
\begin{array}{c}
(2 n)^{2}=\left(2 p^{2}+p+2\right)^{2} . \\
\text { By } \quad 4 n^{2}=4 p^{4}+4 p^{3}+5 p^{2}+2 p+1, \\
4 n^{2}=4 p^{4}+4 p^{3}+4 p^{2}+4 p+4,
\end{array}
$$
(1) This leads to $p^{2}-2 p-3=0$. Solving for the positive root, we get $p=3$.