Question: $p$ 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 .
Select (C).
Since $p$ is a prime number, $p^{4}$ has 5 positive divisors: $1, p, p^{2}, p^{3}, p^{4}$.
According to the problem,

then
$$
1+p+p^{2}+p^{3}+p^{4}=n^{2} \quad(n \in \mathbf{N}) .
$$

then
$$
\begin{aligned}
(2 n)^{2} & >4 p^{4}+4 p^{3}+p^{2}=\left(2 p^{2}+p\right)^{2} . \\
(2 n)^{2} & >4 p^{4}+p^{2}+4+4 p^{3}+8 p^{2}+4 p \\
& =\left(2 p^{2}+p+2\right)^{2},
\end{aligned}
$$

i.e. $\square$
$$
\left(2 p^{2}+p\right)^{2}<(2 n)^{2}<\left(2 p^{2}+p+2\right)^{2} .
$$

Therefore, $(2 n)^{2}=\left(2 p^{2}+p+1\right)^{2}=4 p^{4}+4 p^{3}+5 p^{2}+2 p+1$. 
Combining with $4 n^{2}=4 p^{4}+4 p^{3}+4 p^{2}+4 p+4$, we get $p^{2}-2 p-3=0$, solving which yields $p=3, p=-1$ (discard $p=-1$).