$14.4 .23^{\star \star}$ Find all prime numbers $p$ such that the sum of all divisors of $p^{4}$ is a perfect square.
If $p$ is a prime number, then the sum of all divisors of $p^{4}$ equals $1+p+p^{2}+p^{3}+p^{4}$. If $1+$ $p+p^{2}+p^{3}+p^{1}=n^{2}$, where $n$ is a positive integer, then it is clear that $\left(2 p^{2}+p\right)^{2}<(2 n)^{2}<\left(2 p^{2}+\right.$ $p+2)^{2}$. Therefore, $(2 n)^{2}=\left(2 p^{2}+p+1\right)^{2}$. Thus, $4 n^{2}=4 p^{4}+4 p^{3}+5 p^{2}+2 p+1$. Also, $4 n^{2}=4\left(p^{4}+p^{3}+p^{2}+p+1\right)$, so $p^{2}-2 p-3=0$, yielding $p=3$. Indeed, if $p=3$, then $1+3+3^{2}+3^{3}+3^{4}=11^{2}$, so there is only one prime number $p=3$ that satisfies the given condition.