*For each integer $s, 2 \leqslant s \leqslant 10$, find all prime numbers such that the sum of all its divisors is an integer $s$-th power.
A prime number $p$ has only two factors, namely 1 and $p$. If $1+p=n^{s}$, then $p=n^{5}-1=$ $(n-1)\left(n^{s}+n^{s}+\cdots+1\right)$, since $n>1$, when $s \geqslant 2$, the first factor on the left side of the equation is smaller than the second. Therefore, $n-1=1$, which means $n=2$. Thus, $p=2^{s}-1$. When $s=2,3,5,7$, we get the prime numbers $3,7,31,127$ respectively. When $s=4,6,8,10$, it is clearly impossible to obtain a prime number.