10. Find all positive integers $n>1$, such that $2^{2}+3^{2}+\cdots+n^{2}=p^{k}$, where $p$ is a prime, $k \in \mathbf{N}^{*}$.
10. Solution: When $n=2,3,4,7$, 
$$
\begin{array}{l}
2^{2}+3^{2}+\cdots+n^{2}=2^{2}, 13^{1}, 29^{1}, 139^{1} ; \\
\quad \text{When } n>7, \text{ } 2^{2}+3^{2}+\cdots+n^{2}= \\
\frac{n(n+1)(2 n+1)}{6}-1=\frac{(n-1)\left(2 n^{2}+5 n+6\right)}{6}, \\
\because 2 n^{2}+5 n+6=(n-1)(2 n+7)+13 \\
\therefore\left(2 n^{2}+5 n+6, n-1\right)=(n-1,13)=1
\end{array}
$$

or 13.
(1) If $(n-1,13)=1$, then by $n-1>6,2 n^{2}+5 n+6>6$ we know $\frac{(n-1)\left(2 n^{2}+5 n+6\right)}{6}$ must have two different prime factors, so the equation does not hold;
(2) If $(n-1,13)=13$, if $13^{2} \mid n-1$, then $n \equiv 1(\bmod 169)$,
$\therefore 2 n^{2}+5 n+6 \equiv 2+5+6 \equiv 13(\bmod 169)$, $2 n^{2}+5 n+6=169 m+13=13(13 m+1)$,

i.e., $2 n^{2}+5 n+6$ has a factor other than 13 and greater than 6, $\therefore n-1=13$ or $2 \cdot 13$ or $3 \cdot 13$ or $6 \cdot 13$, i.e., $n=14,27,40,79$, checking shows that the equation does not hold in these cases. Therefore, the positive integers $n$ that satisfy $2^{2}+3^{2}+\cdots+n^{2}=p^{k}$ are $n=2,3,4,7$.