8.37 If $n, p$ are two natural numbers, let
$$
S_{n, p}=1^{p}+2^{p}+\cdots+n^{p} .
$$

Determine the natural number $p$, such that for any natural number $n, S_{n, p}$ is a perfect square.
(French Mathematical Olympiad, 1991)
[Solution] Since for any natural number $n, S_{n, p}$ is a perfect square, in particular, taking $n=2, \quad S_{2, p}$ is also a perfect square. That is, there exists a natural number $x$ satisfying the following equation:
So
$$
\begin{array}{c}
x^{2}=1+2^{p} \quad(x>1) . \\
x^{2}-1=2^{p}, \\
(x-1)(x+1)=2^{p} .
\end{array}
$$

Thus
$$
\left\{\begin{array}{l}
x-1=2^{s}, \\
x+1=2^{t} .
\end{array}\right.
$$

Integers $s, t$ satisfy:
$$
0 \leqslant s<t, \quad s+t=p .
$$

Therefore,
$$
\frac{x+1}{x-1}=2^{t-s} \geqslant 2 \text {. }
$$

So
Thus
$$
\begin{array}{c}
x+1 \geqslant 2 x-2, \\
1<x \leqslant 3 .
\end{array}
$$

When $x=2$,
$$
1+2^{p}=4
$$

There is no solution in this case.
When $x=3$,
$$
\begin{array}{c}
1+2^{p}=9, \\
p=3 .
\end{array}
$$

Also, since
$$
\begin{aligned}
S_{n, 3} & =1^{3}+2^{3}+\cdots+n^{3} \\
& =\left[\frac{n(n+1)}{2}\right]^{2}
\end{aligned}
$$

is a perfect square.
Therefore, the only natural number $p=3$ makes $S_{n, p}$ a perfect square for any natural number $n$.