6.71 Determine all pairs of positive integers $(n, p)$ satisfying:
$p$ is a prime, $n \leqslant 2 p$, and $(p-1)^{n}+1$ is divisible by $n^{p-1}$.
(40th International Mathematical Olympiad, 1999)
[Solution] When $n=1$, since $(p-1)^{n}+1=p$, it is clearly divisible by $1^{p-1}=1$, so $(n, p)=(1, p)$ is a solution.
When $n=2$, since $(p-1)^{2}+1=p^{2}+2 p+2$, if it is divisible by $2^{p-1}$, then $p^{2}$ must be even, and since $p$ is a prime, we have $p=2$,
so $(n, p)=(2,2)$ is another solution.
Now consider the case when $n \geqslant 2, p \geqslant 3$.
When the prime $p \geqslant 3$, $(p-1)^{n}+1$ is odd, if it is divisible by $n^{p-1}$, then $n$ is also odd, $n \neq 2 p$,
thus, $n<2 p$.
Let $q$ be the smallest prime factor of $n$, then by $n^{p-1} \mid(p-1)^{n}+1$, we know
$$
q \mid(p-1)^{n}+1, (p-1)^{n} \equiv-1 \quad(\bmod q),
$$

and
$$
(q, p-1)=1.
$$

By the choice of $q$, we have $(n, p-1)=1$,
so there exist integers $u, v$, such that $u n+v(q-1)=1$.
By Fermat's Little Theorem, $q \mid(p-1)^{q-1}-1$,
thus
$$
\begin{array}{c}
p-1 \equiv(p-1)^{1}=(p-1)^{u n} \\
\equiv(-1)^{n} \cdot 1^{v} \\
q-1 \text { is even, } n \text { is odd, so } \\
p-1 \equiv-1 \quad(\bmod q), \\
p=0 \quad(\bmod q),
\end{array}
$$

This shows that $\quad q \mid p$, and thus $p \mid q$,
which means $q=p$.
Therefore, we have $p^{p-1} \mid(p-1)^{p}+1$
$$
=p^{2}\left(p^{p-2}-C_{p}^{1} p^{p-3}+\cdots+C_{p}^{p-3} p-C_{p}^{p-2}+1\right)
$$

In the expression inside the parentheses, except for the last term which is 1, all other terms are divisible by $p$.
Thus, $p-1 \leqslant 2$, i.e., $p=3$. In this case, $n=3$,
so $(n, p)=(3,3)$ is a solution.
Therefore, the three solutions to this problem are $(n, p)=(1, p),(2,2),(3,3)$.