6.51 Find all positive integers $n>1$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer.
(31st International Mathematical Olympiad, 1990)
[Solution] Clearly, the number $n$ is a positive odd number, so we only need to consider the case where $n \geqslant 3$ and $n$ is odd.
Let $p$ be the smallest prime factor of $n$, then $p \geqslant 3, p \mid 2^{n}+1$.
Let $i$ be the smallest positive integer such that $p \mid 2^{i}+1$, we will prove that $1 \leqslant i0$, we can let $i=r_{1}+b, 1 \leqslant bk+m-\frac{k}{2} \geqslant m+1 \text {. }
$$

If $\beta \geqslant m+2$, then $3^{m+2} \mid 3^{k} C_{n}^{k}$.
Noting that $m \geqslant 2$, so $2 m \geqslant m+2$.
Thus, by (1) we know,
$$
3^{m+2} \mid 3 n \text {. }
$$

Therefore,
$$
3^{m+1} \mid n \text {. }
$$

This contradicts $(c, 3)=1$. Thus, it is proven that $m=1$, i.e.,
$$
n=3 c,(c, 3)=1 \text {. }
$$

Assume $c>1$, and let $q$ be the smallest prime factor of $c$, clearly $q \geqslant 5$, and $q \mid 2^{n}+1$.
Similarly, let $j$ be the smallest positive integer such that $q \mid 2^{j}+1$, then it must be that
$$
1 \leqslant j<q-1 \text {. }
$$

Furthermore, it can be proven that $j \mid n$.
Thus, by the definition of the prime $q$ and $j<q-1$, we have
$$
j \in\{1,3\} \text {. }
$$

Therefore, from $q \mid 2^{j}+1$ we get $q=3$, which contradicts $q \geqslant 5$.
Thus,
$$
\begin{array}{l}
c=1 . \\
n=3 .
\end{array}
$$

So,
it can be verified that
$$
3^{2} \mid 2^{3}+1 \text {. }
$$

Therefore, the only positive integer $n$ that satisfies the requirement is $n=3$.