[Example 3.6.6] Find all positive integers $n>1$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer.
3 is the only solution.
We will prove it in three steps.
Step 1: Prove that $n$ is a multiple of 3, thus $n=3^{m} \cdot c$.
Step 2: Prove that $m=1$, i.e., $n=3 c$.
Step 3: Prove that $c=1$, i.e., $n=3$.
Below is the proof step by step.
Step 1: Prove that $3 \mid n$.
Let $p$ be the smallest prime factor of $n$. Since $n$ is odd, $p \geqslant 3$.
Let the order of 2 modulo $p$ be $l$,
Since
$$
\begin{array}{l}
n^{2} \mid 2^{n}+1, \\
2^{n} \equiv-1 \quad(\bmod p), \\
2^{2 n} \equiv 1 \quad(\bmod p) . \\
2^{2} \equiv 1 \quad(\bmod p),
\end{array}
$$

Also, by Fermat's Little Theorem,
$$
2^{p-1} \equiv 1 \quad(\bmod p) .
$$

From (1), (2), and (3), we have
$$
l|2 n, l| p-1 .
$$

Since $(n, p-1)=1$, then $(2 n, p-1)=2$,
so
$$
l=2 \text {. }
$$

Thus
$$
p \mid 2^{\prime}-1=2^{2}-1=3 .
$$

Since $p$ is the smallest prime divisor of $n$, then $p=3$,
hence $\quad 3 \mid n, n=3^{m} \cdot c$. (where $(c, 3)=1$.)
Step 2: Prove that $m=1$, i.e., $n=3 c$.
By contradiction.
If $m \geqslant 2$, then by the given
and
$$
\begin{array}{c}
n^{2} \mid 2^{n}+1 \\
n^{2}=3^{2 m} \cdot c^{2}, \\
3^{2 m} \mid 2^{n}+1
\end{array}
$$

Then
$$
2^{n}+1=(3-1)^{n}+1 \equiv 3 n+\sum_{k=2}^{2 m-1}(-1)^{k} C_{n}^{k} 3^{k} \quad\left(\bmod 3^{2 m}\right) \text {. }
$$
(Note, here we discard the terms in the binomial expansion $C_{n}^{k} 3^{k}$ where $k \geqslant 2 m$.) From (1) and (2),
$$
3^{2 m} \mid 3 n+\sum_{k=2}^{2 m-1}(-1)^{k} C_{n}^{k} 3^{k} .
$$

Next, we study the highest power of 3 in $C_{n}^{k} 3^{k}$.
Let the highest power of 3 in $k!$ be $\alpha$, then
$$
\begin{array}{c}
\alpha=\sum_{s=1}^{\infty}\left[\frac{k}{3^{s}}\right]m-\frac{k}{2}$.
Thus, the highest power of 3 in $C_{n}^{k} 3^{k}$, denoted by $\beta$, satisfies
$$
\beta>k+m-\frac{k}{2}=m+\frac{k}{2} \geqslant m+1(k \geqslant 2) .
$$

That is
$$
\begin{array}{c}
\beta \geqslant m+2, \\
3^{m+2} \mid C_{n}^{k} 3^{k} .
\end{array}
$$

Thus,
$$
\begin{array}{c}
3^{m+2} \mid C_{n}^{k} \\
m+2
\end{array}
$$

Notice that for (3), we have
For (3), we have
$$
\begin{array}{l}
3^{m+2} \mid 3 n=3 \cdot 3^{m} \cdot c, \\
3^{m+2} \mid 3^{m+1} \cdot c,
\end{array}
$$

That is
$$
3 \mid c \text {. }
$$

This contradicts $(3, c)=1$.
Therefore, $m \geqslant 2$ is not valid, i.e., $m=1$, thus $n=3 c$.
Step 3: Prove that $c=1$.
Again, by contradiction.
If $c>1$, let $q$ be the smallest prime factor of $c$, then by $c$ being odd and $(3, c)=1$, we know $q \geqslant 5$.
Let $l$ be the order of 8 modulo $q$, then
$8^{2}=1(\bmod q)$,
By the given
Then
$n^{2} \mid 2^{n}+1$,
$n \mid 2^{n}+1$,
37 Wu's $q \mid 2^{n}+1=2^{3 c}+1=8^{c}+1$.
From (5), we have
$$
\begin{array}{lll}
\text { From (5), we have } \\
& 8^{c} \equiv-1 \quad(\bmod q), \\
\text { so } & 8^{2 c} \equiv 1 \quad(\bmod q) .
\end{array}
$$

So by Fermat's Little Theorem,
$$
8^{q-1} \equiv 1 \quad(\bmod q) .
$$

Thus, by the definition of order,
$$
\begin{array}{c}
l|q-1, l| 2 c . \\
(2 \bar{c}, q-1)=1, \\
l=1 \text { or } 2 .
\end{array}
$$
$$
\begin{array}{c}
\text { Also, } (2 c, q-1)=1, \\
\text { then } \quad l=1 \text { or } 2 .
\end{array}
$$

When $l=1$, by $8^{\prime} \equiv 1(\bmod q)$ we know $q=1$, which contradicts $q$ being a prime.
When $l=2$, by $8^{\prime} \equiv 1(\bmod q)$ we know $q=7$ (since $\left.q \neq 3\right)$, substituting into $q \mid 8^{c}+1$ gives $7 \mid 8^{c}+1$, which is also a contradiction.
Therefore, $c>1$ is impossible.
Thus, $c=1$.
Combining the above three steps, we conclude that the only $n$ that meets the requirement is 3.