For positive integers $k>1$, define $S_{k}$ as the set of all positive integer triples $(n, a, b)$ that satisfy the following conditions: $n$ is odd, $(a, b)=1, a+b=k$, and $n \mid\left(a^{n}+b^{n}\right)$. Try to find all $k$ such that $S_{k}$ is a finite set.
The 52nd question:
Solution: We discuss in two cases.
(1) If $\mathrm{k}$ has an odd prime factor, let one of them be $\mathrm{p}$. For any positive integer $\alpha$, according to the LTE lemma, we have:
$$
\mathrm{v}_{\mathrm{p}}\left((\mathrm{k}-1)^{\mathrm{p}^{\alpha}}+1\right)=\mathrm{v}_{\mathrm{p}}(\mathrm{k})+\mathrm{v}_{\mathrm{p}}\left(\mathrm{p}^{\alpha}\right)>\alpha
$$

Thus, $\mathrm{p}^{\alpha} \mid\left[(\mathrm{k}-1)^{\mathrm{p}^{\alpha}}+1\right]$, so $\left(\mathrm{p}^{\alpha}, \mathrm{k}-1,1\right) \in \mathrm{S}_{\mathrm{k}}$, hence $\mathrm{S}_{\mathrm{k}}$ is an infinite set.
(2) If $\mathrm{k}$ has no odd prime factors, then $\mathrm{k}$ is a power of 2. Let $\mathrm{k}=2^{\alpha}$.
If there exists $(n, a, b) \in S_{k}$ such that $n>1$. Let $p$ be the smallest prime factor of $n$, then $p$ is an odd prime. Since $\mathrm{n} \mid\left(\mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}\right)$, we have $\mathrm{p} \mid\left(\mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}\right)$. Noting that $(\mathrm{a}, \mathrm{b})=1$, we have $(\mathrm{a}, \mathrm{p})=(\mathrm{b}, \mathrm{p})=1$. Therefore, we know:
$$
\mathrm{p}\left|\left(\mathrm{a}^{\mathrm{n}}+\mathrm{b}^{\mathrm{n}}\right) \Rightarrow \mathrm{p}\right|\left[\left(\mathrm{a} \cdot \mathrm{b}^{-1}\right)^{\mathrm{n}}+1\right]
$$

Let the order of $a \cdot b^{-1}$ modulo $p$ be $\tau$, then $\tau \mid n$. Also, $\tau \mid(p-1)$, and $(n, p-1)=1$, so $\tau=1$, hence
$$
\mathrm{p}\left|\left(\mathrm{a} \cdot \mathrm{b}^{-1}+1\right) \Rightarrow \mathrm{p}\right|(\mathrm{a}+\mathrm{b}) \Rightarrow \mathrm{p} \mid 2^{\alpha}
$$

Since $\mathrm{p}$ is an odd prime, this leads to a contradiction.
From the above analysis, if $(n, a, b) \in S_{k}$, then $n=1$. There are only finitely many pairs of positive integers $(a, b)$ that satisfy $a+b=k$, so $\mathrm{S}_{\mathrm{k}}$ is a finite set.
In summary, the $\mathrm{k}$ that satisfies the condition is $\mathrm{k}=2^{\alpha}$, where $\alpha$ is any positive integer.
Note: From this problem, we can directly derive a selection problem from the 2020 Vietnam national team: Find all positive integers $\mathrm{k}$ such that there are only finitely many positive odd numbers $n$ satisfying $n \mid\left(k^{n}+1\right)$. Clearly, the required $k=2^{c}-1\left(c \in Z^{+}\right)$.