Question 42: Find all pairs of positive integers $(a, b)$ such that for any positive integer $n, \left(a^{n}+n\right)\left(b^{n}+n\right)$ is a perfect square.
Question 42:
Solution: When $a=b$, it is obvious that $\left(a^{n}+n\right)\left(b^{n}+n\right)$ is a perfect square.
Next, we prove that if $a \neq b$, then there exists a positive integer $n$ such that $\left(a^{n}+n\right)\left(b^{n}+n\right)$ is not a perfect square. For this, we only need to prove: there exists a positive integer $n$ and a prime $p$, satisfying $a^{n}+n \equiv p\left(\bmod p^{2}\right)$, while $b^{n}+n \neq 0(\bmod p)$.
First, take any prime $p > \max \{a, b, |a-b|\}$.

According to the Chinese Remainder Theorem, there exists a positive integer $r$ satisfying $\left\{\begin{array}{c}r \equiv 1(\bmod p-1) \\ r \equiv -a(\bmod p)\end{array}\right.$ Thus, $a^{r}+r \equiv a+r \equiv 0(\bmod p)$.

Furthermore, according to the generalized Chinese Remainder Theorem, there exists a positive integer $n$ satisfying: $\left\{\begin{array}{l}n \equiv r(\bmod p(p-1)) \\ n \equiv p-a^{r}(\bmod p^{2})\end{array}\right.$ Thus, according to Euler's theorem, $a^{n}+n \equiv a^{r}+(p-a^{r}) \equiv p(\bmod p^{2})$.
Also, $\left\{\begin{array}{c}n \equiv r \equiv 1(\bmod p-1) \\ n \equiv p-a^{r} \equiv -a^{r}(\bmod p)\end{array}\right.$ According to Fermat's little theorem, $b^{n}+n \equiv b^{r}-a^{r} \equiv b-a \neq 0(\bmod p)$. In summary, the solution is $(a, b)=(m, m)$, where $m$ is any positive integer.