Question 41: Find all pairs of positive integers $(a, b)$, where $a, b \geq 2$, such that for all positive integers $n$ satisfying $a^{n}-n^{a} \neq 0$, we have $\left(a^{n}-n^{a}\right) \mid\left(b^{n}-n^{b}\right)$.
The 41st question:
Proof: We first prove a lemma: If positive integers $\mathrm{m}>\mathrm{n} \geq 3$, then $\mathrm{m}^{\mathrm{n}}\max \left\{\mathrm{a}, \mathrm{b},\left|\mathrm{a}^{\mathrm{b}}-\mathrm{b}^{\mathrm{a}}\right|\right\}$. Let $\mathrm{n}=\mathrm{p}(\mathrm{p}-1)+\mathrm{a}$, according to Fermat's Little Theorem, we know:
$$
\begin{array}{c}
a^{n}-n^{a} \equiv a^{a}-a^{a} \equiv 0(\bmod p) \\
\text { Also, }\left(a^{n}-n^{a}\right) \mid\left(b^{n}-n^{b}\right) \text {, so } p \mid\left(b^{n}-n^{b}\right) .
\end{array}
$$

According to Fermat's Little Theorem, we know:
$$
b^{n}-n^{b} \equiv b^{a}-a^{b}(\bmod p)
$$

Thus, $p \mid\left(b^{a}-a^{b}\right)$. Noting that $p>\left|a^{b}-b^{a}\right|$, it can only be that $a^{b}=b^{a}$.
When $a=2$, then $2^{b}=b^{2}$, we know $b=2, 4$ satisfy the condition.
When $a \geq 3$, according to the lemma, it can only be that $b=a$. And in this case, it is obviously satisfied.
In summary, the pairs $(\mathrm{a}, \mathrm{b})=(2,4)$ or $(\mathrm{k}, \mathrm{k})$, where $\mathrm{k}$ is any positive integer greater than 1, satisfy the condition.