Question 53: Find all prime pairs $(p, q)$ such that $\left(3 p^{q-1}+1\right) \mid\left(11^{p}+17^{p}\right)$.
Question 53:
Solution: If $p=2$, then $\left(3 \cdot 2^{q-1}+1\right) \mid\left(11^{2}+17^{2}\right)$, which means $\left(3 \cdot 2^{q-1}+1\right) \mid 2 \times 5 \times 41$. Upon inspection, there is no solution.
Now assume $p \geq 3$, which is an odd prime number.
Notice that $11^{p}+17^{p} \equiv 11+17 \equiv 4(\bmod 8)$, so $8 \nmid\left(3 p^{q-1}+1\right)$. Therefore, $v_{2}\left(3 p^{q-1}+1\right) \leq 2$.

Take any odd prime factor $r$ of $3 \cdot p^{q-1}+1$, clearly $r \notin\{3,11,17\}$. At this point, $r \mid\left(11^{p}+17^{p}\right)$, which means $r \mid\left(\left(11 \cdot 17^{-1}\right)^{p}+1\right)$. Thus, the half-order $\tau$ of $11 \cdot 17^{-1}$ modulo $r$ satisfies $\tau \left\lvert\,\left(p, \frac{r-1}{2}\right)\right.$, hence $\tau=1$ or $p$. If $\tau=1$, then $r \mid\left(11 \cdot 17^{-1}+1\right) \Rightarrow r \mid (11+17) \Rightarrow r=7$; if $\tau=p$, then $p \left\lvert\, \frac{r-1}{2}\right.$, thus $r \equiv 1(\bmod 2 p)$.
Based on the above analysis, let the standard factorization of $3 \cdot p^{q-1}+1$ be:
$$
3 \cdot p^{q-1}+1=2^{\alpha} \cdot 7^{\beta} \cdot p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p^{\alpha_{k}}
$$

Here $\alpha \in\{1,2\}, \beta \in \mathbb{N}, p_{i} (1 \leq i \leq k)$ are distinct primes not in $\{2,7\}$ and congruent to 1 modulo $2p$, and $\alpha_{i} (1 \leq i \leq k)$ are positive integers.

If $p=7$, then $3 \cdot p^{q-1}+1 \equiv 1(\bmod p)$, thus $\beta=0$; if $p \neq 7$, then by the LTE lemma, $v_{7}\left(11^{p}+17^{p}\right)=v_{7}(11+17)+v_{7}(p)=1$, thus $\beta=v_{7}\left(3 p^{q-1}+1\right) \leq v_{7}\left(11^{p}+17^{p}\right)=1$. In either case, $\beta \leq 1$.

If $q=2$, then by (1), $3p+1=2^{\alpha} \cdot 7^{\beta} \cdot p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}$. Since $p_{i} \geq 2p+1$, it is only possible that $3p+1=2^{\alpha} \cdot 7^{\beta} (\alpha \leq 2, \beta \leq 1)$. Upon inspection, there is no solution in this case.
If $q \geq 3$, then $3 \cdot p^{q-1}+1 \equiv 3+1 \equiv 0(\bmod 4)$, so $\alpha=v_{2}\left(3 \cdot p^{q-1}+1\right)=2$, which means
$$
3 \cdot p^{q-1}+1=2^{2} \cdot 7^{\beta} \cdot p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}
$$

Taking (2) modulo $p$, we get $1 \equiv 2^{2} \cdot 7^{\beta} \equiv 4$ or $28(\bmod p)$, so $p=3$. Thus, $\left(3^{9}+1\right) \mid\left(11^{3}+17^{3}\right)$.
Upon inspection, only $q=3$ satisfies the condition.
In summary, the solution is $(p, q)=(3,3)$.