Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod r)$. Then $r \mid b^{p} N \equiv a^{p}+1(\bmod r)$, where $a=11 b$. Thus $r \mid a^{2 p}-1$, but $r \nmid a^{p}-1$, which means that $\operatorname{ord}_{r}(a) \mid 2 p$ and $\operatorname{ord}_{r}(a) \nmid p$, i.e. $\operatorname{ord}_{r}(a) \in\{2,2 p\}$.
Note that if $\operatorname{ord}_{r}(a)=2$, then $r \mid a^{2}-1 \equiv\left(11^{2}-17^{2}\right) b^{2}(\bmod r)$, which gives $r=7$ as the only possibility. On the other hand, $\operatorname{ord}_{r}(a)=2 p$ implies $2 p \mid r-1$. Thus, all prime divisors of $3 p^{q-1}+1$ other than 2 or 7 are congruent to 1 modulo $2 p$, i.e.

$$
3 p^{q-1}+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}
$$

where $p_{i} \notin\{2,7\}$ are prime divisors with $p_{i} \equiv 1(\bmod 2 p)$.
We already know that $\alpha \leqslant 2$. Also, note that

$$
\frac{11^{p}+17^{p}}{28}=11^{p-1}-11^{p-2} 17+11^{p-3} 17^{2}-\cdots+17^{p-1} \equiv p \cdot 4^{p-1} \quad(\bmod 7)
$$

so $11^{p}+17^{p}$ is not divisible by $7^{2}$ and hence $\beta \leqslant 1$.
If $q=2$, then $(*)$ becomes $3 p+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}$, but $p_{i} \geqslant 2 p+1$, which is only possible if $\gamma_{i}=0$ for all $i$, i.e. $3 p+1=2^{\alpha} 7^{\beta} \in\{2,4,14,28\}$, which gives us no solutions.
Thus $q>2$, which implies $4 \mid 3 p^{q-1}+1$, i.e. $\alpha=2$. Now the right hand side of $(*)$ is congruent to 4 or 28 modulo $p$, which gives us $p=3$. Consequently $3^{q}+1 \mid 6244$, which is only possible for $q=3$. The pair $(p, q)=(3,3)$ is indeed a solution.