Balkan Mathematical Olympiad

For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$. Observe that $N=11^p+17^p\equiv 4(\mathrm{mod}8)$, so $8\nmid 3p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3p^{q-1}+1$. Obviously, $r\notin \{3,11,17\}$. There exist $b$ such that $17b\equiv 1(\mathrm{mod}r)$. Then $r|b^{p}N\equiv a^p+1(\mathrm{mod}r)$,

where $a=11b$. Thus $r|a^{2p}-1$, but $r\nmid a^p-1$, which means that ${\mathrm{ord}}_r(a)|2p$ and ${\mathrm{ord}}_r(a)\nmid p$, i.e. ${\mathrm{ord}}_r(a)\in \{2,2p\}$.

Note that if ${\mathrm{ord}}_r(a)=2$, then $r|a^2-1\equiv (11^2-17^2)b^2(\mathrm{mod}r)$, which gives $r=7$ as the only possibility. On the other hand, ${\mathrm{ord}}_r(a)=2p$ implies $2p|r-1$. Thus, all prime divisors of $3p^{q-1}+1$ other than $2$ or $7$ are congruent to $1$ modulo $2p$, i.e. $$3p^{q-1}+1=2^{\alpha }{7}^{\beta }{p}_1^{{\gamma }_1}{p}_2^{{\gamma }_2}\cdots p_k^{{\gamma }_k},(\ast )$$ where $p_i\notin \{2,7\}$ are prime divisors with $p_i\equiv 1(\mathrm{mod}2p)$.

$$\frac{11^p+17^p}{28}=11^{p-1}-11^{p-2}17+11^{p-3}17^2-\cdots +17^{p-1}\equiv p{4}^{p-1}(\mathrm{mod}7),$$ so $11^p+17^p$ is not divisible by $7^2$ and hence $\beta \le 1$. If $q=2$, then $(\ast )$ becomes $3p+1=2^{\alpha }{7}^{\beta }{p}_1^{{\gamma }_1}{p}_2^{{\gamma }_2}\cdots p_k^{{\gamma }_k}$, but $p_i\ge 2p+1$, which is only possible if ${\gamma }_i=0$ for all $i$, i.e. $3p+1=2^{\alpha }{7}^{\beta }\in \{2,4,14,28\}$, which gives us no solutions. Thus $q>2$, which implies $4|3p^{q-1}+1$, i.e. $\alpha =2$. Now the right hand side of $(\ast )$ is congruent to $4$ or $28$ modulo $p$, which gives us $p=3$. Consequently $3^q+1\equiv 6244(\mathrm{mod}3)$ which is only possible for $q=3$. The pair $(p,q)=(3,3)$ is indeed a solution.