Primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$

Question

From the Balkan Mathematics Olympiad 2018:

Find all prime numbers $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.

I started using Fermat's Little Theorem, which states that $a^{p-1}-1$ is divisible by $p$ if $p$ is a prime. But I could not find any way to solve the problem.

Any suggestions, please.

Answer 1

hint Assume $ p\ne q$ and both $\ne 11,17$.

then by Euler's theorem

$$3p^{q-1}+1 \equiv 4 \mod q$$ and $$11^p+17^p \equiv 11+17 \mod p$$

Answer 2

COMMENT:

$11^p≡11 mod p=k_1 .p + 11$

$17^p≡17 mod p=K_2 .p + 17$

$11^q≡11 mod q=k_3 .q + 11$

$17^p≡17 mod q=K_4 .q + 17$

Summing these relations we get:

$(11^p+17^q)+(11^q+17^p)=56+(k_1+k_2)p+(k_3+k_4)q$

P and q are primes, therefore numbers $(11^p+17^q)$ , $(11^q+17^p)$, $3.p^{q-1} +1$, $(k_1+k_2)$ and $(k_3+k_4)$ must have common divisors like$ 2, 4, 7, 8, 14, 28 and 56$.

Now $3.p^{q-1}+1$ ≤ $(11^p+17^q)= 56 m$; $m∈N$. This is the condition of question. For example with p=q=3 we get 28. I think better question is: Find all possible divisors of $11^p+17^q$ and $3p^{q-1}+1$