Balkan MO :Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$. Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.
This problem was posted 2 times in MSE, but was never solved. Can someone help me?
I am supposed to use orders and mods

Another approach:
Due to Euler's criterion:
$A=3^{\frac{7-1}2=3}+1 \equiv 0 \ mod (7)$
$3\times 3^{q-1}+1\equiv 0 \ mod (7)$
These two relations give $p=q=3$
In this case $A=28$.
We check this with RHS, i.e $B=11^p+17^q$:
$11\equiv 4 \mod (7)$ , $\rightarrow 11^p \equiv 4^p \mod (7)=7a+4^p$
$17 \equiv 3 \ mod (7)$, $\rightarrow 17^q\equiv 3^q \ mod (7)=7b+3^q$
$17^q$ and $3^q$ are odd so $7b$ is even, therefore:
$17^q=14 b_1+3^q$
⇒ $11^p+13^q=7a+7b+4^p+3^q$
$11^p+17^q$ and $4^p+3^q$ are odd so $7a+7b$ must be even; $7b=14b_1$ is even, so $7a$ must also be even, let $7a=14a_1$, so we have:
$11^p+17^q=14(a_1+b_1)+4^p+3^q$
$4^p+3^q=(7-3)^p+3^q= 7t -3^p+3^q$
We need $11^p+11^q\equiv (4^p+3^q)\equiv 0 \ mod (7)$
This is only possible if $p=q$ such that:
$4^p+3^q=(7-3)^p+3^q= 7t +(-3^p+3^q=0)=7t$
Now if $p=q=3$ then $14a_1+14b_1=28 k$, because:
$11^3+17^3=223\times 28$
Hence $A=28\big|B$
That is $p=q=3$ can be a solution.
