Let $2 \leq p < q$ be two prime numbers. Consider $p+1$ and $q+1$.
Observe that $d:=\gcd(p+1,q+1)$ can be 'anything':
(a) If $(p,q)=(2,3)$, then $(p+1,q+1)=(3,4)$, so $d=1$: $p+1,q+1$ are relatively prime.
(b) If $(p,q)=(2,5)$, then $(p+1,q+1)=(3,6)$, so $d=3=p+1$: $p+1$ divides $q+1$.
(c) If $(p,q)=(3,5)$, then $(p+1,q+1)=(4,6)$, so $d=2$: $1 < d < min\{p+1,q+1\}$.
Is it possible to determine when case $(b)$ holds, namely, when $p+1$ divides $q+1$? (also, same question for cases $(a)$ and $(b)$).
I think that when $p=2$ the answer is not difficult (If $p=2$ and $q \geq 3$, then $p+1=3$ and $q+1$ is even, so the answer depends on whether $3$ divides $q+1$ or not; in other words, if $q \cong -1 \mod 3$ then $p+1$ divides $q+1$), so further assume that $p \neq 2$.