Find all primes $(p,q)$ such that $p|q+6$ and $q|p+7$ Find all primes $(p,q)$ such that $p|q+6$ and $q|p+7$
I haven't found any. I initially started from $p,q\gt 3$ since, by a simple substitution you get, if of $p=2$ then $q|8$ and $q=2$, but then $2|9$ and it's a contradiction. Similarly happens with 3. Then, $p,q$ are odd and greater than 3. Also, I tried using   $p,q\equiv \pm1\pmod 6$ and linear combinations, but I haven't gotten anything and I don't know how to proceed.
I would prefer a suggestion rather than an answer, if possible without congruences, thanks beforehand.

Since $p\,|\,q+6$ we have $q+6≥p$.
Since $q\,|\,p+7$ we have $p+7≥q$ or, equivalently $p≥q-7$
We combine these to get $$q+6≥p≥q-7$$
We now distinguish two cases:
Case I:  $q+6>p$
Then, since $p$ is a divisor of $q+6$ we must have $\frac {q+6}2≥p$.  Thus $$\frac {q+6}2≥q-7\implies q+6≥2q-14\implies 20≥q$$
Thus we only have to check $q\in \{2,3,5,7,11,13,17,19\}$  with $p≤\frac {q+6}2$.  By trial and error, there are no solutions of this type.
Case II:  $q+6=p$.
Then we must have $p+7>q$ whence $q≤\frac {p+7}2$ or $2q-7≤p=q+6$ so $q≤13$.  We check the various cases and see that $(p,q)=(19,13)$ is the only solution of this type.
