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.

From $q\mid p+7$ we have $p+7=qk$ for some positive integer $k$.


*

*If $k=1$ then $p+7=q$ so $p,q$ one is even, so $p=2$ and $q=9$. Not good.

*If $k=2$ then $p+7=2q$ and since $p\mid 2q+12$, then $p\mid 19$, so $p=19$ and 
$q=13$. 

*If $k=3$ then $p+7=3q$ so $p,q$ one is even, so $p=2$ and $q=3$ which doesn't work in second relation.

*If $k\geq 4$ then $p+7\geq 4q$. But $p\leq q+6$ so $3q\leq 13\implies q\leq 3$. If $q=2$ then $p=2$ which doesn't work and if $q=3$ then $p= 3$ which also doesn't work.


Conclusion: $p= 19$ and $q=13$.
