Solving $6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$, $a,b,p,q \neq 0$ Is there an efficient way to find solutions to the equation:
$6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$ and $a,b,p,q \neq 0$
If the equation has no solutions, how could you prove that, that is the case?
Edit: The equation can in fact be expressed in the form;
$\frac{(6a+1)(6b+1)}{6}- \frac{1}{6}  - \frac{(36p+13)(36q+19)}{36} - \frac{247}{36} =7$
which is the same as
$6(6a+1)(6b+1)-(36p+13)(36q+19)=11$
but I  am still left wondering how to move forwards.

Partial answer:
You can factor:
$$\frac{1}{6} (6a + 1)(6b + 1) = 36(p + \frac{13}{36})(q + \frac{19}{36}) + \frac{11}{36}$$
$$6(6a+1)(6b+1) = (36p + 13)(36q + 19) + 11$$
Note that $RHS \equiv LHS \equiv 6 \pmod {36}$. Some trial and error with $p, q$ yields $(5, 7, 1, 4), (2, 32, 1, 8), (3, 27, 1, 10), (9, 11, 1, 12)$. With $p = 2$, the first solution is $(15, 5, 2, 5)$. Solutions certainly exist.
