Solving $6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$, $a,b,p,q \neq 0$

Question

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.

Answer 1

Note that $$(6a+1)(6b+1)=6(6ab+a+b)+1$$ and $$(36p+13)(36q+19)=36(36pq+19p+13q)+247$$ Multiply your equation by $36$ and use these formulas to rewrite it as something like $xy-zw=n$ for some number $n$.

Answer 2

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.

Answer 3

It has at least the solutions $(7,5,1,4)$ and $(5,7,1,4)$.