How shall I find all natural numbers p and q such that $$p^2q^2-4(p+q)=a^2$$ for some natural number $a$? Thanks!
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Rewrite as $(pq-a)(pq+a)=4(p+q)$. This means that $pq+a\leq 4(p+q)$, or $(p-4)(q-4) + a \leq 16$.
[Adding fix noted in comments.]
For $p,q\geq 4$, there can only be finitely many such triples.
So you need to only check when at least one of $p,q<4$.
Case $q=0$ is easy.
When $q=1$, the equation becomes $p^2 - 4(p+1) = a^2$. But $p^2-4(p+1)=(p-2)^2 - 8$, so we need: $(p-2-a)(p-2+a) = 8$.
$q=2$ means $4p^2-4(p+2) = (2p-1)^2 - 9 = a^2$, or $(2p-1-a)(2p-1+a)=9$.
$q=3$, then $9p^2-4(p+3) = (3p-1)^2 + (2p-13)$, so $2p-13\leq 0$ which is only true for finitely many $p$.