I recently saw a question that I couldn´t answer, so I decided to do a little C code to test if some solutions were possible but I got nothing. The problem is:
If $m,n, p \in \mathbb{Z}^+$ give the number of solutions of $$4mn-m-n=p^2$$ I couldn't find any answer up to $100$ for $m$ or $n$, and it really surprises me that an equation with so many degrees of freedom apparently doesnt have solutions. Any help is greatly appreciated.
Multiplying your equation by four and regrouping allows us to rewrite it in the form $$ (4m-1)(4n-1)=4p^2+1. $$ Let $q$ be any prime factor of r.h.s. Clearly $q$ is odd. As $$ (2p)^2=4p^2\equiv-1\pmod{q}, $$ we see that $-1$ is a quadratic residue modulo $q$. For odd primes this is known to imply that $q\equiv1\pmod4$. But the left hand side manifestly also has prime divisors congruent to $-1\pmod4$. Both $4n-1$ and $4m-1$ must have at least one such prime factor. This is a contradiction. Therefore there are no solutions with $m,n,p$ positive.