Consider the expression $$ a^2b^2-(p+b)a,\qquad (a,b\in \mathbb{Z}^+), $$$$ a^2b^2-ab-ap\qquad (a,b\in \mathbb{Z}^+), $$ where $p\equiv1\pmod{4}$.
Question. For every prime $p\equiv1\pmod{4}$ do there exist $a,b\in\mathbb{Z}^+$ such that $b\equiv3\pmod{4}$ and the above expression is a perfect square?
Thanks!