Is $ a^2b^2-ab-ap$ a perfect square for suitable $a,b\in \mathbb{Z}^+$? Consider the expression
$$
a^2b^2-ab-ap\qquad (a,b\in \mathbb{Z}^+),
$$
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!
 A: This is just an extended comment. 
The problem is equivalent to finding $b\equiv 3\pmod{4}$ and $c\equiv 0\pmod{4}$ such that
$$(\star)\qquad 4b^2c\mid (b+c+p)^2.$$
In fact, when $(\star)$ holds, setting $a$ equal to the quotient, i.e. $a:=\frac{(b+c+p)^2}{4b^2c}$, gives
$$a^2b^2 - ab - ap = \left(\frac{(p+b)^2 - c^2}{4bc}\right)^2.$$
Now, $(\star)$ implies that $\gcd(b,c)|p$. 
If $\gcd(b,c)=p$, then $b=pb'$ and $c=pc'$, implying that
$$4pb'^2c'\mid (c'+b'+1)^2$$
and thus $b' < c'$ and $4pb'^2c' \leq (c'+b'+1)^2 \leq 4c'^2$, i.e. $pb'^2\leq c'$. On the other hand, since $c'|(b'+1)^2$, we have $pb'^2\leq c'\leq (b'+1)^2$, which is impossible since $p\geq 5$.
Hence, $\gcd(b,c)=1$ and $(\star)$ is equivalent to existence of integers $b,c$ such that
$$\begin{cases}b\mid c+p,\\ 4c\mid (b+p)^2.\end{cases}$$

Let's consider a partial case when $c=4d$ and $d$ is squarefree. Then $(\star)$ implies that $4bd\mid b+4d+p$, and setting $q:=\frac{b+4d+p}{4bd}$ we get
$$(qb-1)(4qd-1) = qp+1.$$
In particular, we have a positive answer to the original question as soon as there is an integer $q>0$ such that $qp+1$ has a divisor $m\equiv -1\pmod{4q}$ and $\frac{m+1}{4q}$ is squarefree.
