For fixed positive integer a and positive $z$ in $$4a x - y^2 + z <0,$$ find how many integer solutions $(x,y)$ there are as a function of $z \in \mathbb{Z}_+$, subject to constraints $x \geq 0$ and $a \geq y \geq 0$.
This form comes from the quadratic polynomial discriminant. I am not sure where to begin as this isn't the usual quadratic Diophantine inequality (or equation for that matter) that I find through search engines.
We can rewrite the starting inequality as $x < \frac{y^2-z}{4a}$. Since we also know that $x \ge 0$, this can be rewritten to $$0 \le x < \frac{y^2-z}{4a}, 0 \le y \le a$$
Note that $y$ can be further restricted to $\sqrt{z} < y \le a$. This is because if $y \le \sqrt{z}$, then the upper bound on $x$ would not be positive.
From here, we can go through the possible $y$ and add: $$\sum_{y= \lfloor \sqrt{z} \rfloor + 1} ^ a \left\lceil \frac{y^2-z}{4a} \right\rceil$$
This is the final answer for the number of lattice points in the region.