Equation $\left\lfloor \frac{y-y^2}{1+\frac{y2}x}\right\rfloor = a$

Question

Let $a \in \mathbb N$. Find all the integer values $x$ for which equation $$\left\lfloor \frac{y-y^2}{1+\frac{y2}x}\right\rfloor = a$$ has solutions in integers.

My work so far:

I proved that $$-2a-1 \le x \le -1$$

Answer 1

This answer proves the following :

For a given $a\ge 1$, $x$ has to satisfy $$-2a\le x\le -1$$

Suppose that $x\gt 0$. Then, since $y-y^2\le 0$ and $y^2\ge 0$ for all $y\in\mathbb Z$, we get $$\frac{y-y^2}{1+\frac{y^2}{x}}\le 0\implies a=\left\lfloor\frac{y-y^2}{1+\frac{y^2}{x}}\right\rfloor\le 0$$ which contradicts that $a\ge 1$. Therefore, we have $x\le -1$.

Suppose that $x\le -2a-1$. Since $1+\frac{y^2}{x}$ has to be negative, we have $$\begin{align}x\le -2a-1&\implies 0\gt 1+\frac{y^2}x\ge 1+\frac{y^2}{-2a-1}\\&\implies \frac{1}{1+\frac{y^2}{x}}\le\frac{1}{1+\frac{y^2}{-2a-1}}\\&\implies a+1\gt\frac{y-y^2}{1+\frac{y^2}{x}}\ge\frac{y-y^2}{1+\frac{y^2}{-2a-1}}\\&\implies (a+1)\left(1+\frac{y^2}{-2a-1}\right)\lt y-y^2\\&\implies (a+1)(-2a-1+y^2)\gt (y-y^2)(-2a-1)\\&\implies ay^2+(-2a-1)y+(a+1)(2a+1)\lt 0\tag1\end{align}$$

Now the discriminant of $ay^2+(-2a-1)y+(a+1)(2a+1)=0$ is $$(-2a-1)^2-4a(a+1)(2a+1)=-(2a+1)(4a^2+a+(a-1))$$ This is negative for $a\ge 1$. Hence, there is no $y$ satisfying $(1)$.

Therefore, we have $-2a\le x\le -1$.