Solve in integers $x,y$ the equation $\left\lfloor\frac{xy-xy^2}{x+y^2} \right\rfloor=a$ 
Solve in integers the equation
  $$\left\lfloor\frac{xy-xy^2}{x+y^2} \right\rfloor=a$$

My work so far:
1) If $a=1$, then $x\in\{-1;-2;-3\}$.
i) $x=-1 \Rightarrow y\ge-3$
ii) $x=-2 \Rightarrow y\ge3$
iii) $x=-3$. No solution.
2) $$a\le\frac{xy-xy^2}{x+y^2}<a+1$$

HINT.-$x+y^2\ne 0$. Consider two cases $x+y^2\gt 0$ and $x+y^2\lt 0$.
$$\boxed{x+y^2\gt 0}$$
You have $$a(x+y^2)\le xy-xy^2\lt a(x+y^2)+(x+y^2)$$ This gives two quadratic inequalities in $y$ 
$$\begin{cases}(a+x)y^2-xy+ax\le 0\\(a+x+1)y^2-xy+ax+x\gt 0\end{cases}$$
Now for $x$ fixed, $(\alpha, \beta)$ and $(\gamma,\delta)$ being the corresponding roots supposed reals and $a+x\gt 0$ you have for the values of $y$ the restrictions $$\alpha\le y\le \beta\\y\in]-\infty,\gamma[\space\cup \space]\delta,+\infty[$$ You can see that the roots play a strong role in this and the discusion of possibilities certainly don't stop here. As you say in your comment "Perhaps the problem is not of a "normal" solutions".
