Solve in integers the equation $\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$ 
Solve in integers the equation 
  $$\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$$

My attempt:
I used http://www.wolframalpha.com/:
$x=-2; y=\{3,4,5,6,7,..\}$ or $x=-1, y=\{-10,-9,...\}$.
1) Let $\lfloor\frac{x^2-y^3}{x+y^2} \rfloor=\frac{x^2-y^3}{x+y^2}-\{\frac{x^2-y^3}{x+y^2}\}$. Then
$$\frac{x^2-y^3}{x+y^2}-\{\frac{x^2-y^3}{x+y^2}\}=1+x-y$$
$$\frac{x^2-y^3}{x+y^2}-1-x+y=\{\frac{x^2-y^3}{x+y^2}\}$$
$$0\le\frac{x^2-y^3}{x+y^2}-1-x+y<1$$
2)
$$\lfloor\frac{x^2-y^3}{x+y^2} \rfloor=\lfloor\frac{x^2+2xy^2+y^4-2xy^2-y^4-y^3}{x+y^2} \rfloor=x+y^2-\lfloor\frac{2xy^2+y^4+y^3}{x+y^2} \rfloor$$

We have
\begin{align}
\lfloor \frac{x^2 - y^3}{x + y^2} \rfloor &= \lfloor \frac{x^2 + xy^2 - xy - y^3}{x + y^2} - \frac{xy^2 - xy}{x + y^2}\rfloor \\
&= \lfloor \frac{(x-y)(x + y^2)}{x + y^2} - \frac{xy^2 - xy}{x + y^2}\rfloor \\
&= \lfloor x - y - \frac{xy^2 - xy}{x + y^2}\rfloor \\
&= x - y + \lfloor - \frac{xy^2 - xy}{x + y^2}\rfloor
\end{align}
Therefore, we further have
$$
\lfloor - \frac{xy^2 - xy}{x + y^2}\rfloor = 1
$$
or say
$$
1 \leq - \frac{xy^2 - xy}{x + y^2} < 2 \tag{1}
$$
Observe that $x$ can not be non-negative because otherwise $- \frac{xy^2 - xy}{x + y^2}$ would be non-positive. We conclude that $x < 0$ and moreover, $x + y^2 > 0$. Therefore, we have
$$
(-x) \cdot \frac{y^2-y}{y^2} < (-x) \cdot \frac{y^2-y}{x + y^2} < 2 \tag{2}
$$
by inequality (1) above. Thus
$$
(-x) < \frac{2y^2}{y^2 - y} \leq 4 \quad\text{(note that $y$ is an integer)}
$$
Finally, we know that
$$
x \in \{-1, -2, -3\}
$$
and enumerating possible $x$ can solve the problem.
