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

Question

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) $$\lfloor\frac{x^2-y^3}{x+y^2} \rfloor=\lfloor\frac{x^2+2xy^2+y^4-2xy^2-y^4-y^3}{x+y^2} \rfloor=x+y^2-\lfloor\frac{2xy^2+y^4+y^3}{x+y^2} \rfloor$$

Answer 1

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.

Answer 2

If $x\ge0$ it is easy to show that the equation cannot hold. If $x=y=0$ that is obvious. So assume $x\ge0,y\ne0$. Then $x+y^2>0$, and we have $(x-y)(x+y^2)-(x^2-y^3)=xy(y-1)\ge0$, so dividing by $x+y^2$ we have $x-y\ge\frac{x^2-y^3}{x+y^2}$. The LHS is an integer, so we have $1+x-y>\lfloor\frac{x^2-y^3}{x+y^2}\rfloor$.

We now need to consider $x\le0$. For $|x|<y^2$, we see, looking carefully at the above, that it proves that $x-y\ge\frac{x^2-y^3}{x+y^2}$. Hence $1+x-y>\lfloor\frac{x^2-y^3}{x+y^2}\rfloor$ and there are no solutions.

For $|x|>y^2$ we need the slightly stronger $2+x-y\le\frac{x^2-y^3}{x+y^2}$ to conclude there are no solutions. We have $x+y^2<0$ so that is equivalent to $(x+y^2)(2+x-y)\ge x^2-y^3$ or $x(y^2-y+2)\ge-2y^2\ (*)$.

This case needs a little more care. If $x\le-3$ then LHS $\ge3(y^2-y+2)$ and so $(*)$ holds since $y^2-3y+6=(y-\frac{3}{2})^2+\frac{3}{2}>0$, so there are no solutions.

If $x=-1$, then the original equation becomes $\lfloor\frac{1-y^3}{y^2-1}\rfloor=-y$. That fails for $y=1,0,-1,-2$ (see Note at end). For $y>1$ we have $1-y^3<-y(y^2-1)$ and hence $\frac{1-y^3}{y^2-1}<-y$, so there are no solutions. For $y\le-3$ we have $y^2+y-2=(y+2)(y-1)>0$, so $1-y^3<-y^3+y^2+y-1$ and hence $\frac{1-y^3}{y^2-1}<-y+1$. But $\frac{1-y^3}{y^2-1}>-y$ and so $\lfloor\frac{1-y^3}{y^2-1}\rfloor=-y$. In other words all $(-1,y)$ are solutions for $y\le-3$.

If $x=-2$, then the original equation becomes $\lfloor\frac{4-y^3}{y^2-2}\rfloor=-y-1$. Again that fails for $y=-1,0,1,2$. For $y\le-2$, we have $y^2-2>0$ and hence $4-y^3>-y(y^2-2)$ implies $\frac{4-y^3}{y^2-2}>-y$ which implies $\lfloor\frac{4-y^3}{y^2-2}\rfloor>-y-1$ and so there are no solutions. For $y>2$ we have $4-y^3<-y^3+2y$ and so $\frac{4-y^3}{y^2-2}<-y$. On the other hand, $y^2-2y+1=(y-1)^2+1>0$, so $4-y^3>-y^3-y^2+2y+2=(y^2-2)(-y-1)$ and hence $\frac{4-y^3}{y^2-2}>-y-1$. Hence $\lfloor\frac{4-y^3}{y^2-2}\rfloor=-y-1$. So all of $(-1,y)$ are solutions for $y\ge3$.

[$\textbf{Note}$ If $x=-y^2$ and $x^2=y^3$, then the floor expression is indeterminate. That corresponds to $(x,y)=(-1,1)$. I do not regard that as a solution.]