# Number of non negative Integral solutions

Need to find the non negative integral solutions for the equation

$x+y+xy=x^{3}+y^{3}$

I have tried various methods for simplifying the RHS and LHS but could not arrive at the solution, so any help will be appreciated

First get a bound for possible solutions. Suppose without loss of generality that $x \leqslant y$. Then you have

$$(y+1)^2 = y^2 + y + y + 1 > xy + y + x = y^3 + x^3 \geqslant y^3.$$

That gives you $y \leqslant 2$. But for $y = 2$, we have $(2+1)^2 = 9$ and $2^3 = 8$, so the equality could only hold if $x = y$, or the first inequality would lose too much, and $x = 0$, or the second inequality would lose too much. So actually $y \leqslant 1$. Then it is a simple exhaustive search that yields the solutions $(0,0),\, (0,1),\, (1,0)$.

• Can you tell me how did you get to the left half of the inequality?
– Bach
Commented Oct 29, 2013 at 14:13
• I assumed $x\leqslant y$, so I replaced all occurrences of $x$ on the LHS with $y$, that yields $y+y+y^2 = y(y+2) = (y+1)^2-1$. The basic insight is that the left hand side grows at most quadratically, while the right grows cubically, so the solutions must all be small, and one uses such an estimate to restrict the search space. Commented Oct 29, 2013 at 14:20
• Ok Thanks, I got it
– Bach
Commented Oct 29, 2013 at 14:51
• Also, @daniel, if the RHS of the question would have been $x^{2}+y^{2}$, then how would I proceed with that?
– Bach
Commented Oct 29, 2013 at 14:58
• Similar. Again suppose $x \leqslant y$. If $x$ and $y$ are close and not too small, the left hand side is $\approx y^2$, and the right hand side $\approx 2y^2$, so that cannot be. If $x$ is much smaller than $y$, then the left hand side is smaller than $y^2$, so that cannot be either. If $x = y$, then $y(y+2) = 2y^2 \Rightarrow 2y = y^2 \Rightarrow y \in \{0,2\}$. If $x = y-1$, then $y(y-1)+y+(y-1) = y^2 + y-1 = y^2 + (y-1)^2 = 2y^2 - 2y + 1$, and that is equivalent to $0 = y^2 - 3y + 2 = (y-1)(y-2)$. If $x \leqslant y-2$, the left hand side is $\leqslant y(y-2) + y + (y-2) = y^2 - 2 < y^2$. Commented Oct 29, 2013 at 15:09