Need to find the non negative integral solutions for the equation


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 $(0,0),\, (0,1),\, (1,0)$.

  • $\begingroup$ Can you tell me how did you get to the left half of the inequality? $\endgroup$ – Bach Oct 29 '13 at 14:13
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Oct 29 '13 at 14:20
  • $\begingroup$ Ok Thanks, I got it $\endgroup$ – Bach Oct 29 '13 at 14:51
  • $\begingroup$ Also, @daniel, if the RHS of the question would have been $x^{2}+y^{2}$, then how would I proceed with that? $\endgroup$ – Bach Oct 29 '13 at 14:58
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    $\begingroup$ 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 $\leqslant y(y-2) + y + (y-2) = y^2 - 2 < y^2$. $\endgroup$ – Daniel Fischer Oct 29 '13 at 15:09

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