<s>[INST] Finding integral solutions of $x+y=x^2-xy+y^2$ 
Find integral solutions of $$x+y=x^2-xy+y^2$$

I simplified the equation down to 
$$(x+y)^2 = x^3 + y^3$$
And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other method to solve this? I am thankful to those who answer!
 [/INST] $x+y=x^2-xy+y^2\to y^2-(x+1) y+x^2-x=0$
$y=\dfrac{1}{2} \left(1+x\pm\sqrt{-3 x^2+6 x+1}\right)$
discriminant must be positive $\Delta=-3 x^2+6 x+1\geq 0\to \dfrac{1}{3} \left(3-2 \sqrt{3}\right)\leq x\leq \dfrac{1}{3} \left(3+2 \sqrt{3}\right)$
which  for integer $x$ means, $0\leq x \leq 2$
For $x=0$ we get $y=0;\;1$ solutions are $\color{red}{(0,0)\;(0;\;1)}$
for $x=1$ we have $y=0;\;y=2$ so  $\color{red}{(1,0)\;(1;\;2)}$
for $x=2$ finally $y=1;\;y=2$ so  $\color{red}{(2,1)\;(2;\;2)}$
hope this helps
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