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!
 A: Write $\Delta\geq0$. 
It must help!
$$x^2-(y+1)x+y^2-y=0,$$
which gives $$(y+1)^2-4(y^2-y)\geq0$$ or
$$3y^2-6y-1\leq0$$ or
$$1-\frac{2}{\sqrt3}\leq y\leq1+\frac{2}{\sqrt3},$$
which gives $$0\leq y\leq2,$$
which gives all solutions:
For $y=0$ we get $x^2-x=0$, which gives $(0,0)$ and $(1,0)$.
For $y=1$ we get $x^2-2x=0$, which gives $(0,1)$ and $(2,1)$.
For $y=2$ we get $x^2-3x+2=0$, which gives $(2,2)$ and $(1,2).$
A: $x+y=x^2-xy+y^2\to y^2-(x+1) y+x^2-x=0$
$y=\dfrac{1}{2} \left(1+x\pm\sqrt{-3 x^2+6 x+1}\right)$
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
A: $x+y=x^2-xy+y^2
$.
Since this is
symmetrical in
$x$ and $y$,
we can assume that
$x \le y$.
If $x=y$,
this becomes
$2x = x^2$,
so
$x=0$ or $x=2$.
If $x < y$,
then
$2y
\gt x+y
=(x-y/2)^2+3y^2/4
\ge 3y^2/4
$.
There is no solution if
$y \le 0$.
If $ > 0$,
$2 \gt 3y/4
$
or
$y \lt 8/3$
so
$y \le 2$.
If $y=2$,
then
$x=0$ or $x=1$;
of there,
only $x=1$ works.
If $y=1$,
then $x=0$ works.
