Find all integer solution for $x^2$ $+$ $y^2$ $=(x-y)^3$ The question is 

Find all integer solution for $x^2$ $+$ $y^2$$=(x-y)^3$ 
  For $x$ and $y $ belongs to integer

I tried this
$(x-y)^2=(x-y)^3-2xy$
Then let $x-y=a$
By solving it further and substituting $y =x-a$
We get $2x^2-2ax+a^2-a^3=0$
Now using quadratic formula 
And little but of working on conditions i derived
$x= \dfrac{m^3+m^2+m+1}4 , \dfrac{-m^3+m^2-m+1}4$
And
 $x-y=\dfrac{m^2+1}2$
Where $m$ belongs to all odd numbers 
 no doubt that this gives all solution but my doubt is 


*

*Are there infinitly many solution .If not how to prove finitely many exists 

*I want to see other elegant solutions because it took me lot of time and computations for getting the above result.
All answers would be appreciated.thanks in advance
Edit : my first querry has been solved but i would like to see some more answers.

$x=a\ GCD(x,y)\ \ ,\ \ y=b\ GCD(x,y)$ where $GCD(a,b)=1$. Now continuing Your work.
$$
\begin{aligned}
(x-y)^{2}&=(x-y)^{3}-2xy\\
2xy&=(x-y)^{2}(x-y-1)\\
\\
2ab\ (GCD(x,y))^{2}&=(GCD(x,y))^{2}(a-b)^{2}((a-b)\ GCD(x,y)-1)\\
\\
2ab&=(a-b)^{2}((a-b)\ GCD(x,y)-1)
\end{aligned}
$$
Now, $(a-b)^{2}|2ab$ but since $GCD(a,b)=1$ then $(a-b)^{2}|2$. The only integer solution is $a-b=1$. Substitute $a=b+1$ and $a-b=1$ back to the previous equation.
$$
\begin{aligned}
2(b+1)b&=GCD(x,y)-1\\
GCD(x,y)&=2(b+1)b+1
\end{aligned}
$$
So the solutions are $x=2(b+1)^{2}b+b+1$ , $y=2(b+1)b^{2}+b$ for all integers $b$. There are infinitely many solutions.
Sorry i missed a possibility that at least one of them can be zero as well. Substitute back to original equation to obtain $(0,0)\ (0,-1)\ (1,0)$.
