Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$. 
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.

I think that $(0,0)$ , $(1,0)$ and  $(0,-1)$ are the only solutions to the above equation, but I'm unable to prove it.
I tried all sorts of things like working $\mod 9$ (but there are just too many cases ), a little bit of algebraic manipulations, tried to determine the parity of $x$ and $y$ etc. But they were to no avail for me.
I tried working modulo $9$ because $a^{3}\equiv 0,1$ or $-1 \pmod 9$.  
The manipulations done by me were as follows:-
         $x^2 + y^2 =(x-y)^3$ implies that by adding and subtracting $2xy$ on LHS we can rewrite the above equation as $(x-y)^2 +2xy=(x-y)^3$  . This can be rewritten as $2xy=(x-y)^3 -(x-y)^2$. This is all I could achieve here. One thing I did here was substitute $x-y=a$ and $x=a+y$ and rewrite the last equation as $2y^2 +2ay+a^2 -a^3=0$ and then I tried to find the roots of this quadratic in $y$ but this didn't work for me(I think there is something wrong with this approach, do tell me if you see it).That is all I could do. 
Another question I would like to ask is do there exist integers $a,b$ and $c$, with none of them equal to zero, which satisfy $a^2 + b^2=c^3$ ? Thank you .
 A: Using a computer , you can find some solutions like :
$$(0 ,-1),(0,0),(1,0),(10,5) , (39,26) , (100,75) , (205,164),(366,305), (595,510),(904,791),(1305,1160),(1810 ,1629)$$
So , it means that your assumption that $(0,0),(1,0)and(0,-1)$ is wrong as their exist infinitely many solutions to the equation.
Also to your second part of question , there also exist infinitely many solutions to:
$$a^2 + b^2 = c^3$$
like $$(2,2,2) , (2,11,5) ,(5,10,5) , (9,46,13) , (10,30,10),(10,198,34)$$etc.
A: To extend your existing approach, multiply by $2$ to obtain :$$0=4y^2+4ay+2a^2-2a^3=(2y+a)^2+a^2(1-2a)$$
To obtain a factorisation with integer coefficients you need $2a-1=b^2$. For convenience, multiply through by $4$ to get $$0=(4y+2a)^2+4a^2(1-2a)=(4y+b^2+1)^2-(b^2+1)^2b^2$$ And the roots are $$4y=-(b^2+1)\pm b(b^2+1)=-(1\pm b)(1+b^2)$$
Now $b$ is odd, so the right-hand side is the product of two even numbers, and any odd value of $b$ will lead to a solution.
For example $b=3$ gives $x=10, y=5$.
A: Letting $x=y+k$, we are looking for the solutions of
$$ 2y^2+2yk+k^2 = k^3 $$
$$ (2y+k)^2+k^2 = 2k^3 $$
which depend on the integer points on the elliptic curve $w^2=2z^3-z^2=z^2(2z-1)$.
We may assume that $z=\frac{q^2+1}{2}$, leading to the solution $k=\frac{q^2+1}{2},w=q\frac{q^2+1}{2},y=(q-1)\frac{q^2+1}{4},x=(q+1)\frac{q^2+1}{4}$.
Of course, in order to have $\frac{q^2+1}{4}\in\mathbb{Z}$ $q$ must be odd, $q=(2t+1)$. These leads to the solutions
$$\boxed{ x = 2t^3+4t^2+3t+1,\qquad y= 2t^3+2t^2+t. }$$
