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 .

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    $\begingroup$ For your last question, $(a, b, c) = (2, 2, 2)$. $\endgroup$ – Theo Bendit Sep 29 at 11:12
  • $\begingroup$ For the different question, you should try to ask it separately. $\endgroup$ – YiFan Sep 29 at 11:12
  • $\begingroup$ Theo Bendit , can we find a solution to the last question if we impose another restriction ,that of them being unequal, on $a,b$ and $c$ . $\endgroup$ – user655800 Sep 29 at 11:15

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 $x=10, y=5$.

  • $\begingroup$ Thank you Mark. $\endgroup$ – user655800 Sep 29 at 11:42
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    $\begingroup$ Note that you can work further on this - setting $b=2c+1$ and separating the choice of signs to get fully parametric solutions. This will also show why $x$ and $y$ typically come with a common factor. $\endgroup$ – Mark Bennet Sep 29 at 11:44
  • $\begingroup$ You also get "mirror image solutions" if you allow both values of the $\pm$ sign. Thus $b=3$ gives both $(x_1,y_1)=(10,5)$ and $(x_2,y_2)=(-y_1,-x_1)=(-5,-10)$. $\endgroup$ – Oscar Lanzi Sep 29 at 12:08

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.


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

$$\boxed{ x = 2t^3+4t^2+3t+1,\qquad y= 2t^3+2t^2+t. }$$


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