# diophantine equation $x(x^2-x)+y(y^2-y)=2xy$

OK so I accidentally posted a wrong equation in my previous question and I didn't realize it after it was solved. Hope it helped someone and sorry. This is the more challenging one I wanted to solve.

-
why don't you provide a link to your referred question? It'll be easier for future reference. –  lab bhattacharjee Aug 3 at 16:38

HINT

$x^3+y^3-(x+y)^2=(x+y)(x^2+y^2-xy-x-y)$

Where the second part has already been discussed in your last question(here).

-
If $x=y$, this becomes $x(x^2-x) = x^2$. A solution is $x = 0$. If $x \ne 0$, $x^2-x = x$ or $x=2$.
If $x > y > 0$, $x(x^2-x) < 2x^2$ or $x^2-x < 2x$ or $x < 3$.
This then reduces to trying $x = 2$, $y=1$ which works.
$x^2-x=x$ for x different than 0? –  Omnitic Aug 3 at 16:36