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.
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 $x &gt; y &gt; 0$,
$x(x^2-x) &lt; 2x^2$
or $x^2-x &lt; 2x$
or $x &lt; 3$.

This then reduces to
trying $x = 2$, $y=1$
which works.