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.
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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.