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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.

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why don't you provide a link to your referred question? It'll be easier for future reference. –  lab bhattacharjee Aug 3 '13 at 16:38
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2 Answers

up vote 5 down vote accepted



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

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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.

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$x^2-x=x$ for x different than 0? –  user4140 Aug 3 '13 at 16:36
Why are you assuming they must be positive? –  Calvin Lin Aug 3 '13 at 16:37
Diophantine equations usually look for positive integer solutions. –  marty cohen Aug 3 '13 at 16:41
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