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My professor gave us a worksheet with diophantine equations including this one he claims that it is one of the easier ones but that it has a unique solution.

Find all integer solutions to the equation $$x^2 − x = y^5 − y$$.

Also my professor will give extra credit for all prime number solutions (x and y).

Jamie Lacey
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  • If the equation was for $y^2$ instead of $y^5$, we would get that $x + y = 1$. – Mr Pie Feb 21 '18 at 04:52
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    Solutions include $y = -1, 0, 1, 2, 3, 30$ corresponding to $x = (0,1)$, $(0,1)$, $(0,1)$, $(-5,6)$, $(-15, 16)$, $(-4929,4930)$ respectively. I suspect those are all, but don't have a proof. The curve has genus $2$, so by Faltings there should be only finitely many rational points. – Robert Israel Feb 21 '18 at 05:01
  • Note that $y^5-y$ is divisible by $5.$ So either $x=5$ or $x\equiv 1\pmod{10}$. Also, $y^5-y$ is divisible by $3$. So $x=5$ doesn't work, so $x\equiv 1\pmod{30}$. – Thomas Andrews Feb 21 '18 at 05:07

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Rearranging the equation gives $$(x^2-y^2)-(x-y)=0$$ Now note that $x^2-y^2=(x-y)(x+y)$.

I'll let you take it from here.

Clive Newstead
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  • Sorry it was supposed to be y^5 not y^2 – Jamie Lacey Feb 21 '18 at 04:48
  • Well this brings forward that $$(x-y)(x+y -1) = 0$$ so either $x - y = 0$ or $x + y = 1$. I assume that $x$ and $y$ are distinct so we don't consider $x - y = 0$, but if there exist prime number solution pairs $(x, y)$, then this is impossible because given $x$ and $y$ are prime, $x + y\neq 1$. The same applies for all integers $x$ and $y$. – Mr Pie Feb 21 '18 at 04:50