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The following problem on Diophantine equation is still solved or not I don't know. However, I found few solutions by trail and error method.

Problem: $X^2 - X = Y^5 - Y$ has integer solutions or not?

Solution: As part of my regular involvement on this problem, I have found the following set of integer solutions by trail and error method or inspection method.

i.e., $(0, -1), (1, -1), (0, 0) =$ trivial solution, $(1, 0),$ $(0, 1)$ and $(1, 1)$.

If there exits still more solutions apart from above, HOW TO FIND SUCH SOLUTIONS? Importantly, is there any method to find solutions instead of trail or inspection method?

with regards, mahoob

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  • $\begingroup$ You might try Y = 2. Or even 3. $\endgroup$
    – The Masked Avenger
    Commented Jun 5, 2015 at 6:32
  • $\begingroup$ @TheMaskedAvenger! I am looking for a method/procedure to find all set of integer solutions, instead of plugging values for $X$ or $Y$. $\endgroup$
    – mahoob
    Commented Jun 5, 2015 at 6:38
  • $\begingroup$ Don't forget $Y = 30$. $\endgroup$
    – Robert Israel
    Commented Jun 5, 2015 at 6:52
  • $\begingroup$ @RobertIsrael! well, how you got the value for $Y = 30?$. I need a traditional method of finding all set of integer solutions, rather than plugging. $\endgroup$
    – mahoob
    Commented Jun 5, 2015 at 6:54
  • $\begingroup$ The curve $X^2-X-Y^5+Y$ has genus $2$ (according to Maple), so there are only finitely many rational points by Faltings's theorem. I don't know if there is an effective way to find all of them. $\endgroup$
    – Robert Israel
    Commented Jun 5, 2015 at 6:56

2 Answers 2

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If one could prove the explicit $abc$-conjecture, i.e., with an explicit constant $C$, then one would have explicit bounds on the heights of the rational points in Falting's theorem. If this is not available, one can try other methods. For example, consider the Fermat quartics $x^4+y^4=c$. For $c=17$ this is Serre's curve. Serre asked whether or not there are rational solutions other than the obvious ones $(\pm 1,\pm 2),(\pm 2,\pm 1)$. This could be proved by Flyn in $2001$ using Chabauty techniques. All other cases $c\le 81$ can be solved easier, i.e., by local methods or by a map onto an elliptic curve of rank $0$.

All integer solutions of $y^2-y=x^5-x$ have beed determined by Yann Bugeaud, Maurice Mignotte, Samir Siksek, Michael Stoll and Szabolcs Tengely in $2008$. They are $$ (x,y)=(-1,0),(-1,1),(0,0),(0,1),(1,0),(1,1),(2,-5),(2,6),(3,-15),(3,16),(30,-4929), (30,4939). $$

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  • $\begingroup$ ! is there any easy method to get the all solutions, which you have given in your post? $\endgroup$
    – mahoob
    Commented Jun 5, 2015 at 8:53
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    $\begingroup$ It is easy to find these solutions, but not easy to show that there are no others. The authors use a "powerful variant of the Mordell–Weil sieve", among other tools. $\endgroup$
    – Dietrich Burde
    Commented Jun 5, 2015 at 9:03
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It is solved: http://msp.org/ant/2008/2-8/ant-v2-n8-p01-p.pdf. Also, the related equation $$x^2-x=y^7-y$$ has been completely solved by Stoll: http://www.mathe2.uni-bayreuth.de/stoll/papers/Kummer-g3-hyp-2014-05-15.pdf. You may also find some results in case of the more general equation $$x^p-x=y^q-y$$ in a paper by Mignotte and Pethő: On the diophantine equation $x^p-x=y^q-y$, Publicacions Matematiques, 43 (1999), 207--216.

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