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I’m blind about integer solutions of a polynomial. I have no number theory background, but I’m curious about how to figure out all integer solutions of a polynomial, for example this question. It is said there are only two solutions, but I can’t give a proof. What I can show is that there are no even solutions. I am actually curious about the general pattern to consider this kind of problem. Can anyone give me some ideas or references? Thanks!


marked as duplicate by Dietrich Burde, punctured dusk, John Gowers, Jonas Meyer, Sujaan Kunalan Apr 3 '15 at 14:52

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    $\begingroup$ I am quite sure this has been asked and answered before. The solution uses a little algebraic number theory. I have a solution on file, and if no one finds the duplicate, I can write out a solution tomorrow. $\endgroup$ – André Nicolas Apr 3 '15 at 5:51
  • $\begingroup$ This equation is a special case of the Mordell equation, $y^2 =x^3 +A$. A tremendous amount of work has been done on the Mordell equation, and solutions have been tabulated for large ranges of values of $A $, for example, here. $\endgroup$ – Alex Ravsky Apr 3 '15 at 5:58
  • $\begingroup$ To add a bit, the obvious solutions $x=\pm 5$, $y=3$ are the only ones, and the proof I am thinking of uses the fact that $\mathbb{Z}[\sqrt{-2}]$ has unique factorization. $\endgroup$ – André Nicolas Apr 3 '15 at 5:59
  • $\begingroup$ @AndréNicolas: Sorry, but how do you get the solutions $(5, 3)$ and $(-5, 3)$? They do not solve the equation $y^{2} = x^{3} - 2$. Should not it be $(3, 5)$? And $(-3, 5)$ does not work because $(-3)^{3} - 2 = -29 \neq 5^{2}$. $\endgroup$ – Benicio Apr 3 '15 at 6:02
  • $\begingroup$ Sorry, I should have written $y=\pm 5$, $x=3$. $\endgroup$ – André Nicolas Apr 3 '15 at 6:05