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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!

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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