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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!

Bart Michels
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    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. – André Nicolas Apr 03 '15 at 05:51
  • 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. – Alex Ravsky Apr 03 '15 at 05:58
  • 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. – André Nicolas Apr 03 '15 at 05:59
  • @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 $(-3)^{3} - 2 = -29 \neq 5^{2}$. – Yes Apr 03 '15 at 06:02
  • Sorry, I should have written $y=\pm 5$, $x=3$. – André Nicolas Apr 03 '15 at 06:05
  • We may first narrow down the solution scope by noting that $x$ can only be integers $\geq 2,$ because if $x \leq 1$ then no real number $y$ can be such that $y^{2} < 0$ and because the function $x \mapsto -\sqrt{x^{2} - 2}$ is meaningful iff $x \geq 2.$ – Yes Apr 03 '15 at 06:06
  • possible duplicate of Solutions to $y^2 = x^3 + k$?, and Diophantine equation: $x^2=y^3-2$ – Dietrich Burde Apr 03 '15 at 09:03
  • @DietrichBurde better link this to this because the one you mentioned asks for a specific proof technique. – Bart Michels Apr 03 '15 at 10:41
  • @barto: I agree. Do you know how I can change this ? I cannot retract my vote because then I am not allowed to vote again for a duplicate. – Dietrich Burde Apr 03 '15 at 10:54
  • I think it's not possible. I found this related discussion: http://meta.stackexchange.com/questions/193654 – Bart Michels Apr 03 '15 at 10:57

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