1

I've recently been wondering how to solve the equation of mordell for $k=9$, namely:

$y^{2}=x^{3}+9.$

It reduced to solving the Thue equation $\lvert\,a^{2}-2b^{3}\rvert=3.$ Interestingly, the equation has several solutions (albeit finitely many, since it is a Thue equation). For instance, $253^{2}=40^{3}+9$ is the biggest one I've found. Can we solve it by somewhat elementary means? I know Tzanakis and de Weger have a method bounding a and b, but I Wanted a clean solutions. Also, if anyone knows where I can find a proof for $k=-17$ (I think it was Nagell who solved it), please post a link in the comments.

JohnColtraneisJC
  • 1,919
user211570
  • 31
  • E_+00009: r = 1 t = 3 #III = 1 E(Q) = <(-2, 1)> x <(0, 3)> R = 0.8146954406 10 integral points 1. (0, 3) = (0, 3) 2. (0, -3) = -(0, 3) 3. (-2, 1) = 1 * (-2, 1) 4. (-2, -1) = -(-2, 1) 5. (3, 6) = (0, -3) - 1 * (-2, 1) 6. (3, -6) = -(3, 6) 7. (6, 15) = (0, -3) + 1 * (-2, 1) 8. (6, -15) = -(6, 15) 9. (40, 253) = -2 * (-2, 1) 10. (40, -253) = -(40, 253) – Will Jagy Mar 21 '15 at 18:59
  • So it really is the biggest solution.Is there any way to solve it without elliptic curves? – user211570 Mar 21 '15 at 19:10
  • 1
    I wouldn't know for sure. Given the amount of effort that goes into integer points on elliptic curves, I suggest it is hard to find them all, perhaps also hard to prove that no elementary trick will do the job, since some Mordell curves fall to completely elementary tricks. – Will Jagy Mar 21 '15 at 19:13
  • It's curious, because I know that Nagell solved the equation for k=-17, giving a total of 16 solutions and he didnt resort to elliptic curves, so I bet one can solve this one in a similar manner.Sadly, I cannot find the paper with the proof. – user211570 Mar 21 '15 at 19:17
  • @user211570 Please ask a new question for $k=-17$. – Bart Michels Apr 10 '15 at 09:18
  • @WillJagy : for $c=+9=+3^2$ or in general: $c=d^2$ being a square I think this reduces to the problem of finiteness of solutions for the perfect powers because we can always write $(y-d)(y+d)=x^3$ and have that the two factors on the lhs have distance $2d$ and have defined and small common factors (usually $2$ and $d$) - so all other involved factors must be perfect third powers. I have not enough concentration at the moment to finish my scribbles but I've put them at https://math.meta.stackexchange.com/a/9386 where it remains for a short time. Perhaps I can come back to this later. – Gottfried Helms Aug 01 '17 at 07:50

0 Answers0