9

I'm trying to find integer solutions to $x^2 + 11 = y^3$. I guess it will work as follows: work in the ring of integers $\mathcal{O}_K$ of the number field $K = \mathbb{Q}(\alpha)$ with $\alpha = \sqrt{-11}$. This ring is $\mathcal O_K = \mathbb{Z}[\frac{1 + \alpha}{2}]$. Here te equation becomes $(x + \alpha)(x-\alpha) = y^3$.

Now I guess we should show that the ideals $(x+\alpha)$ and $(x-\alpha)$ are coprime in $\mathcal{O}_K$. It follows that $(x + \alpha) = I^3$ for some ideal $I\subset \mathcal{O}_K$. I already showed that the class group $Cl(\mathcal O_K)$ is trivial and therefore $I$ is principal. So up to units of $\mathcal O_K$ (but these are just $\pm 1$, so we do not need to worry about them) we have $x + \alpha = \left(a + b\frac{1 + \alpha}{2}\right)^3$ for some $a, b\in \mathbb{Z}$. This equation now yield some condition for $a$ and $b$ and we are able to find the solutions to $x^2 + 11 = y^3$, they are $(\pm 4, 3)$. I have two questions.

  1. Is this proof correct (when given with all the details of course)? Do I miss something?
  2. How do I prove that $(x + \alpha)$ and $(x-\alpha)$ are coprime in $\mathcal O_K$ ? This probably should not be that hard, but I haven't succeeded yet.

Thanks!

ArtW
  • 3,495
Yo.
  • 167
  • 3
    $(\pm 58/15)$ is another solution – Peter Jan 23 '18 at 13:58
  • 3
    @DietrichBurde The question you link to asks about a proof that the Mordell equation has no solutions for certian values of $k$, by using quadratic reciprocity. This question asks about a completely different proof for one specific value of $k$. Though the answers there provide information that is also useful for this question, I wouldn't say they are duplicates. – Servaes Jan 23 '18 at 14:55
  • 2
    @Servaes Well, the duplicate contains several links, in particular the article by K. Conrad, dealing with many of these cases, with and without integer solutions. The above type of proof, using rings of integers is also among them. I have the feeling that very much has been said already on Mordell curves. – Dietrich Burde Jan 23 '18 at 14:58
  • Isn't this an example of a Bachet Equation? Or am I mistaken? Are they the same thing if I am not? – Mr Pie Jan 23 '18 at 14:59
  • 2
    @user477343 Yes, it is also called Mordell-Bachet, see the duplicate, and the Wikipedia link. – Dietrich Burde Jan 23 '18 at 15:00
  • @DietrichBurde oh thank you!! I did not know that :) – Mr Pie Jan 23 '18 at 15:03
  • 1
    You seem to have missed the solutions $x=\pm58$ to the equation $x + \alpha = \left(a + b\frac{1 + \alpha}{2}\right)^3$,corresponding to $a=-1,3$ and $b=2$, – Servaes Jan 23 '18 at 15:07
  • Yes, missed the other solution, thanks! – Yo. Jan 23 '18 at 15:39
  • Reopen votes needed. –  Jan 15 '23 at 22:43

1 Answers1

8

The ideals $(x-\alpha)$ and $(x+\alpha)$ are not necessarily coprime. If $\mathfrak{p}$ is a prime ideal dividing both $(x+\alpha)$ and $(x-\alpha)$, then $2x,2\alpha,x^2+11\in \mathfrak{p}$. Let $p\in \mathbb Z$ lie above $\mathfrak{p}$ (i.e. $\mathfrak{p}\cap \mathbb{Z}=(p)$), then there are two cases.

  • If $p\ne 2$ we must have $\alpha\in \mathfrak{p}$, so that $p=11$. But then $x^2+11$ is a multiple of $11$, so $11\mid x,y$ and we obtain a contradiction because $11^2\mid x^2+11$. Now proceed with your proof to obtain solutions $(\pm 4,3)$. Edit: and $(\pm 58,15)$.
  • $p=2$: we know $x$ is odd by $x^2+11\in (2)$. Check that $2\mathcal{O}_K$ is a prime ideal, so this forces $\mathfrak{p}=2\mathcal{O}_K$. It follows that $(x+\alpha)=2I, (x-\alpha)=2J$ where $I=(\frac{x-1}{2}+\frac{1+\alpha}{2})$, $J=(\frac{x+1}{2}-\frac{1+\alpha}{2})$. Now one can show that $I,J$ are coprime (hint: $x\in I+J$ is odd). Now one can reason analogously as in the previous case to find that $(x+\alpha)=2\beta^3$ for some $\beta\in\mathcal{O}_K$, and here we find the remaining solutions, which I leave to you.

EDIT: As pointed out in the comments, the second case is superfluous by modulo 8 considerations. but I won’t edit it because I believe it gives a good idea of what to do in the general case when you have ideals that are not coprime.

ArtW
  • 3,495
  • 1
    +1 Alternatively one could argue that $2$ cannot divide both $x+\alpha$ and $x-\alpha$ as then $x^2+11\equiv0\pmod4$, which is impossible. This uses the fact that $2\mathcal{O}_K$ is prime. It also shows that the solutions $(\pm58,15)$ come from solving $x + \alpha = \left(a + b\frac{1 + \alpha}{2}\right)^3$. P.S. I corrected the first pair of solutions; they are $(\pm4,3)$. – Servaes Jan 23 '18 at 15:18
  • Actually I think the second case is not necessary, because $x$ can't be odd: If $x$ is odd, then $x^2 \equiv 1 \mod 8$ and thus $x^2 + 11\equiv 4$ mod 8, but there exists no $y\in \mathbb{Z}$ with $y^3 \equiv 4\mod 8$ so we arrive at a contradiction. Or am I mistaken? Thanks for the answer! The rest was clear ! – Yo. Jan 23 '18 at 15:37
  • @Servaes I now see you actually already made my comment. Sorry for that. – Yo. Jan 23 '18 at 15:49
  • Thanks for pointing that out! I haven’t removed the case in the edit however, because it gives a more general strategy. – ArtW Jan 23 '18 at 17:37
  • Excuse for commenting on this but could you please explain clearer why I and J must be coprime? – Mystery girl Feb 12 '24 at 15:03