The Mordell equation $x^2 + 11 = y^3$. 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.


*

*Is this proof correct (when given with all the details of course)? Do I miss something?

*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! 
 A: 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.
