I'm trying to show that the Diophantine equation $x^2+5=y^3$ has no integer solutions using the fact that $\mathbb Z[ \sqrt{-5}]$ has class number two. I think I have the general idea, but I'm having a tough time establishing the statement in bold below. Any ideas?
The equation gives a factorization into ideals
$$
(x+\sqrt{-5})(x-\sqrt{-5})=(y)^3
$$
in $\mathbb Z[ \sqrt{-5}]$. Assuming that $(x+\sqrt{-5})$ and $(x-\sqrt{-5})$ are relatively prime, we know that $(x+\sqrt{-5})=\mathfrak a^3$ and $(x-\sqrt{-5})=\mathfrak b^3$ for some ideals $\mathfrak a, \mathfrak b$ in $\mathbb Z[ \sqrt{-5}]$. Then the classes of $\mathfrak a$ and $\mathfrak b$ have order dividing 3 in the class group, hence they must both be principal, generated by some $\alpha,\beta\in \mathbb Z[\sqrt{-5}]$, respectively, as the class number is 2. Hence $\alpha^3=x+\sqrt{-5}$ up to units ($\pm 1$), and matching up real and imaginary parts then gives a contradiction.
How can I show that the ideals $(x+\sqrt{-5})$ and $(x-\sqrt{-5})$ are relatively prime? I was trying something along the lines of letting $\mathfrak p$ be a prime of $\mathbb Z[\sqrt{-5}]$ diving both ideals, in which case it also divides their sum (gcd). Then $$2\sqrt{-5}=x+\sqrt{-5}-(x-\sqrt{-5})\in (x+\sqrt{-5},x-\sqrt{-5})\subseteq \mathfrak p,$$ so $\mathfrak p\mid (2\sqrt{-5})$… But I'm not totally sure where I'm going with this.
Best Answer
$\mathfrak p\mid(2\sqrt{-5})$, so $\mathfrak p\mid(2)$ or $\mathfrak p\mid(\sqrt{-5})$. Since $(2),(\sqrt{-5})$ are prime, $\mathfrak p$ is equal to one of them. We can't have $\mathfrak p=(2)$, since $2\nmid x+\sqrt{-5}$, so $\mathfrak p=(\sqrt{-5})$, so $\sqrt{-5}\mid x+\sqrt{-5}$. This implies $5\mid x$, and you can derive a contradiction from $x^2+5=y^3$.
EDIT: As Yong Hao Ng notes in the comment, $(2)$ is not a prime ideal - it factors as $(2,1+\sqrt{-5})^2$, so we get that $\mathfrak p=(2,1+\sqrt{-5})$. Then $x+\sqrt{-5}\in(2,1+\sqrt{-5})$, so $x-1\in(2,1+\sqrt{-5})$ and we get $x$ is odd. Now we get a contradiction from $x^2+5\equiv 2\pmod 4$, since $y^3$ can't be even but indivisible by $4$.