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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 $alpha,betain mathbb Z[sqrt{-5}]$, 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 $$2sqrt{-5}=x+sqrt{-5}-(x-sqrt{-5})in (x+sqrt{-5},x-sqrt{-5})subseteq mathfrak p,$$ so $mathfrak pmid (2sqrt{-5})$... But I'm not totally sure where I'm going with this.

$mathfrak pmid(2sqrt{-5})$, so $mathfrak pmid(2)$ or $mathfrak pmid(sqrt{-5})$. Since $(2),(sqrt{-5})$ are prime, $mathfrak p$ is equal to one of them. We can't have $mathfrak p=(2)$, since $2nmid x+sqrt{-5}$, so $mathfrak p=(sqrt{-5})$, so $sqrt{-5}mid x+sqrt{-5}$. This implies $5mid x$, 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 $x-1in(2,1+sqrt{-5})$ and we get $x$ is odd. Now we get a contradiction from $x^2+5equiv 2pmod 4$, since $y^3$ can't be even but indivisible by $4$.

This has been answered many times on this site, so it is certainly a duplicate. On the other hand, it may be still worth to give the link to a very nice lecture note by Keith Conrad here, where he does proofs for many interesting cases of the Mordell equation $y^2=x^3-n$. For $n=5$ there are no integer solutions. More references here are found at OEIS A054504.

In the ideal factorization $(x+sqrt {-5})(x-sqrt {-5})=(y)^3$, "how can I show that the ideals $(x pm sqrt {-5})$ are relatively prime?" Suppose that $P$ is a common prime divisor, then $P^{2a}$ will appear in the LHS of the factorization. But $P^{3b}$ will appear in the RHS because of the cube, hence at least $P^6$ will divide both sides because of uniqueness. Following your argument, $P^3$ must divide $(2)(sqrt {-5})$. In the quadratic field $mathbf Q(sqrt {-5})$, we know that the prime $5$ is totally ramified, and $(5)=(sqrt {-5})^2$ shows that $(sqrt {-5})$ is prime. Besides $(2)=Q^2$, with $Q=(2, 1+sqrt {-5})$ (so the class of $Q$ is the unique class of order $2$ of the class group). It follows again by uniqueness of factorization that $P^3$ cannot divide $(sqrt {-5})Q^2$, and we are done.

If ideals $I,J$ are relatively prime then their sum generates the ring $R$, so we can show this. In particular we can show also that
$$
(a + bsqrt {- 5})(x+ sqrt{- 5}) + (c + dsqrt {- 5})(x - sqrt {- 5}) = 1
$$
for some $a,b,c,d$.

This gives the equations
$$
begin{align}
(a+c)x - 5(b-d) &= 1\
(a-c) + (b+d)x &= 0
end{align}
$$
So a necessary condition is $5$ does not divide $x$. This can be checked to be true from $x^2 + 5 = y^3$, but otherwise the ideals are not necessarily relatively prime in general.

Setting
$$
c = a + (b+d)x
$$
gives us
$$
2ax = (1+5(b-d)) - (b+d)x^2
$$
Taking $pmod 2$ we see that $x$ must be even, so to get an integer solution for $a$ we must have $1+5(b-d)$ divisible by $2x$.

The obvious way is to look at the linear diophantine solutions
$$
xu - 5v = 1
$$
We can always choose $u$ to be even, so now we can set $d=0, v=b$ to get
$$
begin{align}
a &= frac{1+5(b-d)}{2x} - (b+d)frac{x}{2}\
&= frac{1+5v}{2x} - vfrac{x}{2}\
&= frac{u}{2} - vfrac{x}{2}
end{align}
$$
Since $u/2$ is integral we just need $vx$ to be even, which is true based on an earlier assumption: $x$ must be even.

It remains to show that $x$ is indeed even.
If $x=2m+1$ is odd, then $y$ is even and
$$
x^2 + 5 = 4m^2+4m+ 6 equiv 2pmod 4
$$
which contradicts $y^3 equiv 0 pmod 4$. Hence $x$ is even and we are done, finding a solution to the initial equation as
$$
(u/2 -vx/2 + vsqrt {-5})(x + sqrt {-5}) + (u/2 + vx/2)(x - sqrt {-5}) = 1
$$
(Recall that $xu-5v=1$ and $u$ chosen to be even.)

Since the two ideals generate $1$, they are relatively prime.