If $a^2+3b^2$ is a cube in $\mathbb Z$ , then are $a+\sqrt{-3}b$ and $a-\sqrt{-3}b$ both cubes in $\mathbb Z[\sqrt{-3}]$ ?

Question

If $a,b \in \mathbb Z$ are such that g.c.d.$(a,b)=1$ and if $a^2+3b^2$ is a cube in $\mathbb Z$ , then are $a+\sqrt{-3}b$ and $a-\sqrt{-3}b$ both cubes in $\mathbb Z[\sqrt{-3}]$ ? I cannot use factorization in $\mathbb Z[\sqrt{-3}]$ as it is not an UFD . Please help . Thanks in advance

Answer 1

Despite the fact that $\mathbb{Z}[\sqrt{-3}]$ is not a UFD, the ring of Eisenstein integers $\mathbb{Z}[\omega]$ is a unique factorization domain since it is a Euclidean ring. Now it is enough to notice that:

$$ a^2+3b^2 = ((a+b)+2b\omega)((a-b)+2b\omega) \tag{1}$$ since $\sqrt{-3}=2\omega+1$. If $a^2+3b^2$ is an integer cube, we can assume WLOG that $a$ and $b$ do not have the same parity, hence the two terms in the RHS of $(1)$ are coprime and they both need to be cubes in $\mathbb{Z}[\omega]$.