Solutions of $X^2+2=Y^3$ in $\mathbb{Z}[i]$ The Mathematician James T. Cross, on his work "Primitive Pythagorean Triples of Gaussian Integers", displayed a method
for generating all Pythagorean triples in $\mathbb{Z}[i]$. Inspired by his work, I want to find all the gaussian integer solutions of the equation $X^2+2=Y^3$. In $\mathbb{Z}$, the only solution is $x=5,y=3$ or $x=-5,y=3$. A trivial solution in $\mathbb{Z}[i]$ is $x=i$, $y=1$. My question is:
Any other mathematician solved this problem any time before?
 A: I apologize for interchanging $X$ and $Y$, but that makes me more comfortable. No matter what, this is fancy stuff, and is an example of how things that are relatively simple in the quadratic case suddenly become complicated or even intractable in the next case along the way.
You’re dealing with the elliptic curve $Y^2=X^3-2$, and asking for $\Bbb Z[i]$-integral points. The corresponding question for $\Bbb Z$-integral points is covered by Siegel’s Theorem, which says that any elliptic curve over $\Bbb Q$ has only finitely many $\Bbb Z$-points. This is a field that I know only by rumor, and there will certainly be people on MSE who know the field backwards and forwards, much as it can be known. I’d guess that Siegel can be (probably has been) generalized to this case too. This is certain: the $\Bbb Q(i)$-points of the curve form a commutative group that’s finitely generated, that’s the Mordell-Weil Theorem. Finding the generators is doable by fairly advanced methods, but not usually easy.
So your assignment is to go to the ever-reliable Wikipedia and look up elliptic curves, Siegel Theorem (on elliptic curves), and the Mordell-Weil Theorem.
A: I don't see the way of working directly on $\Bbb Z+\Bbb Z[i]$. But the equation in this ring of Gaussian integers is equivalent to the following system of four unknowns in $\Bbb Z$
$$\begin{cases}x^2-y^2+2=z^3-3zw^2\\2xy=3z^2w-w^3\end{cases}$$
 This allows us to find out some infinite sets of solutions. Making, for example, $z=w$ we have $$x^2-y^2+2=-2z^3\\2xy=2z^3$$ therefore
$$x^2-y^2+2+2xy=0\qquad (1)$$ We have in $(1)$ the particular solutions $(x,y)=(\pm1,\mp1),(\pm1,\mp3),(\pm7,\mp3)$
From this one has for $X=x+yi$ and $Y=z+wi$ the solutions of the proposed equation in Gaussian integers
$$(x,y,z,w)=(\pm1,\mp1,t,t),(\pm1,\mp3,t,t),(\pm7,\mp3,t,t)$$ where t is an arbitrary rational integer.
