Solve $y^2 = x^3 - 4$ where $x,y \in \mathbb{Z}$ Solve $y^2 = x^3 - 4$
So far i've tried to look at this:
$x^3 = (y-2i)(y+2i)$.
I think we need to prove that $y+-2i$ are coprime, so suppose not then:
$d = u+iv$ divides both numbers. So also d divides the sum and difference: $d|2y$ and $d|4i$.
If one applies the euclidean norm function we get that $u^2 + v^2|4y^2$ and $u^2 + v^2|16$.
And here i get stuck, any tips?
Kees

Your computation are done in the ring of Gaussian integers, which is a PID. Let $p$ be a prime in this ring dividing both $y+2i$ and $y-2i$. Then $p$ divides $2y$ and $4i$, so it is $1+i$ (up to invertible elements), because this is the only prime dividing $2$.
From $(1+i)(h+ki)=y+2i$ we get $h-k=y$, $h+k=2$, so $h=(2+y)/2$, $k=(2-y)/2$ and $y$ must be even.
So the two elements are indeed coprime if $y$ is odd. In this case $y+2i$ must be a cube times an invertible element, so
$$
y+2i=u(a+bi)^3
$$
In the case $u=1$, we get $y=a(a^2-3b^2)$, $2=b(3a^2-b^2)$. Since $y$ is odd, $a$ must be odd and $b$ even. Thus $b=\pm2$ and $3a^2-4=\pm1$. The only case is $b=-2$, $a=\pm1$. So $y=\pm11$.
A solution has been found: $x=5,y=\pm11$.
In the case $u=i$, we get $y=-b(3a^2-b^2)$, $2=-a(a^2-3b^2)$. So $b$ must be odd and $a$ even. Therefore $a=\pm2$. If $a=2$ we get $4-3b^2=-1$ (no solution). If $a=-2$ we get $4-3b^2=1$, so $b=\pm1$. This corresponds to $y=\pm11$, the same as before.
The cases $u=-1$ and $u=-i$ are the same, because the minus sign can be merged with the cube.

Suppose $y$ is even. Then also $x$ must be even and so we can set $x=2X$ and $y=2Y$, obtaining
$$
2X^3=Y^2+1=(Y+i)(Y-i)
$$
Since $1+i$ is prime and divides the left-hand side, it must divide $Y+i$ or $Y-i$. From $(1+i)(h+ki)=Y+i$ we get $h-k=Y$ and $h+k=1$. Thus $h=(Y+1)/2$ and $Y$ must be odd. Similarly, if $1+i$ divides $Y-i$. If we set $Y=2Z+1$, we get
$$
\frac{2Z+1+i}{1+i}=Z+1-Zi,
\qquad
\frac{2Z+1-i}{1-i}=Z+1+Zi.
$$
and the equation can be written as
$$
X^3=(Z+1-Zi)(Z+1+Zi)
$$
A prime dividing both factors must divide $2(Z+1)$ and $2Zi$. If it divides $Z$, it cannot divide $Z+1$, so it divides $2$. So the only prime that can divide both numbers is $1+i$. However, from
$$
(1+i)(h+ki)=Z+1+Zi
$$
we get $2h=2Z+1$, a contradiction. So the two factors are coprime and therefore they are cubes times an invertible.
Case $Z+1+Zi=(a+bi)^3$. We get $Z+1=a(a^2-3b^2)$ and $Z=b(3a^2-b^2)$.
Therefore $a^3-3ab^2-3a^2b+b^3=1$, or $(a+b)(a^2-4ab+b^2)=1$.
From $a+b=1$ we get $(a+b)^2-6ab=1$, hence or $6ab=0$. Thus $a=0$ or $b=0$ and we get the solutions $a=0$, $b=1$, that is $Z=-1$; $a=1$, $b=0$, that is $Z=0$. This corresponds to $Y=-1$ and $Y=1$.
From $a+b=-1$ we get $-6ab=-2$, that has no solution.
Case $Z+1+Zi=i(a+bi)^3$. $Z+1=-b(3a^2-b^2)$ and $Z=a(a^2-3b^2)$. Here we get $-a^3-3a^2b+3ab^2+b^3=1$, so $(b-a)(a^2+4ab+b^2)=1$ The case $b-a=1$ forces $6ab=0$. The case $b-a=-1$ forces $6ab=2$. Thus we don't find new solutions.
Another solution has been found $x=2$, $y=\pm2$
No other solutions exist
