Without using the Euclidean algorithm, show that in $\mathbb{Z}[i]$ for $x,y \in \mathbb{Z}$ s.t $x^2+9 = y^3$, $x+3i$ and $x-3i$ are relative prime.

Question

I began by assuming there is a $d \in \mathbb{Z}[i]$ (with the goal that $d$ must be a unit) such that:

$d|x+3i$ ,

$d|x-3i$.

Which gives,

$x+3i = \lambda d, (\lambda \in \mathbb{Z}[i]) $

$x-3i = \mu d, (\mu \in \mathbb{Z}[i])$ .

Then combining these two gives,

$6i = (\lambda - \mu)d$

Then factorising $6i$ gives,

$3i(i+1)(1-i)= (\lambda - \mu)d$.

From here I get stuck, I see that:

$d \neq (i+1)(1-i)$, as this would imply $2|x+3i$.

$d \neq 3(i+1)(1-i)$, as this would imply $6|x+3i$.

But I cannot see why:

$d \neq 3, \,3(1+i), \,3(1-i), \,(1+i), \,(1-i).$

So that I can conclude that $d$ must be $\pm1$ or $\pm i$. Hence $x+3i$ and $x-3i$ are relative prime.

Answer 1

$3i(1+i)=-3+3i$, $3i(-1+i)=-3-3i$.