<s>[INST] Show that no GCD exists for $4$ and $2 \cdot (1 + \sqrt{-3})$ in $\{ x + y \sqrt{-3} | x, y \in \mathbb{Z}\}$ I need to show that there is no greatest common divisor for $4$ and $2 \cdot(1 + \sqrt{-3})$ in the ring $R := \{ x + y \cdot\sqrt{-3} \,\,|\,\, x, y \in \mathbb{Z}\}$.
I found that both $2$ and $(1 + \sqrt{-3})$ divide $4$ as well as $2 \cdot(1 + \sqrt{-3})$. Therefore if $x$ was a greatest common divisor of the two, it has to be divisible by both $2$ and $(1 + \sqrt{-3})$. 
How can I proceed from here?
 [/INST] Let the common divisor of $4$ and $2(1+\sqrt{-3})$ be $x+y\sqrt{-3}$. Then we have
\begin{align}
(x+y\sqrt{-3})(a+b\sqrt{-3})&=4,\\
(x+y\sqrt{-3})(c+d\sqrt{-3})&=2(1+\sqrt{-3}),
\end{align}
where $x,y,a,b,c,d\in\mathbb{Z}$. Multiplying $x-y\sqrt{-3}$ on both sides of the equations, one obtains
\begin{align}
a&=\frac{4x}{x^2+3y^2},\quad
b=-\frac{4y}{x^2+3y^2},\\
c&=\frac{2(x+3y)}{x^2+3y^2},\quad
d=\frac{2(x-y)}{x^2+3y^2}.
\end{align}
The denominators have higher degrees than the numerators, so the number of integer pairs of $(x,y)$ that makes $|a|\geq1$ or $|b|\geq 1$ or $|c|\geq 1$ or $|d|\geq 1$ is finite. One can therefore enumerate all such cases. After ruling out them all, the only possibility left is $a=b=c=d=0$, which cannot hold either (not allowed to have $x=y=0$ as they appear in the denominators), thus completing the proof.
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