I'm solving the following problem. Unfortunately, I don't know any source of the problem because it's a question from another school.
Let $R=\{a+b\alpha|a, b\in \Bbb{Z}, \alpha=\frac{1+\sqrt{-19}}{2}\}$. Assume that $R$ is a Euclidean domain with a Euclidean norm $\psi$. Choose $a\in$$R$ such that $\psi(a)$ is the smallest integer in {$\psi(x)$|$x$$\notin$$R^{\mathsf{x}}\cup\{0\}$}. (Here, $R^{\mathsf{x}}$ is the set of all units.) Show that there exist no $q$ and $r$ such that $2=aq+r$ with $r=0$ or $\psi(r)<\psi(a)$. Deduce that $R$ is not Euclidean.
My Attempt
First, I know that both $2$ and $3$ are irreducible over $R$ and $R^{\mathsf{x}}=\{\pm1\}$.
Suppose that there are such $q$ and $r$.
If $r$$\notin$$R^{\mathsf{x}}\cup\{0\}$, then $\psi(r)<\psi(a)$ and this is impossible by the minimality. Thus, $r$$\in$$R^{\mathsf{x}}\cup\{0\}$.
Case1 ($r=0$)
If $r=0$, then $2=aq$. By the irreduciblility of $2$, we get $a=\pm1$, $q=\pm2$ or $a=\pm2$, $q=\pm1$.
Case2 ($r=1$)
If $r=1$, then $1=aq$. Since $a$ is a unit, we obtain $a=\pm1$.
Case3 ($r=-1$)
If $r=-1$, then $3=aq$. By the irreduciblility of $3$, we get $a=\pm1$, $q=\pm3$ or $a=\pm3$, $q=\pm1$.
I don't know what to do next. I think $a$$\notin$$R^{\mathsf{x}}\cup\{0\}$. If so, the possibility of $a$ is $\pm2$ or $\pm3$. I'm not sure if $a$$\notin$$R^{\mathsf{x}}\cup\{0\}$.Please help me.
