TITLE: So what *is* the Euclidean function for $\mathcal{O}_{\mathbb{Q}(\sqrt{69})}$?
QUESTION [12 upvotes]: It's my understanding that $\mathcal{O}_{\mathbb{Q}(\sqrt{69})}$ is UFD, PID, and Euclidean, but not norm-Euclidean.
If it were norm-Euclidean, there would be a solution to $28 = q \left(\frac{5}{2} + \frac{\sqrt{69}}{2}\right) + r$ with $q \neq 0$ and $abs(N(r)) < 11$.
Now I get the feeling I'd have infinitely many values of $q$ to check, so instead I'll check all the suitable values for $r$ and see if there is a case in which $28 - r$ or $28 + r$ has a norm divisible by $11$.

$1$ gives us $27$ and $29$
$\left(\pm\frac{9}{2} \pm \frac{\sqrt{69}}{2}\right)$ gives us $\left(\pm\frac{47}{2} \pm \frac{\sqrt{69}}{2}\right)$, $\left(\pm\frac{65}{2} \pm \frac{\sqrt{69}}{2}\right)$
$2$ gives us $26$ and $30$
$\pm 8 \pm \sqrt{69}$ gives us $\pm 20 \pm \sqrt{69}$, $\pm 36 \pm \sqrt{69}$
$3$ gives us $25$ and $31$
$\left(\pm\frac{75}{2} \pm \frac{\sqrt{69}}{2}\right)$ gives us, you get the idea

But now I get this sinking feeling that maybe I do have infinitely many more values of $r$ to check, and that maybe one of those does lead to a suitable $r$...
Regardless of that, it seems to me like trying to implement the Euclidean algorithm with the absolute value of the norm function in this domain is quite futile. So does anyone know what the Euclidean function is for this domain? Is there more than one such function?
And for a real ring that is Euclidean for the absolute value of the norm, like, say, $\mathcal{O}_{\mathbb{Q}(\sqrt{73})}$, how do you more effectively find the appropriate values of $q$ and $r$? In $\mathcal{O}_{\mathbb{Q}(\sqrt{69})}$, I tried division: $$\frac{28}{\left(\frac{5}{2} + \frac{\sqrt{69}}{2}\right)} = \left(\frac{-70}{11} + \frac{14 \sqrt{69}}{11}\right)$$ but couldn't figure out how to round that to a suitable algebraic integer.

REPLY [8 votes]: As suggested in the comments, the norm function needs indeed very slight adjustments. 
Define 
$
\phi(\pi) =
\left\{
 \begin{array}{ll}
  N(\pi)  & \mbox{if } \pi \neq 10+3\alpha \\
  23 & \mbox{if } \pi = 10+3\alpha
 \end{array}
\right.
$
where $\alpha=\frac{1+\sqrt{69}}{2}$ and $N$ denotes the absolute value of the norm.
$\phi$ is an Euclidean function for $\mathcal{O}_{\mathbb{Q}(\sqrt{69})}$.
For the proof, check David Clark's paper.