I know that $\mathbb{Z}\left[\sqrt{6}\,\right]$ is a Euclidean domain with respect to the absolute valued norm map $x+y\sqrt{6} \mapsto |x^2-6y^2|$. I think I proved this result with some common techniques, but the proof is a bit sloppy and it requires a lot of cases. (Basically, I checked that for every $z \in \mathbb{Q}(\sqrt{6})$ there is $\gamma \in \mathbb{Z}[\sqrt{6}]$ such that $|N(z-\gamma)|<1$.)
Does there exist a short proof for this result, with less cases or a more, say, enlightening method?
Many thanks.
Best Answer
this question has been answered here: https://math.stackexchange.com/a/124573/4997
this is a picture of one of the "starfish" - a region bounded by two hyperbolas.
Reading this review by Franz Lemmermeyer, The Euclidean Algorithm in Algebraic Number Fields
Fortunately, in the internet age. Oppenheim's 1934 paper Quadratic Fields with and Without Euclid's Algorithm is online. The preview page is all you need.
Forget about the case $m \equiv 1 \mod 4$ (since $m = 6$), we want to find $m$ such that for any rational point $(a,b)$ we can find $(x,y) \in \mathbb{Z}^2$ such that $|(x-a)^2 - m (y-b)^2 | < 1 $
Oppenheim proceeds by process of eliination: if we can find $(a,b) \in \mathbb{Q}^2$ such that for all integers $(x,y) \in \mathbb{Z}^2$ either
Then we have ruled out the possibility of $m$ as a Euclidean domain (with this particular norm). This then becomes a quadratically constrained quadratic programming problem. Let's assume $0 < a,b < \tfrac{1}{2}$.
Examining the constraints around the points $(0,0), (1,0), (-1,0)$ one can show $mb^2 \geq 1 + (1 + a)^2 \geq 2$ and $m \geq 8$.
$m = 2,3,\mathbf{6},7$ work.
To get $m = 5,13,17,21,29$, a modified Euclidean algorithm gives the same constraints with norm $|(x + \tfrac{1}{2}y)^2 - \tfrac{1}{4}my^2 | $
I read this proof and would like to understand better how these constraints dictate problems with factorization.