# $\mathbb{Z}[\sqrt{11}]$ is norm-euclidean

I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$

By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in \mathbb{Z}(\sqrt{11}):|N(n-x)| < 1$

For the analogous statement for $\mathbb Z [\sqrt6]$, it worked by considering different cases, so I tried to do the same thing here. Here is what I did so far:

Let $x+y\sqrt{11} \in \mathbb Q (\sqrt{11})$

Case 1: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \frac{1}{\sqrt{11}}$, then we can choose such a $b$ and a $a \in \mathbb Z$ s.t. $|x-a| \leq \frac{1}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \frac{1}{\sqrt{11}}$

Case 2: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{5}{44}}$ Then we have $1 < 11 (y-b)^2 < \frac{5}{4}$, so we can choose $a \in \mathbb Z$ such that $\frac{1}{2} \leq |x-a| \leq 1$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \sqrt{\frac{5}{44}}$

Case 3: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{2}{11}}$ Then we can choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on, we may suppose that $|y-b| > \sqrt{\frac{2}{11}}$.

This is where I'm stuck. I tried choosing $b \in \mathbb Z$ s.t. $\frac{1}{2} \geq |y-b| > \sqrt{\frac{2}{11}}$, but then I run into problems, whether I choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$ or s.t. $\frac{3}{2} \leq |x-a| \leq 2$

• Look at math.stackexchange.com/questions/124484/… It appears that 11 will require at least 12, maybe 16 "starfish" – Will Jagy Apr 3 '17 at 0:04
• same history as 6, proved first by Perron in 1932, then Oppenheim 1934. – Will Jagy Apr 3 '17 at 0:07