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My question has connection with this question. Let $k>0$ be an integer without square factor. We consider the ring $\mathbb{Z}[\sqrt{k}]$. Let $N(a+b\sqrt{k}):= |a^2-b^2k|$ for $(a,b) \in \mathbb{Q}^2$. We suppose $(\mathbb{Z}[\sqrt{k}],N)$ is not euclidian.

Let $F:=\{(x,y) \in \mathbb{Q}^2 | N(x+y\sqrt{k})\geq 1\}$.

We consider the set $E=\bigcap_{(a,b)\in \mathbb{Z}^2} (F+(a,b)).$

The complement of $E$ in $\mathbb{Q}^2$ is dense.

Is $E$ locally finite ?

Remark: If $(\mathbb{Z}[\sqrt{k}],N)$ is euclidian, $E$ is empty. Because if $\frac{a}{d}+\frac{b}{d}\sqrt{k} \in \mathbb{Q}[\sqrt{k}]$, we have the euclidian division $a+b\sqrt{k}=(q_1+q_2\sqrt{k})d+r_1+r_2\sqrt{k}$ with $N(r_1+r_2\sqrt{k}). So $(\frac{a}{d},\frac{b}{d}) \notin (F+ (q_1,q_2))$

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    Sorry, I made a mistake. I have edited. It's $N(x+y \sqrt{k})\geq 1$ in the definition of $F$.2012-07-03

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