Chapter 9

Let \(d \in \mathbb{Z}\) be such that \(\sqrt{d} \notin \mathrm{Q} .\) In \(\mathbb{Z}[\sqrt{d}]\) let $$ v(z)=|z \bar{z}|=\left|a^{2}-d b^{2}\right|, \text { where } z=a+b \sqrt{d} $$ (a) (Pell's equation) Show that \(a+b \sqrt{d}\) is a unit in \(\mathbb{Z}[\sqrt{d}]\) if and only if \(a^{2}-d b^{2}=\pm 1\) (b) Assume that for any rational numbers \(x\) and \(y\) there exist integers \(n\) and \(m\) such that \(I(x-n)^{2}-d(y-m)^{2} \mid<1\). Show that in this case \(\mathbb{Z}[\sqrt{d}]\) is a Euclidean domain $$ \text { with } v(a+b \sqrt{d})=\left|a^{2}-d b^{2}\right| \text { . } $$

In \(\mathbb{Z}[\sqrt{2}]\) show that (a) \(1+\sqrt{2}\) is a unit. (b) \(\pm(1+\sqrt{2})^{n}\) for \(n \in \mathbb{Z}\) is a unit. (c) \(\pm(1+\sqrt{2})^{n}\) for \(n \in \mathbb{Z}\) are all the units.