## Characterize the units in QQ(sqrt(5))

Characterize the units in $\mathbb{Q}(\sqrt{5})$.

Note that every element of $\mathbb{Q}(\sqrt{5})$ is real; in particular, every element is either positive or negative. Note also that $\lambda = (1+\sqrt{5})/2$ is a unit with inverse $(-1+\sqrt{5})/2$.

We begin with a lemma.

Lemma: No unit $\zeta$ in $\mathbb{Q}(\sqrt{5})$ satisfies $1 < \zeta < (1+\sqrt{5})/2$. Proof: Suppose to the contrary that $\zeta = x+y\sqrt{5}$ is such a unit. Now $N(\zeta) = x^2 - 5y^2 = \pm 1$, so that $x - y\sqrt{5} = \pm 1/(x+y\sqrt{5})$. Since $x+y\sqrt{5} > 1$, we have $-1 < x-y\sqrt{5} < 1$. Adding these inequalities, we have $0 < 2x < 3/2 + \sqrt{5}/2$, and thus $0 < x < 3+\sqrt{5}$. So $0 < x < 5.23 + \varepsilon$. Since $x \in \mathbb{Z}$, we have $x \in \{1,2,3,4,5\}$. Consider again the inequality $1 < x+y\sqrt{5} < (1+\sqrt{5})/2$; for $x$ in this range, we have no integer solutions $y$. $\square$

Now let $\eta \in \mathbb{Q}(\sqrt{5})$ be a unit; say $\eta > 0$. Since $\lambda = (1+\sqrt{5})/2 > 1$, there exists $k \in \mathbb{Z}$ such that $\lambda^k \leq \eta < \lambda^{k+1}$. Now $1 \leq \eta\lambda^{-k} < \lambda$, and by the lemma (since $\eta\lambda^{-k}$ is a unit), $\eta = \lambda^k$.

Thus every unit in $\mathbb{Q}(\sqrt{5})$ is an integer power of $(1+\sqrt{5})/2$.