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.

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