Let $K$ be the quadratic field $\mathbb{Q}(\sqrt d)$ where $d\in\mathbb{N}_{\geq1}$ is square-free.
Is $d=2$ the only value for $d$ such that the fundamental unit of $K$ is of the form $1+\sqrt d$?
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Sign up to join this communityLet $K$ be the quadratic field $\mathbb{Q}(\sqrt d)$ where $d\in\mathbb{N}_{\geq1}$ is square-free.
Is $d=2$ the only value for $d$ such that the fundamental unit of $K$ is of the form $1+\sqrt d$?
$$\mathrm{Nm}(1+\sqrt{d}) = (1 + \sqrt{d})(1 - \sqrt{d}) = 1-d$$
For this element to even be a unit, let alone a fundamental one, you must have $1 - d = \pm 1$.