Is the fundamental unit of $\mathbb{Q}(\sqrt d)$ special, in the following sense?

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 $1+\sqrt d$?

• See also Pell's equation – reuns Aug 11 '17 at 4:11
• I had that question myself some time ago. I thought about fundamental units of the form $2 + \sqrt d$, then $3 + \sqrt d$ and $4 + \sqrt d$. The pattern became obvious to me then. – Bill Thomas Aug 11 '17 at 21:39

$$\mathrm{Nm}(1+\sqrt{d}) = (1 + \sqrt{d})(1 - \sqrt{d}) = 1-d$$
For this element to even be a unit, let alone a fundamental one, you must have $1 - d = \pm 1$.