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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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  • $\begingroup$ See also Pell's equation $\endgroup$
    – reuns
    Aug 11, 2017 at 4:11
  • $\begingroup$ 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. $\endgroup$
    – Bill Thomas
    Aug 11, 2017 at 21:39

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$$\mathrm{Nm}(1+\sqrt{d}) = (1 + \sqrt{d})(1 - \sqrt{d}) = 1-d$$

For this element to even be a unit, let alone a fundamental one, you must have $1 - d = \pm 1$.

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