$\varepsilon=\frac12(t+u\sqrt d)$ is a fundamental unit of $\mathbb Q(\sqrt d)$ or the square of one

Let $d > 0$ be a fundamental discriminant and let $K = \mathbb Q(\sqrt d)$, presented as a quadratic number field $K$ with an element $\alpha$ satisfying $\alpha^2 = d$. Let $t>0$, $u>0$ be the solution of $t^2 - du^2 = 4$ with $u$ minimal and put $\varepsilon = \frac12(t + u\alpha)\in K$. Then there is a unit $\eta \in \mathcal O_K^\times$ such that

1. $\eta$ is a fundamental unit: every unit of $\mathcal O_K$ equals $\eta^n$ or $-\eta^n$ for some $n\in\mathbb Z$;
2. $\varepsilon = \eta$ or $\varepsilon = \eta^2$.

Formalization Note A fundamental unit is taken here to be any generator of the unit group modulo $\{\pm1\}$, without the normalisation $\eta > 1$ under a real embedding. The choice of square root $\alpha$ of $d$ replaces $\varepsilon$ by its conjugate $\varepsilon^{-1}$, and the statement holds for either choice.