Units of Z[√5] are a proper subgroup of the units of its maximal order

The maximal order is ${\mathcal{O}}_K=\mathbb{Z}[\epsilon ]$ with $\epsilon =(1+\sqrt{5})\mathrm{/}2$, and $\mathbb{Z}[\sqrt{5}]⊊{\mathcal{O}}_K$; its elements are the $(x+y\sqrt{5})\mathrm{/}2$ with $x\equiv y(\mathrm{m}\mathrm{o}\mathrm{d}2)$, of nonzero norm when the element is nonzero because the norm is a product of the two embeddings.

The element $\epsilon$ is a unit of ${\mathcal{O}}_K$: its norm is $N_{K\mathrm{/}\mathbb{Q}}(\epsilon )=\frac{1+\sqrt{5}}{2}\cdot \frac{1-\sqrt{5}}{2}=\frac{1-5}{4}=-1,$ so [F2] applies; also $\epsilon >1$ because $\sqrt{5}>1$, and direct multiplication gives ${\epsilon }^2=\frac{(1+\sqrt{5}{)}^2}{4}=\frac{3+\sqrt{5}}{2},{\epsilon }^3=\epsilon \cdot {\epsilon }^2=\frac{(1+\sqrt{5})(3+\sqrt{5})}{4}=\frac{8+4\sqrt{5}}{4}=2+\sqrt{5}\mathrm{.}$

The units of the Pell order are $\mathbb{Z}[\sqrt{5}{]}^{\times }=\pm ⟨2+\sqrt{5}⟩=\pm ⟨{\epsilon }^3⟩$: an element $x+y\sqrt{5}$ of $\mathbb{Z}[\sqrt{5}]$ is a unit of that order exactly when $x^2-5y^2=\pm 1$ by [F3], the norm-one solutions are $\pm {\epsilon }_5^k$, and $2+\sqrt{5}$ is the least positive norm-$(-1)$ solution, so every norm-$\pm 1$ solution is $\pm (2+\sqrt{5}{)}^n=\pm {\epsilon }^{3n}$.

The maximal order has ${\mathcal{O}}_K^{\times }=\{\pm {\epsilon }^n:n\in \mathbb{Z}\}$: by [F5] there is a unit $\gamma >1$ with ${\mathcal{O}}_K^{\times }=\{\pm {\gamma }^n:n\in \mathbb{Z}\}$, and then every unit $v>1$ is ${\gamma }^n$ with $n\ge 1$, hence $v\ge \gamma$, so $\gamma$ is the least unit $>1$ of ${\mathcal{O}}_K$; by [F4] the element $\epsilon >1$ is the least unit $>1$ of ${\mathcal{O}}_K$, so $\gamma =\epsilon$.

The inclusion of unit groups is proper: the element $\epsilon$ lies in ${\mathcal{O}}_K^{\times }$ by step 1.2, while $\epsilon =(1+\sqrt{5})\mathrm{/}2$ is not of the form $x+y\sqrt{5}$ with $x,y\in \mathbb{Z}$ and so does not lie in $\mathbb{Z}[\sqrt{5}]$, hence not in $\mathbb{Z}[\sqrt{5}{]}^{\times }$; therefore $\mathbb{Z}[\sqrt{5}{]}^{\times }=\pm ⟨{\epsilon }^3⟩⊊\pm ⟨\epsilon ⟩={\mathcal{O}}_K^{\times }$.

The index is $3$: the assignment $n\mapsto {\epsilon }^n$ is an isomorphism $\mathbb{Z}\to ⟨\epsilon ⟩\subseteq {\mathcal{O}}_K^{\times }$, since $\epsilon >1$ forces ${\epsilon }^{n-m}=1$ only for $n=m$. It maps $3\mathbb{Z}$ onto $⟨{\epsilon }^3⟩$, so $[⟨\epsilon ⟩:⟨{\epsilon }^3⟩]=[\mathbb{Z}:3\mathbb{Z}]=3$; multiplying by the common sign group $\{\pm 1\}$ does not change the index, hence $[{\mathcal{O}}_K^{\times }:\mathbb{Z}[\sqrt{5}{]}^{\times }]=3$.

Conclusion: the maximal order ${\mathcal{O}}_K$ of $K=\mathbb{Q}(\sqrt{5})$ has unit group $\{\pm {\epsilon }^n\}$ generated modulo its sign subgroup by $\epsilon =(1+\sqrt{5})\mathrm{/}2$, while the Pell order $\mathbb{Z}[\sqrt{5}]$ has unit group $\{\pm (2+\sqrt{5}{)}^n\}=\{\pm {\epsilon }^{3n}\}$, a proper subgroup of index $3$. The fundamental Pell solution ${\epsilon }_5=9+4\sqrt{5}={\epsilon }^6$ generates the norm-one Pell subgroup of the order up to sign; the maximal-order fundamental unit is $\epsilon$. The counterexample is the sharpened form of the design's warning for this pair: it identifies both groups and the exact index rather than merely exhibiting one missing unit. Choice is used only through the rank-one structure [F5]; all arithmetic in $\mathbb{Z}[\sqrt{5}]$ and the comparisons of the two generators are elementary.