[proofplan] We regard a positive Pell solution $(x,y)$ as the norm-one unit $\alpha=x+y\sqrt D$ in the ordered real quadratic ring $\mathbb Z[\sqrt D]$. Since $\varepsilon_1>1$, the powers of $\varepsilon_1$ form a strictly increasing sequence, so we can place $\alpha$ between two consecutive powers. Dividing by the lower power gives a norm-one element lying in the interval $[1,\varepsilon_1)$, and the defining minimality of the fundamental unit forces this element to be $1$. Uniqueness follows from strict monotonicity of the positive powers of $\varepsilon_1$. [/proofplan] [step:Convert the Pell solution into a positive norm-one unit] By a positive solution, we mean a pair $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ with $x>0$ and $y>0$. Let $(x,y)$ be such a positive solution satisfying \begin{align*} x^2 - D y^2 = 1. \end{align*} Define \begin{align*} \alpha := x + y\sqrt D \in \mathbb{Z}[\sqrt D]. \end{align*} By the definition of the norm map, \begin{align*} N(\alpha)=N(x+y\sqrt D)=x^2-Dy^2=1. \end{align*} Since $x>0$, $y>0$, and $D\geq 1$, we have \begin{align*} \alpha=x+y\sqrt D>1. \end{align*} Thus $\alpha$ is a real norm-one element of $\mathbb{Z}[\sqrt D]$ greater than $1$. [/step] [step:Place $\alpha$ between two consecutive powers of $\varepsilon_1$] Let $\mathbb{R}$ denote the ordered field of [real numbers](/page/Real%20Numbers), and regard $\mathbb{Z}[\sqrt D]$ as a subring of $\mathbb{R}$ through the positive real square root $\sqrt D$. Because $\varepsilon_1>1$, the sequence $(\varepsilon_1^k)_{k\geq 0}$ is strictly increasing and unbounded in $\mathbb{R}$. Hence there is a unique integer $n\geq 0$ such that \begin{align*} \varepsilon_1^n \leq \alpha < \varepsilon_1^{n+1}. \end{align*} Since $\varepsilon_1^0=1$ and $\alpha>1$, the integer $n$ is allowed to be $0$ at this stage only if $\alpha<\varepsilon_1$. [guided] We need to compare $\alpha$ with the discrete ordered list of powers \begin{align*} 1=\varepsilon_1^0,\quad \varepsilon_1,\quad \varepsilon_1^2,\quad \dots . \end{align*} The inequality $\varepsilon_1>1$ implies strict monotonicity: if $k<\ell$, then \begin{align*} \varepsilon_1^\ell=\varepsilon_1^k\varepsilon_1^{\ell-k}>\varepsilon_1^k. \end{align*} Let $\mathbb{R}$ denote the ordered field of real numbers, and regard $\mathbb{Z}[\sqrt D]$ as a subring of $\mathbb{R}$ through the positive real square root $\sqrt D$. The same inequality also implies unboundedness of the powers in $\mathbb R$. Therefore the set \begin{align*} A:=\{k\in\mathbb N\cup\{0\}:\varepsilon_1^k\leq \alpha\} \end{align*} is nonempty, because $0\in A$, and finite, because the powers eventually exceed $\alpha$. Let $n:=\max A$. Then $\varepsilon_1^n\leq \alpha$, while maximality of $n$ gives $\alpha<\varepsilon_1^{n+1}$. Strict monotonicity also gives uniqueness of this integer $n$. [/guided] [/step] [step:Divide by the lower power and use minimality of the fundamental unit] Define \begin{align*} \beta := \alpha\,\varepsilon_1^{-n}. \end{align*} Since $\varepsilon_1=x_1+y_1\sqrt D$ and $N(\varepsilon_1)=x_1^2-Dy_1^2=1$, direct multiplication gives \begin{align*} \varepsilon_1(x_1-y_1\sqrt D)=(x_1+y_1\sqrt D)(x_1-y_1\sqrt D)=x_1^2-Dy_1^2=1. \end{align*} Thus $\varepsilon_1^{-1}=x_1-y_1\sqrt D\in\mathbb{Z}[\sqrt D]$, so $\varepsilon_1^{-n}\in\mathbb{Z}[\sqrt D]$ and hence $\beta\in\mathbb{Z}[\sqrt D]$. We also record the norm product calculation used here: for arbitrary $a,b,c,d\in\mathbb{Z}$, \begin{align*} N((a+b\sqrt D)(c+d\sqrt D)) &=N((ac+bdD)+(ad+bc)\sqrt D)\\ &=(ac+bdD)^2-D(ad+bc)^2\\ &=(a^2-Db^2)(c^2-Dd^2)\\ &=N(a+b\sqrt D)N(c+d\sqrt D). \end{align*} Applying this identity repeatedly to $\beta=\alpha\varepsilon_1^{-n}$ gives \begin{align*} N(\beta)=N(\alpha)N(\varepsilon_1)^{-n}=1\cdot 1^{-n}=1. \end{align*} Dividing the inequalities \begin{align*} \varepsilon_1^n \leq \alpha < \varepsilon_1^{n+1} \end{align*} by the positive real number $\varepsilon_1^n$ gives \begin{align*} 1\leq \beta < \varepsilon_1. \end{align*} If $\beta>1$, then $\beta$ is a norm-one element of $\mathbb{Z}[\sqrt D]$ strictly between $1$ and $\varepsilon_1$, contradicting the defining minimality of $\varepsilon_1$. Therefore $\beta=1$, and hence \begin{align*} \alpha=\varepsilon_1^n. \end{align*} [guided] The purpose of introducing $\beta$ is to remove the largest power of $\varepsilon_1$ that does not exceed $\alpha$. Define \begin{align*} \beta := \alpha\,\varepsilon_1^{-n}. \end{align*} We must verify that $\beta$ is still an admissible competitor in the definition of the fundamental unit. First, $\beta$ lies in $\mathbb{Z}[\sqrt D]$. Since $\varepsilon_1=x_1+y_1\sqrt D$ and $N(\varepsilon_1)=x_1^2-Dy_1^2=1$, direct multiplication gives \begin{align*} \varepsilon_1(x_1-y_1\sqrt D)=(x_1+y_1\sqrt D)(x_1-y_1\sqrt D)=x_1^2-Dy_1^2=1. \end{align*} Therefore $\varepsilon_1^{-1}=x_1-y_1\sqrt D\in\mathbb{Z}[\sqrt D]$, and hence $\varepsilon_1^{-n}\in\mathbb{Z}[\sqrt D]$. Since $\alpha\in\mathbb{Z}[\sqrt D]$, closure under multiplication gives $\beta\in\mathbb{Z}[\sqrt D]$. Second, $\beta$ has norm $1$. We justify the norm product identity rather than treating it as a black box. For arbitrary $a,b,c,d\in\mathbb{Z}$, multiplication in $\mathbb{Z}[\sqrt D]$ gives \begin{align*} (a+b\sqrt D)(c+d\sqrt D)=(ac+bdD)+(ad+bc)\sqrt D. \end{align*} Therefore \begin{align*} N((a+b\sqrt D)(c+d\sqrt D)) &=N((ac+bdD)+(ad+bc)\sqrt D)\\ &=(ac+bdD)^2-D(ad+bc)^2\\ &=a^2c^2+2abcdD+b^2d^2D^2-Da^2d^2-2abcdD-Db^2c^2\\ &=(a^2-Db^2)(c^2-Dd^2)\\ &=N(a+b\sqrt D)N(c+d\sqrt D). \end{align*} Applying this identity repeatedly to $\beta=\alpha\varepsilon_1^{-n}$ gives \begin{align*} N(\beta)=N(\alpha\varepsilon_1^{-n}) =N(\alpha)N(\varepsilon_1)^{-n} =1\cdot 1^{-n} =1. \end{align*} Third, $\beta$ lies in the interval $[1,\varepsilon_1)$. The real number $\varepsilon_1^n$ is positive, so dividing \begin{align*} \varepsilon_1^n \leq \alpha < \varepsilon_1^{n+1} \end{align*} by $\varepsilon_1^n$ preserves the inequalities and gives \begin{align*} 1\leq \alpha\varepsilon_1^{-n}<\varepsilon_1. \end{align*} Thus \begin{align*} 1\leq \beta<\varepsilon_1. \end{align*} Now suppose $\beta>1$. Then $\beta$ is a real norm-one element of $\mathbb{Z}[\sqrt D]$ with \begin{align*} 1<\beta<\varepsilon_1, \end{align*} which contradicts the fact that $\varepsilon_1$ is the least such element greater than $1$. Hence $\beta$ cannot exceed $1$. Since already $\beta\geq 1$, we obtain $\beta=1$. Multiplying by $\varepsilon_1^n$ gives \begin{align*} \alpha=\varepsilon_1^n. \end{align*} [/guided] [/step] [step:Exclude the zeroth power and prove uniqueness] The equality $\alpha=\varepsilon_1^n$ has been obtained for some integer $n\geq 0$. Since $\alpha>1$ while $\varepsilon_1^0=1$, we must have $n\geq 1$, so $n\in\mathbb N$. It remains to prove uniqueness. Suppose $m,n\in\mathbb N$ and \begin{align*} \varepsilon_1^m=\varepsilon_1^n. \end{align*} If $m<n$, then $\varepsilon_1^{n-m}>1$, and multiplying by the positive real number $\varepsilon_1^m$ gives \begin{align*} \varepsilon_1^n=\varepsilon_1^m\varepsilon_1^{n-m}>\varepsilon_1^m, \end{align*} contradicting $\varepsilon_1^m=\varepsilon_1^n$. The case $n<m$ is identical with $m$ and $n$ interchanged. Therefore $m=n$. This proves both existence and uniqueness of $n\in\mathbb N$ such that \begin{align*} x+y\sqrt D=\varepsilon_1^n. \end{align*} [/step]