[proofplan] Minkowski's Theorem applied to rectangular boxes $S_t = \{|y_1| \leq t,\ |y_2| \leq |D_L|^{1/2}/t\}$ in the Minkowski image $\sigma(\mathcal{O}_L) \subset \mathbb{R}^2$ produces, for each $t > 0$, a non-zero $\alpha_t \in \mathcal{O}_L$ with small norm ($1 \leq |N(\alpha_t)| \leq |D_L|^{1/2}$). Varying $t$ over a strictly decreasing sequence $t_1 > t_2 > \cdots$, and using that the rectangles cover progressively smaller neighbourhoods of the $y_2$-axis, we extract infinitely many distinct elements $\alpha_j \in \mathcal{O}_L$ all with norms in the fixed finite set $\{m \in \mathbb{Z} : 1 \leq |m| \leq |D_L|^{1/2}\}$. Two pigeonhole arguments — first on the norm values and then on residues modulo $m\mathcal{O}_L$ — isolate a common residue class, and the ratios $\alpha_j/\beta$ become norm-one elements of $\mathcal{O}_L$ (hence units), and because the $\alpha_j$ are distinct there are infinitely many such units. [/proofplan] [step:Set up the Minkowski embedding and the rectangular bodies $S_t$] Let $L = \mathbb{Q}(\sqrt{d})$ with $d > 1$ square-free, so $L$ is a real quadratic field of signature $(r, s) = (2, 0)$ and degree $n = 2$. The Minkowski embedding specialises to \begin{align*} \sigma: L &\to \mathbb{R}^2 \\ \alpha &\mapsto (\sigma_1(\alpha), \sigma_2(\alpha)) \end{align*} where $\sigma_1, \sigma_2$ are the two real embeddings ($\sigma_1(\sqrt{d}) = \sqrt{d}$, $\sigma_2(\sqrt{d}) = -\sqrt{d}$). By [Covolume of Ideal Lattices](/theorems/1609), applied to $\mathfrak{a} = \mathcal{O}_L$: hypotheses are satisfied (this is a non-zero ideal of $\mathcal{O}_L$), and we obtain \begin{align*} \operatorname{covol}(\sigma(\mathcal{O}_L)) &= 2^{-s} |D_L|^{1/2} N(\mathcal{O}_L) = |D_L|^{1/2}. \end{align*} For each $t > 0$, define the closed rectangle \begin{align*} S_t &= \{(y_1, y_2) \in \mathbb{R}^2 : |y_1| \leq t,\ |y_2| \leq |D_L|^{1/2}/t\}. \end{align*} $S_t$ is closed, convex, and symmetric about $\mathbf{0}$. Its area is \begin{align*} \operatorname{vol}(S_t) &= (2t)(2 |D_L|^{1/2}/t) = 4|D_L|^{1/2} = 2^2 \operatorname{covol}(\sigma(\mathcal{O}_L)). \end{align*} [guided] **Signature of real quadratic fields.** A real quadratic field $L = \mathbb{Q}(\sqrt{d})$ with $d > 1$ has exactly two embeddings into $\mathbb{C}$, both real: $\sigma_1 : \sqrt{d} \mapsto +\sqrt{d}$ and $\sigma_2 : \sqrt{d} \mapsto -\sqrt{d}$. Hence $(r, s) = (2, 0)$ and the Minkowski embedding map is \begin{align*} \sigma : L \to \mathbb{R}^2, \qquad \alpha = x + y\sqrt{d} \mapsto (x + y\sqrt{d}, x - y\sqrt{d}). \end{align*} **Covolume via Theorem 1609.** The hypothesis of [Theorem 1609](/theorems/1609) is that $\mathfrak{a}$ is a non-zero ideal of $\mathcal{O}_L$. Taking $\mathfrak{a} = \mathcal{O}_L$ (the unit ideal, non-zero), the formula gives $\operatorname{covol}(\sigma(\mathcal{O}_L)) = 2^{-0} |D_L|^{1/2} \cdot 1 = |D_L|^{1/2}$ since $s = 0$ and $N(\mathcal{O}_L) = 1$. **Choice of $S_t$.** We want a closed, convex, symmetric body whose volume is exactly $2^n \operatorname{covol}(\sigma(\mathcal{O}_L)) = 4|D_L|^{1/2}$, so we can apply the closed-body case of [Minkowski's Theorem](/theorems/1608). Rectangles with side lengths $2t$ and $2|D_L|^{1/2}/t$ have area $4|D_L|^{1/2}$ independently of $t$, so varying $t$ gives a one-parameter family of rectangles all satisfying the volume equality. **Verifying $S_t$ is closed, convex, symmetric.** - *Closed:* $S_t$ is the intersection of two closed half-planes in each coordinate, hence closed. - *Convex:* a product of intervals is convex. - *Symmetric:* $|-y_i| = |y_i|$, so $S_t$ is invariant under $(y_1, y_2) \mapsto -(y_1, y_2)$. **Why rectangles rather than the Minkowski $\ell^1$-body.** Either choice works to find an element of small norm. Rectangles have the practical advantage that varying $t$ sweeps out a different non-zero lattice point each time (the rectangle becomes thinner in one axis and longer in the other), allowing us to generate infinitely many distinct elements rather than a single one. [/guided] [/step] [step:Extract a non-zero algebraic integer of bounded norm from each $S_t$] Fix $t > 0$. By [Minkowski's Theorem](/theorems/1608) part (2) with the closed-body case: the hypotheses are $S_t$ closed, convex, symmetric, and $\operatorname{vol}(S_t) \geq 2^n \operatorname{covol}(\sigma(\mathcal{O}_L))$. The last holds with equality, so the theorem applies and produces \begin{align*} \gamma_t \in \sigma(\mathcal{O}_L) \cap S_t, \qquad \gamma_t \neq 0. \end{align*} Write $\gamma_t = \sigma(\alpha_t) = (\sigma_1(\alpha_t), \sigma_2(\alpha_t))$ with $\alpha_t \in \mathcal{O}_L \setminus \{0\}$ (injectivity of $\sigma$: if $\sigma(\alpha) = 0$ then $\sigma_1(\alpha) = 0$, so $\alpha = 0$). By [Norm and Trace via Embeddings](/theorems/1577), applied to the real quadratic field $L$ (hypothesis: $L$ is a number field of degree $2$): \begin{align*} N_{L/\mathbb{Q}}(\alpha_t) &= \sigma_1(\alpha_t) \sigma_2(\alpha_t). \end{align*} Since $\alpha_t \in \mathcal{O}_L \setminus \{0\}$ and $\alpha_t \neq 0$, its norm is a non-zero integer by [Integrality and the Characteristic Polynomial](/theorems/1574). Moreover, both coordinates of $\gamma_t$ are non-zero: if $\sigma_1(\alpha_t) = 0$ or $\sigma_2(\alpha_t) = 0$, the product $N(\alpha_t) = 0$ would contradict $\alpha_t \neq 0$ and the non-zero-norm property. Finally, bounding: \begin{align*} 1 \leq |N(\alpha_t)| = |\sigma_1(\alpha_t)| \cdot |\sigma_2(\alpha_t)| \leq t \cdot \frac{|D_L|^{1/2}}{t} = |D_L|^{1/2}. \end{align*} [guided] **Applying Minkowski's Theorem to $S_t$.** Step 1 verified the hypotheses: $S_t$ is closed, convex, symmetric about $\mathbf{0}$, and $\operatorname{vol}(S_t) = 2^n \operatorname{covol}(\sigma(\mathcal{O}_L))$. The closed-body case of [Theorem 1608](/theorems/1608) part (2) then yields a non-zero $\gamma_t \in \sigma(\mathcal{O}_L) \cap S_t$. **Lifting to an element of $\mathcal{O}_L$.** The Minkowski map $\sigma : L \to \mathbb{R}^2$ restricts to an injective map $\mathcal{O}_L \hookrightarrow \sigma(\mathcal{O}_L) \subset \mathbb{R}^2$. Injectivity: $\sigma_1$ is an injective field embedding, so $\sigma(\alpha) = 0$ entails $\sigma_1(\alpha) = 0$ and hence $\alpha = 0$. Thus $\gamma_t = \sigma(\alpha_t)$ uniquely determines a non-zero $\alpha_t \in \mathcal{O}_L$. **Norm as a product.** [Theorem 1577](/theorems/1577) says the norm is the product of embeddings: $N(\alpha) = \prod_{i=1}^n \sigma_i(\alpha)$. For $n = 2$ with both embeddings real: $N(\alpha_t) = \sigma_1(\alpha_t) \sigma_2(\alpha_t)$. **Integrality of the norm.** By [Theorem 1574](/theorems/1574), $\alpha_t \in \mathcal{O}_L$ implies $N(\alpha_t) \in \mathbb{Z}$. Since $\alpha_t \neq 0$ and $L$ is a field, $N(\alpha_t) \neq 0$. Hence $|N(\alpha_t)| \geq 1$. **Upper bound from $S_t$.** Since $\gamma_t \in S_t$, both coordinates satisfy $|\sigma_1(\alpha_t)| \leq t$ and $|\sigma_2(\alpha_t)| \leq |D_L|^{1/2}/t$. Multiplying: \begin{align*} |N(\alpha_t)| = |\sigma_1(\alpha_t)| \cdot |\sigma_2(\alpha_t)| \leq t \cdot |D_L|^{1/2}/t = |D_L|^{1/2}. \end{align*} **Summary.** For each $t > 0$, we have a non-zero $\alpha_t \in \mathcal{O}_L$ with $1 \leq |N(\alpha_t)| \leq |D_L|^{1/2}$. [/guided] [/step] [step:Construct a sequence of infinitely many distinct $\alpha_j \in \mathcal{O}_L$ via shrinking rectangles] [claim:Distinctness Sequence] There exists a strictly decreasing sequence $t_1 > t_2 > \cdots > 0$ and corresponding elements $\alpha_j = \alpha_{t_j} \in \mathcal{O}_L$ produced by Step 2 such that the $\alpha_j$ are pairwise distinct. [/claim] [proof] We build $t_j$ inductively. Start with any $t_1 > 0$; Step 2 produces $\alpha_1 = \alpha_{t_1}$. Suppose $t_1 > t_2 > \cdots > t_k > 0$ and the corresponding $\alpha_1, \ldots, \alpha_k$ are pairwise distinct. We choose $t_{k+1}$. Consider any $\beta \in \mathcal{O}_L \setminus \{0\}$ with $\sigma_1(\beta) \neq 0$ (this holds for all non-zero $\beta$, since $\sigma_1$ is injective). Define \begin{align*} t(\beta) &= |\sigma_1(\beta)|. \end{align*} For $t < t(\beta)$, $\sigma(\beta) \notin S_t$, because the first-coordinate constraint $|\sigma_1(\beta)| \leq t$ fails. In particular, choosing \begin{align*} t_{k+1} &:= \frac{1}{2} \min\bigl(t_k,\ \min\{|\sigma_1(\alpha_j)| : j = 1, \ldots, k\}\bigr) > 0, \end{align*} we have $t_{k+1} < t_k$ and $t_{k+1} < |\sigma_1(\alpha_j)|$ for each $j \leq k$. Thus $\sigma(\alpha_j) \notin S_{t_{k+1}}$ for any $j \leq k$. Now apply Step 2 to $t = t_{k+1}$ to produce $\alpha_{k+1} := \alpha_{t_{k+1}} \in \mathcal{O}_L \setminus \{0\}$ with $\sigma(\alpha_{k+1}) \in S_{t_{k+1}}$. Because $\sigma(\alpha_j) \notin S_{t_{k+1}}$ for $j \leq k$, $\alpha_{k+1} \neq \alpha_j$ for any $j \leq k$. Hence $\alpha_1, \ldots, \alpha_{k+1}$ are pairwise distinct. By induction, we obtain an infinite sequence $\alpha_1, \alpha_2, \ldots$ of pairwise distinct non-zero elements of $\mathcal{O}_L$, each satisfying $1 \leq |N(\alpha_j)| \leq |D_L|^{1/2}$. [/proof] [guided] **Why the rectangles thin out.** The rectangles $S_t$ have area $4|D_L|^{1/2}$ for every $t$, but as $t \to 0^+$, $S_t$ becomes very thin in the $y_1$-direction (width $2t \to 0$) and very tall in the $y_2$-direction (height $2|D_L|^{1/2}/t \to \infty$). A lattice point $\sigma(\alpha) = (\sigma_1(\alpha), \sigma_2(\alpha))$ with $\sigma_1(\alpha) \neq 0$ eventually fails the first-coordinate constraint once $t < |\sigma_1(\alpha)|$. **Construction of the sequence.** Choose $t_1 > 0$ arbitrary and $\alpha_1 \in \mathcal{O}_L \setminus \{0\}$ from $S_{t_1}$. Inductively, once $\alpha_1, \ldots, \alpha_k$ have been chosen, select $t_{k+1}$ small enough that $\sigma(\alpha_j) \notin S_{t_{k+1}}$ for every $j \leq k$: \begin{align*} t_{k+1} &:= \frac{1}{2} \min\bigl(t_k, |\sigma_1(\alpha_1)|, \ldots, |\sigma_1(\alpha_k)|\bigr). \end{align*} This is a positive real number: all $|\sigma_1(\alpha_j)| > 0$ because $\sigma_1$ is injective and $\alpha_j \neq 0$, and $t_k > 0$ inductively. Moreover $t_{k+1} < t_k$ and $t_{k+1} < |\sigma_1(\alpha_j)|$ for each $j \leq k$, so $\sigma(\alpha_j) \notin S_{t_{k+1}}$. **Applying Step 2 again.** The non-zero $\alpha_{k+1}$ produced from $S_{t_{k+1}}$ satisfies $\sigma(\alpha_{k+1}) \in S_{t_{k+1}}$, and therefore $\alpha_{k+1} \neq \alpha_j$ for $j \leq k$ (otherwise $\sigma(\alpha_j) = \sigma(\alpha_{k+1}) \in S_{t_{k+1}}$). **Outcome.** We obtain an infinite sequence $(\alpha_j)_{j \geq 1}$ of pairwise distinct non-zero elements of $\mathcal{O}_L$, each with bounded norm $1 \leq |N(\alpha_j)| \leq |D_L|^{1/2}$. [/guided] [/step] [step:Pigeonhole on norm values to fix $N(\alpha_j) = m$ for infinitely many $j$] The set of possible norms $\{m \in \mathbb{Z} : 1 \leq |m| \leq |D_L|^{1/2}\}$ is finite, with cardinality at most $2\lfloor |D_L|^{1/2} \rfloor$. The sequence $(N(\alpha_j))_{j \geq 1}$ takes values in this finite set. By the pigeonhole principle, there exists a fixed integer $m$ with $1 \leq |m| \leq |D_L|^{1/2}$ such that the subsequence \begin{align*} J_1 &:= \{j \geq 1 : N(\alpha_j) = m\} \end{align*} is infinite. [guided] **Finite range.** The norm values $N(\alpha_j)$ are integers (by [Theorem 1574](/theorems/1574)) lying in $[-|D_L|^{1/2}, -1] \cup [1, |D_L|^{1/2}]$, a finite set of size at most $2\lfloor |D_L|^{1/2} \rfloor$. **Pigeonhole.** An infinite sequence taking values in a finite set has at least one value attained infinitely often. Pick $m$ to be such a value. The infinite index set $J_1 = \{j : N(\alpha_j) = m\}$ indexes a subsequence $(\alpha_j)_{j \in J_1}$, still pairwise distinct and with common norm $N(\alpha_j) = m$. **Fixing the sign too.** Although "norm equal to $m$" might seem weaker than "norm equal to $|m|$," we use the signed value here: $N(\alpha_j) = m$ includes the sign. This will matter in Step 6 when we divide, since $N(\alpha_j)/N(\beta)$ needs to equal $1$ (not $\pm 1$) for a unit. [/guided] [/step] [step:Pigeonhole on residues modulo $m\mathcal{O}_L$ to isolate a common class] The quotient ring $\mathcal{O}_L / m\mathcal{O}_L$ is finite: since $\mathcal{O}_L$ is a free $\mathbb{Z}$-module of rank $2$ (by [Existence of an Integral Basis](/theorems/1581)), the quotient by the subgroup $m\mathcal{O}_L = m\mathbb{Z}^2$ is $(\mathbb{Z}/m\mathbb{Z})^2$, of order $|m|^2$. The sequence $(\alpha_j \mod m\mathcal{O}_L)_{j \in J_1}$ is an infinite sequence in the finite set $\mathcal{O}_L/m\mathcal{O}_L$. By pigeonhole, there exists $\beta \in \mathcal{O}_L$ and an infinite subset $J_2 \subseteq J_1$ such that \begin{align*} \alpha_j &\equiv \beta \pmod{m\mathcal{O}_L} \quad \text{for all } j \in J_2. \end{align*} Since each $\alpha_j \neq 0$ and the $\alpha_j$ for $j \in J_2$ are pairwise distinct, replacing $\beta$ by any $\alpha_{j_0}$ with $j_0 \in J_2$ ensures $\beta \neq 0$ and $N(\beta) = m$. [guided] **Finiteness of $\mathcal{O}_L/m\mathcal{O}_L$.** The ring of integers $\mathcal{O}_L$ is a free $\mathbb{Z}$-module of rank $n = 2$ by [Theorem 1581](/theorems/1581) (the integral basis theorem). Choose an integral basis $\omega_1, \omega_2$, identifying $\mathcal{O}_L \cong \mathbb{Z}^2$ as abelian groups. Multiplication by $m$ scales each $\mathbb{Z}$-coordinate by $m$, so $m\mathcal{O}_L \cong m\mathbb{Z}^2$. The quotient is $\mathbb{Z}^2/m\mathbb{Z}^2 \cong (\mathbb{Z}/m\mathbb{Z})^2$, of order $m^2$. **Pigeonhole on residues.** The infinite sequence $(\alpha_j)_{j \in J_1}$ projects to an infinite sequence in the finite set $\mathcal{O}_L/m\mathcal{O}_L$ (of size $m^2$). By pigeonhole, some residue class is hit infinitely often. Pick any representative $\beta \in \mathcal{O}_L$ of this class, and let $J_2 \subseteq J_1$ be the infinite set of indices $j$ with $\alpha_j \equiv \beta \pmod{m\mathcal{O}_L}$. **Ensuring $\beta \neq 0$ and $N(\beta) = m$.** To simplify later calculations, we may replace $\beta$ by any $\alpha_{j_0}$ for some $j_0 \in J_2$. This choice is in the same residue class as all $\alpha_j$ for $j \in J_2$, and satisfies $\beta \neq 0$, $N(\beta) = m$, and $\beta \in \mathcal{O}_L$. [/guided] [/step] [step:Convert the common residue class into infinitely many units] For $j \in J_2$, write $\alpha_j = \beta + m \delta_j$ with $\delta_j \in \mathcal{O}_L$. Let $\bar\beta = \sigma_2(\beta)$ denote the Galois conjugate of $\beta$, i.e., its image under the non-trivial embedding. By [Norm and Trace via Embeddings](/theorems/1577), $N(\beta) = \beta \bar\beta$, so (using $N(\beta) = m$): \begin{align*} \beta \bar\beta &= m \implies \frac{m}{\beta} = \bar\beta \in \mathcal{O}_L. \end{align*} Consider \begin{align*} \varepsilon_j &:= \frac{\alpha_j}{\beta} = \frac{\beta + m\delta_j}{\beta} = 1 + \frac{m}{\beta} \delta_j = 1 + \bar\beta \delta_j. \end{align*} Since $\bar\beta \in \mathcal{O}_L$ and $\delta_j \in \mathcal{O}_L$, $\varepsilon_j \in \mathcal{O}_L$. Computing the norm: \begin{align*} N(\varepsilon_j) &= \frac{N(\alpha_j)}{N(\beta)} = \frac{m}{m} = 1. \end{align*} By [Units via Norm](/theorems/1578), $\varepsilon_j \in \mathcal{O}_L^\times$. (The hypotheses of Theorem 1578 — that $\varepsilon_j \in \mathcal{O}_L$ and $N(\varepsilon_j) = \pm 1$ — are satisfied.) Finally, the $\varepsilon_j$ for $j \in J_2$ are pairwise distinct: $\varepsilon_j = \varepsilon_k$ gives $\alpha_j = \alpha_k$ (by multiplying by $\beta \neq 0$), which by distinctness of the $\alpha_j$ forces $j = k$. Hence we obtain infinitely many distinct units $\varepsilon_j \in \mathcal{O}_L^\times$. Each unit $\varepsilon_j = x_j + y_j \sqrt{d}$ satisfies $N(\varepsilon_j) = x_j^2 - d y_j^2 = \pm 1$ (using $N_{L/\mathbb{Q}}(x + y\sqrt{d}) = x^2 - dy^2$, which is $\pm 1$ by $|N(\varepsilon_j)| = 1$). Therefore there are infinitely many solutions $(x, y) \in \mathbb{Q}^2$ with $x^2 - dy^2 = \pm 1$ and $x + y\sqrt{d} \in \mathcal{O}_L$. [guided] **Constructing $\varepsilon_j$.** From $\alpha_j \equiv \beta \pmod{m\mathcal{O}_L}$ we write $\alpha_j = \beta + m\delta_j$ with $\delta_j \in \mathcal{O}_L$. The ratio $\varepsilon_j := \alpha_j/\beta$ seems like a fraction, but we show it is integral by rewriting: \begin{align*} \varepsilon_j = \frac{\beta + m\delta_j}{\beta} = 1 + \frac{m}{\beta} \delta_j. \end{align*} We must show $m/\beta \in \mathcal{O}_L$. **Key identity $m/\beta = \bar\beta$.** The norm is $N(\beta) = \sigma_1(\beta) \sigma_2(\beta) = \beta \bar\beta$ (identifying $L$ with $\sigma_1(L) \subset \mathbb{R}$, so $\beta = \sigma_1(\beta)$ and writing $\bar\beta = \sigma_2(\beta)$). Since $N(\beta) = m$, we have $\beta\bar\beta = m$, so $m/\beta = \bar\beta$. **Integrality of $\bar\beta$.** The Galois conjugate $\bar\beta = \sigma_2(\beta)$ is another element of $\mathcal{O}_L$ (since $\beta \in \mathcal{O}_L$, integrality is preserved by ring embeddings — $\bar\beta$ satisfies the same monic integer polynomial as $\beta$, namely the conjugate of the minimal polynomial, which is the same polynomial because $\beta$'s minimal polynomial has integer coefficients). Hence $\bar\beta \in \mathcal{O}_L$. **Integrality of $\varepsilon_j$.** From $\varepsilon_j = 1 + \bar\beta \delta_j$ with $1, \bar\beta, \delta_j \in \mathcal{O}_L$, we conclude $\varepsilon_j \in \mathcal{O}_L$. **Norm of $\varepsilon_j$.** Multiplicativity of the norm: $N(\alpha_j/\beta) = N(\alpha_j)/N(\beta) = m/m = 1$. **Applying [Units via Norm](/theorems/1578).** The theorem states that $\varepsilon \in \mathcal{O}_L$ is a unit if and only if $N(\varepsilon) = \pm 1$. Hypothesis: $\varepsilon_j \in \mathcal{O}_L$, which we checked; $N(\varepsilon_j) = 1 = \pm 1$, so the hypothesis is satisfied. Conclusion: $\varepsilon_j$ is a unit. **Distinctness.** If $\varepsilon_j = \varepsilon_k$, then $\alpha_j/\beta = \alpha_k/\beta$ and hence $\alpha_j = \alpha_k$; by the pairwise-distinctness from Step 3, this forces $j = k$. Hence the infinite sequence $(\varepsilon_j)_{j \in J_2}$ of units is pairwise distinct. **Translation to Pell's equation.** Writing $\varepsilon_j = x_j + y_j\sqrt{d}$ with $x_j, y_j \in \frac{1}{2}\mathbb{Z}$ (or $\mathbb{Z}$, depending on the residue of $d$ modulo $4$, per [Integers of Quadratic Fields](/theorems/1575)), the norm is \begin{align*} N(x + y\sqrt{d}) = (x + y\sqrt{d})(x - y\sqrt{d}) = x^2 - dy^2, \end{align*} so $x_j^2 - dy_j^2 = N(\varepsilon_j) = 1 = \pm 1$. This is Pell's equation (with $\pm 1$ on the right). **Summary.** The infinitely many distinct units $\varepsilon_j \in \mathcal{O}_L^\times$ give infinitely many distinct solutions $(x_j, y_j)$ of $x^2 - dy^2 = \pm 1$ with $x + y\sqrt{d} \in \mathcal{O}_L$. [/guided] [/step]