Pell's equation¶

Pell's equation

Find all integer solutions to \(x^2 - dy^2 = n\)(\(d\) is square free and \(d \geqslant 2\)).

First consider some special cases.

Question

Is there a solution to \(x^2 - dy^2 = 1\)?

Observe that \(x^2 - dy^2 = 1\) is equivalent to \((x + y\sqrt{d})(x - y\sqrt{d}) = 1\). So we can define the norm \(N(\alpha) = (x + y \sqrt{d})(x - y \sqrt{d}) = x^2 - dy^2, \alpha = x + y\sqrt{d}\). So \(N(\alpha \beta) = N(\alpha)N(\beta)\).

Definition

We call \(x^2-dy^2 = 1\) Standard Pell's equation, \(x^2-dy^2 = -1\) Negative Pell's equation and \(x^2-dy^2 = n\) Generalized Pell's equation.

Standard Pell's equation¶

Main Theorem

(Lagrange) Standard Pell's equation has a non-trivial solution.

Lemma

There are infinitely many postitive integers \(x, y\) such that

\[ \lvert x - y\sqrt{d} \rvert < \frac{1}{y}. \]

Proof

\(\{x\}\): decimal part of \(x\). \(\{x\} \in [0, 1)\).
Consider \(0, \{\sqrt{d}\}, \{2\sqrt{d}\}, \cdots, \{m\sqrt{d}\}, \cdots\). (\(m\) is large enough).
Use Pigeonhole principle: Devide \([0, 1)\) into \(m\) intervals \([\frac{i}{m}, \frac{i+1}{m})\). So there exists \(\{k\sqrt{d}\}, \{l\sqrt{d}\}\) falling in the same interval. So \(\lvert \{k\sqrt{d}\} - \{l\sqrt{d}\} \rvert < \frac{1}{m}(0 \leqslant l < k \leqslant m)\). Which means \(\lvert (k\sqrt{d} - [k\sqrt{d}]) - (l\sqrt{d} - [l\sqrt{d}]) \rvert < \frac{1}{m}\). Set \(x = [k\sqrt{d}] - [l\sqrt{d}], y = k - l\). So \(\lvert x - y\sqrt{d} \rvert < \frac{1}{m} \leqslant \frac{1}{k - l} = \frac{1}{y}\). Then we can use this to generate new \(m\). So there are infinitely many \(x, y\) satisfying the condition.

Collary

\(\exists M\) such that \(x^2 - dy^2 = M\) has infinitely many solutions.

Proof

Find \((x_n, y_n)\), \(\lvert x_n - y_n\sqrt{d} \rvert < \frac{1}{y_n}\).
So we have
(a) \(x_n = x_n - y_n\sqrt{d} + y_n\sqrt{d} < \frac{1}{y_n} + y_n\sqrt{d}\).
(b) \(\lvert x_n^2 - dy_n^2 \rvert = \lvert x_n - y_n\sqrt{d} \rvert \cdot \lvert x_n + y_n\sqrt{d} \rvert < \frac{1}{y_n}(\frac{1}{y_n} + 2y_n\sqrt{d}) < 1 + 2\sqrt{d}\).
So \(\exists M >0\), such that \(\lvert x^2 - dy^2 \rvert = M\) has infinitely many solutions.

Proof of Main Theorem

\(\exists M, \lvert M \rvert < 1 + 2\sqrt{d}\), such that \(x^2 - dy^2 = M\) has infinitely many solutions. So we have \(x_1^2 - dy_1^2 = M, x_2^2 - dy_2^2 = M\).
By the fitness of \(\mathbb{Z}/\lvert M \rvert \mathbb{Z} \times \mathbb{Z}/\lvert M \rvert \mathbb{Z}\), we assume that

\[ \begin{gather} x_1 \equiv x_2 \pmod {\lvert M \rvert}, x_2 = x_1 + M \cdot k \\ y_1 \equiv y_2 \pmod{\lvert M \rvert}, y_2 = y_1 + M \cdot l \end{gather} \]

So \(x_2 + y_2\sqrt{d} = x_1 + y_1\sqrt{d} + (k + l\sqrt{d})M\). And \(x_1^2 - dy_1^2 = M\). So \(x_2 + y_2\sqrt{d} = (x_1 + y_1\sqrt{d})(1 + (k + l\sqrt{d})(x_1 - y_1\sqrt{d}))\). Set \(\gamma = 1 + (k + l\sqrt{d})(x_1 - y_1\sqrt{d})\), then \(x_2 + y_2\sqrt{d} = (x_1 + y_1\sqrt{d}) \cdot \gamma\). So \(N(x_2 + y_2\sqrt{d}) = N(x_1 + y_1\sqrt{d}) \cdot N(\gamma) \Rightarrow M = M \cdot N(\gamma)\), \(N(\gamma) = 1\).

Theorem

(i) \(N(\gamma) = 1\) has a fundamental solution \(\alpha\). Every other solution is equal to \(\pm \alpha^k, k \in \mathbb{Z}\).
(ii) The fundamental solution is the one with \(y > 0\) and minimal.

Lemma

\(x^2 - dy^2 = a^2 - db^2 = 1\), \((x, y) \neq (a, b), x, y, a, b > 0\). The following statements are equivalent:
(i) \(x+y\sqrt{d} > a+b\sqrt{d}\).
(ii) \(x > a\) and \(y > b\).
(iii) \(x > a\) or \(y > b\).

Negative Pell's equation¶

Theorem

Consider negative Pell's equation \(x^2 - dy^2 = -1\)(\(N(\gamma) = -1\)). Assume \(x^2 - dy^2 = -1\) has a solution.
(i) There exists a fundamental solution \(\gamma_1 = x_1 + y_1\sqrt{d}\), \(\gamma_1 > 1\) and is minimal.
(ii) \(N(\gamma_1) = -1 \Rightarrow N(\gamma_1^2) = 1\). Then \(\gamma_1^2\) is the fundamental solution of \(x^2 - dy^2 = 1\).
(iii) Every solution of \(x^2 - dy^2 = -1\) has the form \(\gamma = \gamma_1^{2k+1}, k \in \mathbb{Z}\).

Generalized Pell's equation¶

Theorem

Let \(u = a + b\sqrt{d}\) be the fundamental solution of the standard Pell's equation \(a^2 - d \cdot b^2 = 1\). Then for each \(n \in \mathbb{Z}, n \neq 0\), every solution of \(x^2 - dy^2 = n\) has the form \(x + y\sqrt{d} = (x' + y' \sqrt{d} \cdot u^k)\), where

\[\begin{gather} \lvert x' \rvert \leqslant \dfrac{\sqrt{\lvert n \rvert}(\sqrt{u} + \sqrt{u}^{-1})}{2} \\ \lvert y' \rvert \leqslant \dfrac{\sqrt{\lvert n \rvert}(\sqrt{u} + \sqrt{u}^{-1})}{2\sqrt{d}} \end{gather}\]

We also have a more precise bound for \(y'\):

\[ \lvert y' \rvert \leqslant \dfrac{\sqrt{\lvert n \rvert}(\sqrt{u} - \sqrt{u}^{-1})}{2\sqrt{d}} \]

This allows us to find the solutions of \(x^2 - dy^2 = n\) with fewer trials.