Rational solutions of Pell's equation

Question

1) $D$ is a positive integer, find all rational solutions of Pell's equation

$$x^2-Dy^2=1$$

2) What about $D\in\Bbb Q$ ?

Answer 1

The following theorem is due to David Hilbert:

Let $K/\mathbf Q$ be a cyclic Galois extension, with Galois group generated by $\sigma$. Then every element of norm $1$ of $K$ is of the form $\sigma(x)/x$ for some $x \in K^\times$.

Applying this to $K=\mathbf Q(\sqrt D)$ and noticing that $N(x+\sqrt D y) = x^2-Dy^2$, it follows that every solution to $N(x+\sqrt D y) = 1$ is of the form

$$x+\sqrt Dy = \frac{u - \sqrt D v}{u + \sqrt D v} = \frac{u^2+Dv^2}{u^2-Dv^2} + \sqrt D \frac{-2uv}{u^2-Dv^2},$$

i.e.

$$(x,y) = \left(\frac{u^2+Dv^2}{u^2-Dv^2}, \frac{-2uv}{u^2-Dv^2}\right), \qquad (u, v) \in \mathbf Q,\qquad u^2+v^2 \neq 0.$$

Answer 2

$$(\frac{x}y)^2-(\frac{1}y)^2=D,\\(\frac{x}y+\frac{1}y)(\frac{x}y-\frac{1}y)=D,\\\frac{x}y+\frac{1}y=t,\frac{x}y-\frac{1}y=\frac{D}t,\\\frac{x}y=\frac{1}2(t+\frac{D}{t}),\frac{1}y=\frac{1}2(t-\frac{D}{t}),\\x=\dfrac{t^2+D}{t^2-D},y=\frac{2t}{t^2-D},(t\in \mathbb Q,t^2\neq D).$$

Answer 3

The difference equation given by initial conditions $a_0=0$, $b_0=1$ and the recursion: $a_{n+1}=A\times a_n+B\times b_n$ and $b_{n+1}=N\times B\times a_n+A\times b_n$

Gives solutions to: $(b_n)^2-N\times (a_n)^2=(A^2-N*B^2)^n$

so for example $A=2, B=3, N=1/3$ gives solutions to $(b_n)^2-(1/3)\times (a_n)^2=1$ because $A^2-(1/3)\times B^2=1$