Pell's equation

 

정의1

Let $D$ be a positive interger such that $\sqrt{D} \notin \mathbb{N}$.

Then $x^{2} - Dy^{2} = 1$ is called a Pell equation.

 

 

 

정리1

Let $D \in \mathbb{N}$  s.t.  $\sqrt{D} \notin \mathbb{N}$.

Then, for the Pell equation $x^{2} - Dy^{2} = 1$ the following hold

1. There exists positive intergers and $x_{0}$  and  $y_{0}$  s.t.  $x{_{0}}^{2} - Dy{_{0}}^{2} = 1$

 

2. Let $(x_{1}, y_{1})$ be a pair of positive intergers s.t.  $x{_{1}}^{2} - Dy{_{1}}^{2} = 1$  and  $x_{1}$ is the smallest among $x \in \mathbb{N}$ which satisfies $x^{2} - Dy^{2} = 1$ for some $y \in \mathbb{N}$.

Then $u^{2} - Dv^{2} = 1$ for some $u, v \in \mathbb{N}$.

$\Leftrightarrow$  $u+v\sqrt{D} = (x_{1} + y_{1}\sqrt{D})^{n}$ for some $n \in \mathbb{N}$.

 

pf)

1.

hard

 

2.

Similar to the case when $D=2$

 

 

 

Why $\sqrt{D} \notin \mathbb{N}$?

Suppose 1 : $\sqrt{D} \in \mathbb{N}$ (i.e. $D = A^{2}$  $(A \in \mathbb{N})$),  $x^{2} - Dy^{2} = 1$ for some $x,y \in \mathbb{N}$

$x^{2} - A^{2}y^{2} = 1$

$\Leftrightarrow$  $(x+Ay)(x-Ay) = 1$

 

Since $x, y, A \in \mathbb{N}$,

so $x+Ay \geq 2$  and  $0< x-Ay = \frac{1}{x+Ay} < 1$

Since $x, y, A \in \mathbb{N}$,

so $x-Ay$ must be a interger.

Contradiction

$\therefore$  $x^{2} - A^{2}y^{2} = 1$ has no interger solutions.