Pell’s equation and simple continued fractions

Theorem 1.

Let $d$ be a positive integer which is not a perfect square, and let $\mathrm{(}x\mathrm{,}y\mathrm{)}$ be a solution of $x^{\mathrm{2}}\mathrm{-}d\mathit{}{y}^{\mathrm{2}}\mathrm{=}\mathrm{1}$. Then $\frac{x}{y}$ is a convergent in the simple continued fraction expansion of $\sqrt{d}$.

Proof.

Suppose we have a non-trivial solution $x,y$ of Pell’s equation, i.e. $y\ne 0$. Let $x,y$ both be positive integers. From

$${\left(\frac{x}{y}\right)}^2=d+\frac{1}{y^2}$$

we see that ${\left(\frac{x}{y}\right)}^2>d$, hence $\frac{x}{y}>\sqrt{d}$. So we get

$\left|{\displaystyle \frac{x}{y}}-\sqrt{d}\right|={\displaystyle \frac{1}{y^2\left(\frac{x}{y}+\sqrt{d}\right)}}$

This implies that $\frac{x}{y}$ is a convergent of the continued fraction of $\sqrt{d}$. ∎

Title	Pell’s equation and simple continued fractions
Canonical name	PellsEquationAndSimpleContinuedFractions
Date of creation	2013-03-22 13:21:04
Last modified on	2013-03-22 13:21:04
Owner	Thomas Heye (1234)
Last modified by	Thomas Heye (1234)
Numerical id	9
Author	Thomas Heye (1234)
Entry type	Theorem
Classification	msc 11D09