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Pell's equation and simple continued fractions (Theorem)
Theorem 1   Let $ d$ be a positive integer which is not a perfect square, and let $ (x,y)$ be a solution of $ x^2 -dy^2 =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 \neq 0$. Let $ x,y$ both be positive integers. From
$\displaystyle \left(\frac{x}{y}\right)^2 =d +\frac{1}{y^2}$
we see that $ \left(\frac{x}{y}\right)^2 > d$, hence $ \frac{x}{y} > \sqrt{d}$. So we get
$\displaystyle \left\vert \frac{x}{y} -\sqrt{d}\right\vert =\frac{1}{y^2\left(\frac{x}{y} +\sqrt{d}\right)}$   $\displaystyle < \frac{1}{y^2\left(2\sqrt{d}\right)}$  
    $\displaystyle < \frac{1}{2y^2}.$  

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



"Pell's equation and simple continued fractions" is owned by Thomas Heye.
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Cross-references: continued fraction, implies, Pell's equation, simple continued fraction, convergent, solution, perfect square, integer, positive
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This is version 6 of Pell's equation and simple continued fractions, born on 2003-01-04, modified 2006-10-10.
Object id is 3870, canonical name is PellsEquationAndSimpleContinuedFractions.
Accessed 4100 times total.

Classification:
AMS MSC11D09 (Number theory :: Diophantine equations :: Quadratic and bilinear equations)

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