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Pell's equation and simple continued fractions
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(Theorem)
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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 $$\left(\frac{x}{y}\right)^2 =d +\frac{1}{y^2}$$ we see that $\left(\frac{x}{y}\right)^2 > d$ hence $\frac{x}{y} > \sqrt{d}$ So we get \begin{eqnarray*} \left\vert \frac{x}{y} -\sqrt{d}\right\vert =\frac{1}{y^2\left(\frac{x}{y} +\sqrt{d}\right)} & < \frac{1}{y^2\left(2\sqrt{d}\right)} \\ & < \frac{1}{2y^2}. \end{eqnarray*}This implies that $\frac{x}{y}$ is a convergent of the continued fraction of $\sqrt{d}$ 
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"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
There is 1 reference to this entry.
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 4565 times total.
Classification:
| AMS MSC: | 11D09 (Number theory :: Diophantine equations :: Quadratic and bilinear equations) |
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Pending Errata and Addenda
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