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If √N is co...

Question

If $\sqrt{N}$ is converted into a continued fraction, and if n is the number of quotients in the period, show that $q_{2 n} = 2 p_{n} q_{n}, p_{2 n} = 2 p_{n}^{2} + (- 1)^{n + 1}$ .

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Solution

It is a general thesis and can be proved that $(2 n) t h$ convergent is
$\frac{p_{2 n}}{q_{2 n}} = \frac{1}{2} (\frac{p_{n}}{q_{n}} + \frac{N q_{n}}{p_{n}})$
$= \frac{p_{n}^{2} + N q_{n}^{2}}{2 p_{n} q_{n}}$
And hence by comparison, it can be shown that;-
$q_{2 n} = 2 p_{n} q_{n}$ and $p_{2 n} = p_{n}^{2} + N q_{n}^{2}$
Also it is known that ;-
$a_{1} p_{n} + p_{n - 1} = N q_{n}; a_{1} q_{n} + q_{n - 1} = p_{n}$
$∴ p_{n}^{2} - N q_{n}^{2} = p_{n} q_{n - 1} - p_{n - 1} q_{n} = {(- 1)}^{n}$
$N q_{n}^{2} = p_{n}^{2} + {(- 1)}^{n + 1}$
and therefore;-
$p_{2 n} = 2 p_{n}^{2} + {(- 1)}^{n + 1}$

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