If $\sqrt{a^{2} + 1}$ be expressed as a continued fraction, show that $2 (a^{2} + 1) q_{n} = p_{n - 1} + p_{n + 1}, 2 p_{n} = q_{n - 1} + q_{n + 1}$ .

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Solution

We can write $\sqrt{a^{2} + 1} = a + \sqrt{a^{2} + 1} - a = a + \frac{1}{\sqrt{a^{2} + 1} + a}$
$\sqrt{a^{2} + 1} + a = 2 a + (\sqrt{a^{2} + 1} - a) = 2 a + \frac{1}{\sqrt{a^{2} + 1} + a}$
The complete quotient at any stage is always $\sqrt{a^{2} + 1} + a$ ; hence,
$\sqrt{a^{2} + 1} = \frac{(\sqrt{a^{2} + 1} + a) p_{n} + p_{n - 1}}{(\sqrt{a^{2} + 1} + a) q_{n} + q_{n - 1}}$
Multiplying up, and equating rational and irrational parts,
$(a^{2} + 1) q_{n} = a p_{n} + p_{n - 1} ⟶ 1$
$a q_{n} + q_{n - 1} = p_{n} ⟶ 2$
Now, $\sqrt{a^{2} + 1} = a + \frac{1}{2 a +} \frac{1}{2 a +} \frac{1}{2 a +} . . . .$
Therefore, $p_{n + 1} = 2 a p_{n} + p_{n - 1}; q_{n + 1} = 2 a p_{n} + q_{n - 1}$
From $1, 2 (a^{2} + 1) q_{n} = 2 a p_{n} + 2 p_{n - 1} = p_{n + 1} + p_{n - 1}$
From $2, 2 p_{n} = 2 a q_{n} + 2 q_{n - 1} = q_{n + 1} + q_{n - 1}$

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