<s>[INST] continued fraction for $\sqrt{n^2 − 1}$ Question:

Find the continued fraction for $\sqrt{n^2 − 1}$, where $n \ge 2$ is
  an integer.

My attempt:
$n - 1 = \sqrt{n^2} - 1 \lt \sqrt{n^2 − 1} \lt \sqrt{n^2}$
So far, $[n-1; ]$
$\sqrt{n^2 − 1} = n - 1 + \frac{1}{x}$ 
$\to \frac{1}{x} = \sqrt{n^2 - 1} - (n-1) \to  \frac{1}{x^2} = n^2-1-n^2-1+2n-2\times(n-1)\times\sqrt{n^2-1} = 2\times(n-1)\times(1-\sqrt{n^2-1})$
But, it doesn't really help to find continued fraction.
 [/INST] Use that $$x = \frac{1}{\sqrt{n^2-1} - (n-1)} = 1+ \frac{\sqrt{n^2-1} - (n-1)}{2(n-1)} = 1 + \frac{1}{2(n-1)x} $$
to conclude that $$\sqrt{n^2-1} = [n-1; \overline{1, 2(n-1)}]$$
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