<s>[INST] continued fraction development of the numbers of the following form, $\sqrt{n^2-1}$ 
Determine the continued fraction development of the numbers of the
  following form, $\sqrt{n^2-1}$, with $n>1$ an integer.

I wasn't sure how to tackle this, so I just tried to write it out and tried to find $a_1...a_n$.
$$\alpha_0=\sqrt{n^2-1}=n-1 +\sqrt{n^2-1}+n-1$$where $n-1$ is $a_0$ and $\sqrt{n^2-1}+n-1$ is {$\alpha_0$}.
This brings us to $\alpha_1=\frac{1}{\{\alpha_0\}}=\frac{1}{\sqrt{n^2-1}+n-1}=\frac{\sqrt{n^2-1}-n+1}{n^2-1-(n-1)^2}=\frac{\sqrt{n^2-1}-n+1}{2n-2}$
But $a_1$ would be $0$ here, so I don't think I'm on the right track.
 [/INST] $a_0=\lfloor \sqrt {n^2-1} \rfloor = n-1$
The repeating part is $1,2(n-1)$.  We know when we get to $2a_0$ in a square root we have hit the repeat.  You can write 
$$\sqrt{n^2-1}=n-1+(\sqrt {n^2-1}-(n-1))\\=n-1+\frac 1{\frac 1{\sqrt {n^2-1}-(n-1)}}\\=n-1+\frac 1{\frac {\sqrt{n^2-1}+n-1}{n^2-1-(n-1)^2}}\\=n-1+\frac 1{\frac {\sqrt{n^2-1}+n-1}{2n-2}}$$
and the first $1$ comes out in the continued fraction.  Keep going and you will get $2n-2$ at the next step.
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