determine the continued fraction of $\sqrt{n^2 + 2}$ for $n \in \mathbb{N}$.
For rationals it is rather easy to do this and i know the algorithm, i only get stuck a lot by irrational number such as these. I tried this:
$\sqrt{n^2 + 2} = 1 + (\sqrt{n^2 + 2}) -1 = 1 + \frac{1- n^2}{1+\sqrt{n^2 + 2}}$. I almost get the form that i want, but the $n^2$ bothers me, how can i get rid of it?
Kees
$$\sqrt{n^2+2}-n = \frac{2}{\sqrt{n^2+2}+n} = \frac{2}{2n+\frac{2}{n+\sqrt{n^2+2}}} = \frac{1}{n+\frac{1}{n+\sqrt{n^2+2}}}$$ hence it follows that: $$ \sqrt{n^2+2} = [n;\overline{n,2n}].$$ On the other hand, if $\alpha=[n;n,n,n,\ldots]$, we have $\alpha=n+\frac{1}{\alpha}$, from which $\alpha^2-n\alpha-1=0$ and $\alpha=\frac{n+\sqrt{n^2+4}}{2}$. If $\beta=[n,2n,n,2n,n,2n,\ldots]$, in a similar way $\beta=n+\frac{1}{2n+\frac{1}{\beta}}$, hence $\beta=\frac{1}{2}\left(n+\sqrt{n^2+2}\right)$, $\frac{1}{\beta}=-n+\sqrt{n^2+2}$ and $\sqrt{n^2+2}=[n;\overline{n,2n}]$ as stated before.