I have to write $x= \sqrt{n^2+2}$ as a continued fraction, where $n \in N^*$.
I tried something like this: $$n< \sqrt{n^2+2}<n+1 \text{ so } [a_{0}]=n\\ x_1= \frac{1}{x-a_0}=\frac{1}{2}(\sqrt{n^2+2}+n), [a_1]=n $$
but for $x_2$ I obtained $\frac{2}{\sqrt{n^2+2}}$. From here I don't know how to proceed.
Consider instead $y=x-n$ then we have \begin{align} y &=\sqrt{n^2+2}-n\\ &=\frac{\left(\sqrt{n^2+2}-n\right)\left(\sqrt{n^2+2}+n\right)}{\sqrt{n^2+2}+n}\\ &=\frac2{\sqrt{n^2+2}+n}\\ &=\frac1{n+\frac{y}2}\\ &=\frac1{n+\frac{1/(n+y/2)}2}\\ &=\frac1{n+\frac1{2n+y}}\\ \end{align} Applying this final recurrence should give a continued fraction representation of $$y=[0;\overline{n,2n}]$$ and so we have $$x=[n;\overline{n,2n}]$$ $x$ is then irrational because any rational number has a finite continued fraction representation.