I am reading a book that says that if $N = y^2+1$ then the continued fraction of $\sqrt{N}$ has period 1, i.e $\sqrt{N} = [q_0;\overline{q_1}]$ or similarly $$x=q_0+\cfrac{1}{q_1+\cfrac{1}{q_1+\cfrac{1}{ q_1+ \cdots}}}$$
What I've tried to do was to compute the continued fraction of $\sqrt{y^2+1}$ directly: Therefore $q_0 = \lfloor \sqrt{y^2+1} \rfloor = y$ and I tried to compute $$\frac{1}{\sqrt{y^2+1}-y} = \frac{\sqrt{y^2+1}-y}{2y^2+1-2y\sqrt{y^2+1}} = \frac{\sqrt{y^2+1}-y}{-2y(\sqrt{y^2+1}-y)+1} < \frac{\sqrt{y^2+1}-y}{-2y(\sqrt{y^2+1}-y)}$$ But here I am stuck and I guess that I made an stupid error because there's no way that $$\frac{\sqrt{y^2+1}-y}{-2y(\sqrt{y^2+1}-y)+1} < \frac{1}{-2y}$$
$\begin{array}\\ \sqrt{y^2+1}-y &=(\sqrt{y^2+1}-y)\dfrac{\sqrt{y^2+1}+y}{\sqrt{y^2+1}+y}\\ &=\dfrac{1}{\sqrt{y^2+1}+y}\\ \end{array} $
so $\sqrt{y^2+1} =y+\dfrac{1}{y+\sqrt{y^2+1}} $.