If $N = y^2+1$ then the continued fraction of $\sqrt(N)$ has period 1. 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}}
$.
