Let $d \in \mathbb{N}$ such that $d$ not a square number. Now show that the continued fractions for $\sqrt{d} + \lfloor\sqrt{d}\rfloor$ and $\frac{1}{(\sqrt{d} - \lfloor\sqrt{d}\rfloor)}$ are purely periodic.
Furthermore, show that there exists $m \in N$ so that $\sqrt{d}$ is periodic from the second number, i.e. $[a_0$; $\overline > {a_1a_2,...,a_m}]$ .
I tried to solve this, but I do not have an idea where to start. Some of my attempts:
$1)$ For irrational number a of a degree $2$, and assuming $a'$ is the other root of the minimal polynomial of $a$, that $-1 < a' < 0$.
$~~1.1)$ Define following series $x_n$ for $a =[a_0$; ${a_1,a_2,...}$], then, $x_{n+1}$ is the real number such that $a =[a_0$; ${a_1,a_2,..., x_{n+1}}$] and that $x_n$ = $a_n$ + $\frac{1}{x_{n+1}}$.
$~~1.2)$ Show that for any $n \in\mathbb{N}$, $-1 < x_{n+1} < 0$ and that $a_n=\lfloor$$\frac{-1}{x_{n+1}}$$\rfloor$.
I've shown $1.1$ and $1.2$
Thank you.
