How to show that the continued fraction of $\sqrt{a^2+1}$ is $[a,2a,2a,2a,\dots]$?

Question

I thought about the following, as $a^2\leq a^2+1 \leq (a+1)^2$, then we have $a\leq \sqrt{a^2+1}\leq a+1$ so $\lfloor \sqrt{a^2+1}\rfloor=a$ but I am having trouble doing the same for the $2a,2a,\dots$ and even If I am able to do that, I don't know what I'd use to prove that all other elements are equal to $2a$.

Answer 1

Note that $(\sqrt{a^2+1} +a) (\sqrt{a^2+1} -a)=1$. This can be rearranged to give \begin{eqnarray*} \sqrt{a^2+1}=a+ \frac{1}{a+\sqrt{a^2+1}}. \end{eqnarray*}

Answer 2

You get then that the fractional part of $\sqrt{a^2+1}$ is $\sqrt{a^2+1}-a,$ multiply and divide by $\sqrt{a^2+1}+a$ you get $$\sqrt{a^2+1}=a+\frac{1}{\sqrt{a^2+1}+a}=a+\frac{1}{2a+(\sqrt{a^2+1}-a)},$$ so you will always get the same expression for the integer part.