Calculating $\Big\lfloor\underbrace{\sqrt{2017+\sqrt{2017+...+\sqrt{2017}}}}_\text{2017 roots}\Big\rfloor$ 
Could you please help me computing 
  $\Big\lfloor\underbrace{\sqrt{2017+\sqrt{2017+...+\sqrt{2017}}}}_\text{2017 roots}\Big\rfloor$?

 A: Looking at the sequence $x_{n+1}=\sqrt{x_n+2017}$ it's easy to see that if $x_n\geq 8$ then $x_{n+1}\geq 45$, and you can prove that $x_n<46$ by induction easily since $x_{n+1}=\sqrt{x_n+2017}<\sqrt{2017+46}<46=\sqrt{46^2}=\sqrt{45^2+91}=\sqrt{2116}$
Side note: This is a problem from the 2017 Serbian high school republic contest (this is how I solved the problem, I think the official solution was similar might take a look if I still have them). 

The official solution (translated):

Look at the sequence given by $a_1=\sqrt{2017}$ and $a_{n+1}=\sqrt{2017+a_n}$. The value we are looking for is $\lfloor a_{2017}\rfloor$. Since $44^2=1936<2017<2025=45^2$, we have $44<a_1<45$, and from there it follows that $\lfloor a_1 \rfloor=44$.
Similarly, since $45^2=2025<2061<2017+a_1<2062<46^2=2116$ and $a_2=\sqrt{2017+a_1}$, we have $45<a_2<46$, and from there it follows that $\lfloor a_2 \rfloor=45$. With induction we'll prove for all $n\geq 2$ it holds that $\lfloor a_n\rfloor=45$, then we have $45\leq a_n< 46$ and $45^2=2025<2062\leq 2017+a_n<2063<46^2=2116$ and $a_{n+1}=\sqrt{2017+a_n}$ so we conclude that $45<a_{n+1}<46$ from where it follows that $\lfloor a_{n+1} \rfloor=45$. The induction proof is done and we immediately conclude $\lfloor a_{2017}\rfloor =45$.
A: With infinite nested square roots, the bracketed expression equals the positive solution of $x^2=2017+x$, namely $\frac{1+\sqrt{8069}}{2}=45.4138\ldots$. It follows that the answer is $\color{red}{45}$, and I leave to you to fill the missing details, i.e. proving that $2017$ nested square roots or an infinity of them do not make much of a difference. Sketch: $f(x)=\sqrt{2017+x}$ is a contraction on $[0,+\infty)$, hence by the Banach fixed point theorem the sequence defined by $a_0=\sqrt{2017}, a_{n+1}=f(a_n)$ converges to the aforementioned limit, and $|a_{n+1}-a_n|$ can be bounded in terms of $\sup_{x\in\mathbb{R}^+}\left|f'(x)\right|$, proving that $a_{2017}-a_{\infty}$ is very small. 
Without invoking heavy artillery (which, by the way, is pretty useful for tackling similar problems), one may simply notice that the sequence $\{a_n\}_{n\geq 0}$ is increasing and bounded above by its limit. Since both $a_1$ and $a_\infty$ lie in the interval $(45,46)$, the outcome is clear.
