For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$

Here's a more pedestrian solution that does not rely on any number-theoretic result. We show the inequalities $$\lfloor \sqrt k +\sqrt{k+1} \rfloor  \leq \sqrt{4k+2} < \lfloor \sqrt k +\sqrt{k+1} \rfloor+1.$$
Note that $\sqrt k + \sqrt{k+1}\leq \sqrt{4k+2} \iff 2\sqrt k \sqrt{k+1} \leq 2k+1 \iff 0\leq 1$, hence $$\lfloor \sqrt k +\sqrt{k+1} \rfloor  \leq \sqrt{4k+2}.$$
For the remaining strict inequality, note by squaring and comparing integers that
$\begin{align}
\sqrt{4k+2} < \lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 
&\iff 4k+2 < \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor + 1   \\
&\iff 4k+3 \leq \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor + 1 \\
&\iff 4k+2 \leq \lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor.
\end{align}$
The inequality is true for $k\in \{0,1\}$ so we assume in what follows that $k\geq 2$. Since $\lfloor z \rfloor > z-1$ and $\sqrt k +\sqrt{k+1}-1\geq 2\sqrt k -1\geq 0$,we have
$$\begin{align}
\lfloor \sqrt k +\sqrt{k+1}\rfloor^2 + 2\lfloor \sqrt k +\sqrt{k+1}\rfloor 
&\geq (\sqrt k +\sqrt{k+1}-1)^2 + 2(\sqrt k +\sqrt{k+1}-1)\\
&\geq (2\sqrt k -1)^2 + 2(\sqrt k +\sqrt{k+1}-1)\\
&=4k + 1 + 2(\sqrt{k+1}-1)\\
&\geq 4k + 1 + 2(\sqrt{3}-1)\\
&\geq 4k+2.
\end{align}$$
