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: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} = \Bigl\lfloor{\sqrt{4n+2}\Bigl\rfloor}$$

Taking the listed approach, we get
$$[\sqrt{n^2+k}+\sqrt{n^2+k+1}]=[\sqrt{4n^2+4k +2}]$$
As $k$ relates to $n$, we have a floor-function boundary for the RHS when $k\in\{-1,0\}$ and also at $k\in\{n-1,n\}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well.  For $k\in\{-1,0\}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $k\in\{n-1,n\}$ by comparing with $2n+1$ from the RHS:
$$\left[\sqrt{n^2+n-1}+\sqrt{n^2+n}\right]\, R\, 2n+1 \\ \text{ vs. }\\ \left[\sqrt{n^2+n}+\sqrt{n^2+n+1}\right]\, R\, 2n+1$$
Squaring yields
$$\left[2n^2+2n-1+2\sqrt{(n^2+n)(n^2+n-1)}\right]\, R\, 4n^2+4n+1\\
\left[2\sqrt{(n^2+n)(n^2+n-1)}\right]\, R\, 2n^2+2n+2$$
and it should be clear that $[\sqrt{n^2+n-1}+\sqrt{n^2+n}]\lt 2n+1$.  Next we have
$$\left[2n^2+2n+1+2\sqrt{(n^2+n)(n^2+n+1)}\right]\, R\, 4n^2+4n+1\\
\left[2\sqrt{(n^2+n)(n^2+n+1)}\right]\, R\, 2n^2+2n$$
and it should also be clear that $[\sqrt{n^2+n}+\sqrt{n^2+n+1}]\ge 2n+1$, completing the proof.
