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

First a kooky Lemma:
For $x > 1$ then $\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$
[$\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$ iff
$1 - \frac 1 x + 1 + \frac 3 x + 2\sqrt{(1 - \frac 1 x)(1 + \frac 3 x)} > 4$ iff
$2  + \frac 2 x + 2\sqrt{(1 + \frac 2 x -\frac 2 {x^2}} > 4$ iff
$1 + \frac 2 x -\frac 2 {x^2} > 1$ iff
$\frac 2 x -\frac 2 {x^2} > 0$ ]
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So now to answer your question:
Let $n^2 \le x < (n+1)^2$ or in other words $n^2 \le x \le n^2 + 2n$.
It's obvious:
$2n < \sqrt x + \sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = \{2n, 2n + 1\}$.
Likewise $2n = \sqrt{4n} < \sqrt{4x + 2} \le \sqrt{4n^2 + 8n + 2} < \sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $\lfloor  \sqrt {4x + 2} \rfloor = \{2n, 2n + 1\}$.
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Case 1: 
$n^2 \le x < x + 1 \le n^2 + n < n^2 + n + \frac 1 4 = (n + \frac 1 2)^2$
Then $ \sqrt x + \sqrt {x + 1} < n + \frac 1 2 + n + \frac 1 2 < 2n + 1$
So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n$.
Likewise $ \sqrt{4x + 2} \le \sqrt{4n^2 + 4(n -1) + 2} < \sqrt{4n^2 + 4n + 1} = 2(n + \frac 1 2) = 2n + 1$.
So $\lfloor  \sqrt {4x + 2} \rfloor = 2n = \lfloor \sqrt x + \sqrt {x + 1} \rfloor$.
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Case 2: 
$n^2 + 2n + 1 \ge x + 1 > x \ge n^2 + n$.
Then $\sqrt{4x + 2} \ge \sqrt{4n^2 + 4n + 2} = 2(n + \frac 12) = 2n + 1$.
So $\lfloor  \sqrt {4x + 2} \rfloor = 2n + 1$.
Now $\sqrt x + \sqrt {x + 1} \ge \sqrt{n^2 + n} + \sqrt{n^2 + n + 1} = \sqrt{n^2 + n + \frac 1 4 - \frac 1 4} + \sqrt{n^2 + n + \frac 1 4 + \frac 3 4}$
$= (n + \frac 1 2)[\sqrt{1 - \frac 1 {4(n + \frac 1 2)^2}} + \sqrt{1 + \frac 3 {4(n + \frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + \frac 1 2)2 = 2n + 1$.
So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n + 1 = \lfloor \sqrt{4x +2} \rfloor$.
