Is my approach correct to this equation? The problem is the following:

Does $a \in \mathbb{R}$ exist such that $[a + \sqrt{2n + 1}] = [a + \sqrt{2n + 2}]$ for all $n \in \mathbb{N}$? ($[x]$ denotes the whole part of $x$).

Note: I will also use $\{x\}$ to denote the fractional part of $x$. Also that $[x] + \{x\} = x$.
My approach is the following:  
There is no such that $a$. 
Proof: For the sake of simplicity, let's assume $a = k + \alpha$, $k \in \mathbb{Z}$ and $0 \le \alpha \lt 1$. Because of $[x + k] = [x] + k$ for any $k \in \mathbb{Z}$, and substituting $a = k + \alpha$, our equation $[k + \alpha + \sqrt{2n + 1}] = [k + \alpha + \sqrt{2n + 2}]$ becomes $[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}]$. Now our task is to find $\alpha \in [0, 1)$ that satisfies the equation for any $n \in \mathbb{N}$.  
First, let's find a formula for $\alpha$ for given $n$. There are two cases:  
1. $[\sqrt{2n + 1}] = [\sqrt{2n + 2}] = m$, $m \in \mathbb{N}$. This happens when $2n + 2$ isn't a perfect square $\iff$ $2n + 2 \ne p^2 \iff n \ne \frac {p^2} {2} - 1$, $p \in \mathbb{N}$.  
$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}] \iff [\alpha + [\sqrt{2n + 1}] + \{\sqrt{2n + 1}\}] = [\alpha + [\sqrt{2n + 2}] + \{\sqrt{2n + 2}\}] \iff [\alpha + \{\sqrt{2n + 1}\} + m] = [\alpha + \{\sqrt{2n + 2}\} + m] \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha + \{\sqrt{2n + 2}\}]$.
From this we get that $\alpha \in [0, 1 - \{\sqrt{2n + 2}\}) \tag {1'}$
Edit #1 (no longer actual beacuse of Edit #2): took out the maximum function from $(1')$ for obvious reason: the fractional part function is monotonically growing on its period, so if $[x] = [y]$, then $\{x\} \lt \{y\}$ only if $x \lt y$.
Edit #2: $(1')$ is only a partial solution for the 1. case, it is indeed a solution if both sides of the latest equation is equal to $0$. But they can both equal to $1$, so we have to consider that $\alpha \in [1 - \{\sqrt{2n + 1}\}, 1)$. 1. case overall:
$\alpha \in [0, 1 - \{\sqrt{2n + 2}\}) \cup [1 - \{\sqrt{2n + 1}\}, 1) \tag 1$
2. $[\sqrt{2n + 1}] \ne [\sqrt{2n + 2}] \implies [\sqrt{2n + 1}] = m, [\sqrt{2n + 2}] = m + 1$, $m \in \mathbb{N}$. This happens when $2n + 2$ is a perfect square $\iff$ $2n + 2 = p^2 \iff n = \frac {p^2} {2} - 1$, $p \in \mathbb{N}$. Now $[\sqrt{2n + 2}] = \sqrt{2n + 2} \implies \{\sqrt{2n + 2}\} = 0$.  
$[\alpha + \sqrt{2n + 1}] = [\alpha + \sqrt{2n + 2}] \iff [\alpha + [\sqrt{2n + 1}] + \{\sqrt{2n + 1}\}] = [\alpha + [\sqrt{2n + 2}] + \{\sqrt{2n + 2}\}] \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha + \{\sqrt{2n + 2}\}] + 1 \iff [\alpha + \{\sqrt{2n + 1}\}] = [\alpha] + 1 \iff [\alpha + \{\sqrt{2n + 1}\}] = 1$
From this we get that $\alpha = 1 - \{\sqrt{2n + 1}\} \tag 2$ 
Now we prove the following: (this is where it gets really ambiguous)
There does exists $n_1, n_2 \in \mathbb{N}$, $n_1 \ne \frac {{p_1}^2} {2} - 1$, $n_2 = \frac {{p_2}^2} {2} - 1$, $p_1$, $p_2 \in \mathbb{N} \iff [\sqrt{2n_1 + 1}] = [\sqrt{2n_1 + 2}]$, $[\sqrt{2n_2 + 1}] \ne [\sqrt{2n_2 + 2}]$ such that $ 1 - \{\sqrt{2n_1 + 2}\} \le 1 - \{\sqrt{2n_2 + 1}\} \lt 1 - \{\sqrt{2n_1 + 1}\} \tag 3 \iff$ there is no $\alpha$ satisfying both $(1)$ and $(2)$.
$\implies \{\sqrt{2n_1 + 1}\} \lt \{\sqrt{2n_2 + 1}\} \le \{\sqrt{2n_1 + 2}\} \tag 4$
(this is where it gets really-really-really ambiguous) 

If we substitute $n_1 = 3$ and $n_2 = 1$ in $(4)$ we get the following:
$\{\sqrt{7}\} \lt \{\sqrt{3}\} \le \{\sqrt{8}\} \implies \approx 0.64 \lt 0.73 \le 0.82$ wich is indeed true. So we found two exceptions for $n$ and we should be done with the proof. 

This part is strongly wrong! I made a mistake. $(4)$ is wrong. In fact, I can't find $n_1$ and $n_2$ satisfying $(3)$. Now I believe that I am wrong and there does exist a desired $a$. 
Note: I made a lot of edits. My proof changed since I posted the question. Please take a look at the edits.  
 A: Beautiful proof that there is a solution:

You can choose $a=\frac{1}{2}$. Then the equality holds for all $n \in \mathbb{N}$. We can prove it by contradiction. Assume there exists $n$ such that we don’t have equality. Then we can find $q \in \mathbb{N}$ such that
  $\displaystyle \frac{1}{2} + \sqrt{2n+1} < q \leq \frac{1}{2} + \sqrt{2n+2}$
  This is equivalent to
  $\displaystyle 8n+4 < (2q-1)^2 \leq 8n+8$
  In particular this implies that one of the four numbers $8n+5, 8n+6, 8n+7, 8n+8$ must be equal to $(2q-1)^2$.
  What is the remainder when we divide $(2q-1)^2$ by $8$? Well, we have
  $\displaystyle (2q-1)^2 = 4q^2-4q+1 = 4q(q-1)+1.$
  Since one of the numbers $q-1$ and $q$ is even, the first summand is divisible by $8$ and hence $(2q-1)^2\mod 8 =1$. But the other four numbers obviously leave remainders $5,6,7,$ and $8,$ respectively.
  So there we have our contradiction, meaning that no such $q$ can exist.  

Proof by Christian Schmidt on Quora. Link to my question there.
A: Your idea of looking at $\alpha=\{a\}$ is a good one. 
If $\alpha \lt 2-\sqrt 3\approx 0.268$ we choose $n=1$ and note that $\sqrt {2n+1}+\alpha=\sqrt 3+\alpha$ has floor $1$ while $\sqrt {2n+2}=\sqrt 4=2$ 
If $\alpha \ge 2-\sqrt 3$ we want to choose $n$  so that $$\lfloor\alpha+\sqrt {2n+2}\rfloor = k \gt \lfloor \alpha +\sqrt {2n+1} \rfloor\\
\lfloor \alpha^2+2\alpha\sqrt{2n+2}+2n+2 \rfloor=k^2\gt \lfloor \alpha^2+2\alpha\sqrt{2n+1}+2n+1\rfloor \\
\lfloor \alpha^2+2\alpha\sqrt{2n+2}+1 \rfloor=k^2\gt \lfloor \alpha^2+2\alpha\sqrt{2n+1}\rfloor $$ 
We note that when $2n$ is a square $\sqrt {2n+1}\approx \sqrt{2n}(1+\frac 1{2n})$ (rounding up the Taylor series) so if we choose $\sqrt{2n} \gt \frac 1\alpha$ and $2n$ a square it will fail.
