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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.

Your idea of looking at $alpha={a}$ is a good one.

If $alpha lt 2-sqrt 3approx 0.268$ 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 $$lflooralpha+sqrt {2n+2}rfloor = k gt lfloor alpha +sqrt {2n+1} rfloor\
lfloor alpha^2+2alphasqrt{2n+2}+2n+2 rfloor=k^2gt lfloor alpha^2+2alphasqrt{2n+1}+2n+1rfloor \
lfloor alpha^2+2alphasqrt{2n+2}+1 rfloor=k^2gt lfloor alpha^2+2alphasqrt{2n+1}rfloor $$
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 $sqrt{2n} gt frac 1alpha$ and $2n$ a square it will fail.

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 $(2q-1)^2mod 8 =1$. 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.