Integer part equation with sum and radicals 
For a fixed $n$, natural number greather or equal than $2$, solve the following equation for real $x$.
$$\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}$$

I have noticed that all but the last terms in the equations are integers, so should be the last one, so the modulus under the radical is an integer's perfect square. So, it's formula is:
$$x = \pm \sqrt{a^2 + 1}$$
Where $a$ is an integer. So I should proceed in finding $a$, but I could only prove that $a = 1$ is a solution.

As you noted, we must have $|x^2-1|=a^2$ for some integer $a$.

Consider two cases . . .

Case $(1)$:$\;|x|\le 1$.

From $|x|\le 1$ we get $|x^2-1|=1-x^2$, hence
\begin{align*}
&
|x^2-1|=a^2
\\[4pt]
\implies\;\;&
1-x^2=a^2
\\[4pt]
\implies\;\;&
1-a^2=x^2
\\[4pt]
\implies\;\;&
1-a^2\ge 0
\\[4pt]
\implies\;\;&
a^2\le 1
\\[4pt]
\implies\;\;&
a^2=0\;\text{or}\;a^2=1
\\[4pt]
\implies\;\;&
x^2=1\;\text{or}\;x^2=0
\\[4pt]
\implies\;\;&
x\in\{\pm 1,0\}
\\[4pt]
\end{align*}
but then the equation 
$$
\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}
$$
fails since, for $x\in\{\pm 1,0\}$, the $\text{LHS}$ evaluates to $0$, while for $x=\pm 1$, the $\text{RHS}$ evaluates to $n-1$, and for $x=0$, the $\text{RHS}$ evaluates to $n$.

Thus for case $(1)$, there are no solutions.

Case $(2)$:$\;|x| > 1$.

From $|x| > 1$ we get $|x^2-1|=x^2-1 > 0$, hence $|a|\ge 1$.

Then we have
$$
|a|=\sqrt{x^2-1} < \sqrt{x^2+1}=\sqrt{a^2+2}\le\sqrt{a^2+2|a|} < \sqrt{a^2+2|a|+1}=|a|+1
$$
which implies $\lfloor \sqrt{x^2 + 1} \rfloor=|a|$.

But then for the equation 
$$
\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor = (n - 1) \lfloor \sqrt{x^2 + 1} \rfloor + \sqrt{|x^2 - 1|}
$$
the $\text{RHS}$ evaluates to $n|a|$ which is at least $n$, whereas for the $\text{LHS}$, noting that
$$
t-1 < \lfloor t \rfloor \le t
$$
holds for all real $t$, we get
\begin{align*}
&
\lfloor n^2 x \rfloor -\sum_{k = 1}^n \lfloor (2k -1)x \rfloor
\\[4pt]
 < \;&
n^2 x -\sum_{k = 1}^n \Bigl((2k -1)x -1\Bigr)
\\[4pt]
=\;&
n^2 x -(n^2x-n)
\\[4pt]
=\;&
n
\\[4pt]
\end{align*}
so the $\text{LHS}$ is less than the $\text{RHS}$, contradiction.

Thus for case $(2)$, there are no solutions.


Hence the given equation has no solutions.

