find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$ 
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to largest by $x_1,x_2,\cdots$. Provide a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}.$

First, I'm not even sure why there are infinitely many solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$. One inequality that might be useful is the Cauchy-Schwarz inequality. There's also the AM-GM inequality. We have $ac + bd \leq \sqrt{a^2+b^2}\sqrt{c^2+d^2}$ for all real numbers $a,b,c,d$ where $ac,bd\ge 0$ with equality iff $(a,b),(c,d)$ are proportional vectors in $\mathbb{R}^2$. Observe that $\lfloor x\rfloor$ always has the same sign as $x$, so $\lfloor x\rfloor \lfloor x^3\rfloor$ has the same sign as $x^4 \ge 0$. It might be useful to substitute $\{x\} = x-\lfloor x\rfloor$ into the original equation and simplify the result somehow. Also, it could be possible to write the given sum as a telescoping sum.
 A: Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$\frac{\{x\}}{\lfloor x\rfloor}
= \frac{\{x^3\}}{\lfloor x^3\rfloor}$
or
$$\lfloor x^3\rfloor = x^2 \lfloor x\rfloor. \tag{1}$$
Clearly, $x=1, 2, \cdots$ are solutions of (1).
Let $k \in \mathbb{Z}_{>0}$.
Consider the solutions of (1) in $(k, k + 1)$ i.e. $k < x < k + 1$. We have $\lfloor x \rfloor = k$.
Let $u = k(x^2 - k^2)$.
We have $x = k\sqrt{1 + u/k^3}$
and $x^3 = (k^3 + u)\sqrt{1 + u/k^3}$.
We have
$x^2\lfloor x \rfloor = k^3 + u$.
Thus, $u$ is an positive integer.
(1) If $u \ge 2$, we have
$$x^6 - (k^3 + u + 1)^2
= \frac{(u-2)k^6 + [2(u-2)^2 + 6(u-2) + 3]k^3 + u^3}{k^3} > 0$$
which results in
$\lfloor x^3\rfloor \ge k^3 + u + 1 > k^3 + u =  x^2 \lfloor x\rfloor$.
(2) If $u = 1$ i.e. $x = k \sqrt{1 + 1/k^3}$, we have $x^3 = (k^3 + 1)\sqrt{1 + 1/k^3}$.
It is easy to prove that
$\lfloor x^3 \rfloor = k^3 + 1$.
Also, $x^2 \lfloor x\rfloor = k^3 + 1$. Thus, $x = k \sqrt{1 + 1/k^3}$
is a solution of (1).
Thus, we have $x_{2k-1} = k$ and $x_{2k} = k\sqrt{1 + 1/k^3}$ for $k = 1, 2, \cdots$.
A: Let $u$ be a noncubic integer with $\lfloor\sqrt[3]{u}\rfloor = n \in \mathbb{Z}^+$. Assuming $\sqrt[3]{u} \le x < \sqrt[3]{u+1}$, we can reformulate the equation as follows.
$$ \sqrt{nu} + \sqrt{(x-n)(x^3-u)} = x^2 $$
Squaring both sides, we have
$$(\sqrt{n} x - \sqrt{u})^2 = 0$$
Hence the equation holds for $x = \sqrt{u/n}$ if $\sqrt[3]{u} \le \sqrt{u/n} < \sqrt[3]{u+1}$. Since the first inequality always holds, we only need to check which $u$ satisfies the second inequality. After calculating some examples, it is easy to guess that
$$u = n^3, n^3+1$$
are only $u$ we find. You can show this guess by substituting $u$ by $n^3+k$ for $k=0,1,2$, and checking that
$$f(u) = u^3 - n^3(u+1)^2$$
is increasing for $u \ge n^3+1$.
