find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$

Question

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.

Answer 1

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 $k = 1, 2, \cdots$.

Answer 2

This is a corrected version of my initial post, which solved an other problem. Please consider accepting the good and first answer of River Li, it is all what is needed for the solution.

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Well, we have to find all solutions to the given peculiar equation, then compute the series. Let $$ x = n+s\ ,\qquad\ n\in\Bbb Z_{\ge 1}\ ,\ s\in[0,1)\ , $$ be a solution of the given equation. Observe that $$s=0\ ,\qquad x=n\ ,$$ a natural number, is always leading to a solution. Which are the other solutions, i.e. those with $$s\in(0,1)\ ?$$ As mentioned by OP and River Li, using the Cauchy-Schwarz inequality for $a,b,c,d$ being the square roots of $[x]$, $[x^3]$, $\{x\}$, $\{x^3\}$, we obtain: $$ x^4=(ac+bd)^2\le (a^2+b^2)(c^2+d^2)=([x]+\{x\})([x^3]+\{x^3\}) =x\cdot x^3=x^4\ , $$ so we have equality, so the vectors involved are proportional, $$ \tag{$*$} \frac{\{x\}}{[x]} = \frac{\{x^3\}}{[x^3]}\ . $$ Let us see with a pedestrian approach which are the possible solutions $x=1+s\in(1,2)$, $s\in (0,1)$. Then $x^3\in (1,8)$, so $[x^3]$ is among $1,2,\dots,7$, so one of the following cases occurs from $(*)$:

• $1>s=(1+s)^3-1>0$, no solution, zero excluded,
• $1>2s=(1+s)^3-2>0$, so $2(1+s)=(1+s)^3$, and $s=\sqrt 2-1$ is the/a solution,
• $1>3s=(1+s)^3-3>0$, so $3(1+s)=(1+s)^3$, but $s=\sqrt 3-1$ makes $3s>1$,
• $1>4s=(1+s)^3-4>0$, so $4(1+s)=(1+s)^3$, but $s=\sqrt 4-1=1$ is too big,
• $1>5s=(1+s)^3-5>0$, so ... and $s$ is too big,
• $1>6s=(1+s)^3-6>0$, so ... and $s$ is too big,
• $1>7s=(1+s)^3-7>0$, so ... and $s$ is too big.

Continuing the pedestrian approach, let us find solutions $x=2+s$ in $(2,3)$. Then $(*)$ leads to one of the cases:

• $1>\frac 12\cdot 8s=(2+s)^3-8>0$, no solution, zero was excluded,
• $1>\frac 12\cdot 9s=(2+s)^3-9>0$, so $\frac 92(2+s)=(2+s)^3$, and $\sqrt {9/2}-2$ is a solution,
• $1>\frac 12\cdot 10s=(2+s)^3-10>0$, so $\frac {10}2(2+s)=(2+s)^3$, but the solution $s=\sqrt{10/2}-2\approx0.236067977\dots$ is not satisfying $1>\frac12 \cdot 10s$, so we reject it,
• $1>\frac 12\cdot 11s=(2+s)^3-11>0$, so $\frac {11}2(2+s)=(2+s)^3$, but the bigger solution $s=\sqrt{11/2}-2\approx0.34520787991\dots$ is not satisfying $1>\frac12 \cdot 11s$, so we reject it, too, and for the same reason the further cases also lead to invalid values for $s$.

So let us work in the general case. Denote by $m$ the number $[x^3]$ for a solution $x=n+s$, $s\in(0,1)$. So $n^3\le m<(n+1)^3$. Then $(*)$ becomes $$ \frac sn = \frac{(n+s)^3-m}m\ . $$ Equivalently: $$ 1>\frac 1n sm = (n+s)^3-m >0\ . $$ Add $m$ in the equality, so $\frac 1n(n+s)m=(n+s)^3$, so $\frac 1nm=(n+s)^2$, so $s=\sqrt{\frac mn}-n$. And we check the needed condition $1>\frac 1nsm$, which becomes: $$ \begin{aligned} n&>sm\ ,\\ n&>\left(\sqrt{m/n}-n\right)m\ ,\\ (m+1)n&>m\sqrt{m/n}\ ,\\ n^3&>\frac{m^3}{(m+1)^2}\ . \end{aligned} $$ Considered as a function of $m$, the R.H.S. is increasing, for $m=n^3$ and $m=n^3+1$ the inequality is true, but starting with $n^3+2$ we have the opposite inequality: $$ n^3(n^3+3)^2=n^9 + 6n^6 + 9n^3<n^9+6n^6 + 12n^3 + 8 = (n^3+2)^3\ . $$

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In conclusion, the solutions are $1,2,3,4,\dots$, the natural numbers $\ge 1$, and intercalated between $\sqrt{2/1}=\sqrt{(1+1^3)/1}$, $\sqrt{9/2}=\sqrt{(1+2^3)/2}$, $\sqrt{28/3}=\sqrt{(1+3^3)/3}$, $\sqrt{65/4}=\sqrt{(1+4^3)/4}$, and so on. So the question wants us to compute the sum $$ \begin{aligned} S(n) &= \sum_{1\le k\le n} \frac1{\frac 1k(1+k^3) - k^2} = \sum_{1\le k\le n} k =\frac 12n(n+1)\ . \end{aligned} $$

Answer 3

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 $u \ge n^3+1$.