Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$

Question

Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$

$$$$ Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor$$ I'm not sure about what to do with the $3\lfloor 2x\rfloor $ term; I'd prefer to resolve it in terms of $\lfloor x\rfloor $ but am not able to. All that struck me was using the identity for $\lfloor nx\rfloor, n\in \Bbb Z$. However on first glance, it did not strike me as particularly useful.$$$$ I would be grateful for any help. Many thanks!

Answer 1

$$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}\tag1$$

We have $\lfloor x\rfloor$ and $\lfloor 2x\rfloor$, so one way is to separate it into two cases :

Case 1 : $x=n+\alpha$ where $n\not=0\in\mathbb Z,0\le\alpha\lt 1/2$

Case 2 : $x=n+\alpha$ where $n\not=0\in\mathbb Z,1/2\le\alpha\lt 1$

For case 1, $$\begin{align}(1)&\implies \frac 1n+\frac{1}{2n}=\alpha+\frac 13\\&\implies \alpha=\frac{9-2n}{6n}\\&\implies0\le \frac{9-2n}{6n}\lt \frac 12\\&\implies (n,\alpha)=(2,5/12),(3,1/6),(4,1/24)\end{align}$$

I think that you can do for case 2 similarly.

Answer 2

This is a horrible question!

For $x<0$, the LHS is negative and the RHS is positive, so there are no solutions for $x<0$. For $0\le x< 2$ the LHS is at least $1+\frac{1}{3}$ whereas the RHS is always $<1+\frac{1}{3}$, so there are no solutions for $x<2$.

Now consider the range $2\le x<2.5$. The LHS is $0.75$. So this equals the RHS for $x=2\frac{5}{12}$. There are clearly no further solutions for $x<3$ because the LHS is decreasing and the RHS increasing.

$3\le x<3.5$. The LHS is $0.5$, so this equals RHS for $x=3\frac{1}{6}$. Again there are no further solutions for $x<4$.

$4\le x<4.5$. The LHS is $0.375$, so this equals RHS for $x=4\frac{1}{24}$. Again there are no further solutions for $x<5$.

$x\ge5$. The LHS $\le\frac{3}{10}<\frac{1}{3}\le$ RHS, so there are no further solutions.

Answer 3

I decided to do @mathlove's case $2$.

CASE $2$

We see that $n \gt 0$.

$$\dfrac{1}{n} + \dfrac{1}{2n + 1}= \delta + \frac 13$$

$$\dfrac{3n+1}{2n^2+n} - \dfrac 13 = \delta$$

$$ \delta = \frac{-2n^2 + 8n + 3}{6n^2 + 3n}$$

\begin{align} 0 \le \frac{-2n^2 + 8n + 3}{6n^2 + 3n} \lt 1 \\ 0 \le -2n^2 + 8n + 3 \lt 6n^2 + 3n \\ \end{align}

\begin{align} 0 &\le -2n^2 + 8n + 3 \\ 2n^2 - 8n - 3 &\le 0 \\ n^2 - 4n &\le \frac 32 \\ (n - 2)^2 &\le \frac{11}{2} \\ (n - 2)^2 &\le 4 \\ n &\in \{1,2,3,4\} \end{align}

\begin{align} -2n^2 + 8n + 3 &\lt 6n^2 + 3n \\ 8n^2 - 5n - 3 &\gt 0 \\ 256n^2 - 160n &\gt 96 \\ (16n - 5)^2 &\gt 96 \\ 16n - 5 &\gt 10 \\ 16n &\gt 15 \\ n &\ge 1 \end{align}

$n= -1 \implies \delta = -\frac 73$

$n=1 \implies \delta = 1$

$n=2 \implies \delta = \frac{11}{30} \implies x = 2\frac{11}{30}$

$n=3 \implies \delta = \frac 17 \implies x = 3\frac 17$

$n=4 \implies \delta = \frac{1}{36} \implies x = 1\frac{1}{36}$

$n=5 \implies \delta = - \frac{7}{165}$

So $n \in \{ 2\frac{11}{30}, 3\frac 17, 1\frac{1}{36}\}$