Equation involving floor function and fractional part function How to solve $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = \{x\} + \frac{1}{3}$ , where $\lfloor  \rfloor$ denotes floor function and {} denotes fractional part.  I did couple of questions like this by solving for {x} and bounding it from 0 to 1. So here we will have, $$0\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} - \frac{1}{3}\lt1$$  adding throughout by $\frac{1}{3}$ $$\frac{1}{3}\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} \lt\frac{4}{3}$$  Now I am stuck.

Notice that (1) the values $\lfloor x \rfloor, \lfloor 2 x \rfloor$ depend only on the value of the integer $\lfloor 2 x \rfloor$ and (2) $x \geq 1$. On the other hand, for, e.g., $x \geq 5$, we have $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2 x \rfloor} \leq \frac{1}{5} + \frac{1}{10} = \frac{3}{10} < \frac{1}{3}$, leaving only finitely many integer values $\lfloor 2 x \rfloor$ to check.
