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.
 A: Let $x = n+z$ where $n$ is an integer and $0\leq z <1,$ and break into two cases.
Case 1:  If $z<1/2$ the equation becomes
$$\frac{1}{n}+\frac{1}{2n} = z+\frac{1}{3},$$
which leads to 
$$\frac{1}{3} \leq \frac{3}{2n} <\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$$
which is equivalent to $1.8 < n \leq 4.5.$  So $n = 2, 3,$ or $4.$
Plug each of these into the original equation to get $z = 5/12, 1/6, 1/24,$ respectively, giving 3 solutions:  $2 \frac{5}{12}, 3\frac{1}{6}, 4\frac{1}{24}$.
Case 2:  If $z\geq 2$, the equation becomes
$$\frac{1}{n}+\frac{1}{2n} = z+\frac{1}{3},$$
and since $1/2 \leq z <1$ we have
$$\frac{5}{6} \leq \frac{1}{n}+\frac{1}{2n} < \frac{4}{3},$$
which has no integer solutions.  
A: 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.
A: Let $x=n+f$, where $f$ is the fractional part.
If $f<\dfrac12$,
$$0\le f=\frac1n+\frac1{2n}-\frac13<1.$$
This is achieved for $n=2,3,4.$
If $f\ge\dfrac12$,
$$0\le f=\frac1n+\frac1{2n+1}-\frac13<1,$$ which is not possible.
