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

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!

Solution 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.

Solution 2:

We start by rewriting $\left\lfloor{x}\right\rfloor = k$ and $\{x\} = d$, with $k \in \Bbb{N}$ and $d \in [0, 1)$, where we exclude negative integers thanks to a comment on the OP. Now there are two cases: either $d \in [0, 0.5)$ or $d \in [0.5, 1)$. Notice that if $d \in [0.5, 1)$ then the integer part of $2x$ is not $2k$ but $2k + 1$.

Let us assume $d \in [0, 0.5)$ The equation then becomes

$$\frac1k + \frac1{2k} = d + \frac13 \iff \frac3{2k} = \frac{3d + 1}3 \iff$$

$$\iff \frac9{3d + 1} = 2k \iff k = \frac9{2(3d + 1)}$$

But given that $k$ is an integer, we must have that $6d + 2$ divides 9. Now you are left with very few possibilities that can be tested separately. Can you take it from here?

Also, do you think you can mimic this for the case $d \in [0.5, 1)$ making the necessary changes for the denominator of the second fraction?

EDIT As taking a similar path would make you go a looong way around, we try this other approach:

If $d \in [0.5, 1) $ then $\frac46 \leq d + \frac13 < \frac43$ therefore

$$\frac1k + \frac1{2k +1} \in [\frac46, \frac43) $$

Writing the two inequalities and multiplying by 6 one gets

$$4 \leq \frac6k + \frac6 {2k+1} < 8$$

One can easily spot that $k = 3$ is too big already thus you are left with a couple of values to try. If there is some integer solution for those inequalities, one then must find out how much $d $ would have to be. If after solving that $d \in [0.5, 1) $ you found another solution. Otherwise you did not.

Solution 3:

Realize that RHS is positive so must be the LHS
So if $ \left \lfloor x\right \rfloor\geq 5$ then $$\frac{1}{\left \lfloor x\right \rfloor}+\frac{1}{\left \lfloor 2x\right \rfloor}< \frac{1}{3}$$ So then you just check the cases

Solution 4:

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.