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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!

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  • $\begingroup$ Well if you look closely to your rearranged equation, you will notice that the $LHS$ is an integer. While the only thing in the $RHS$ which may or may not be an integer is the $3[x]\{x\}$. check for which cases, it will be an integer. $\endgroup$ – Abhijit A J Jun 10 '16 at 14:56
  • $\begingroup$ @AbhijitAJ Could you please show me how to check? The only way I can think of is $\{3\lfloor x\rfloor\{x\}-2\lfloor x\rfloor\}$ which doesn't get me anywhere. $\endgroup$ – user342209 Jun 10 '16 at 15:07
  • $\begingroup$ I'll give you all the hints. All we need to check when $3[x]\{x\}$ is an integer. there are three possibilities for this. (1) as $[x]$ is integer $3\{x\}$ is an integer. Thus $\{x\}$ must be of the form $\frac{k}{3}$, where $k$ is a positive integer less than $3$. (2) As $3$ is an integer $[x]\{x\}$ must be an integer.(3) $\{x\}$ must be zero. $\endgroup$ – Abhijit A J Jun 10 '16 at 15:25
  • $\begingroup$ @AbhijitAJ Sir, from the third case, I got $x=0$ as the only solution. Could you please show me how to evaluate the solutions for the second case? For $I \in \Bbb z$ I reached $0<\frac{I}{[x]}<1$. I'm not sure about how to proceed further. I still have to work on the first case. $\endgroup$ – user342209 Jun 10 '16 at 16:19
  • $\begingroup$ Firstly $x=0$ cannot be a solution if you substitute that in the original equations, you will have a $0$ in the denominator. In the third case. Let $[x]=k$, then $\{x\}$ must be $\frac{n}{k}$ for their product to and integer. Where, $n$ is an integer less that $k$. If you are finding my method confusing, you must refer to the solution my "mathlove".P.S. Please don't call me "sir" :) $\endgroup$ – Abhijit A J Jun 10 '16 at 16:38
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$$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}\tag1$$

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.

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  • $\begingroup$ Sir, could you please show me how you went from $0\le \frac{9-2n}{6n}\lt \frac 12$ to $(n,\alpha)=(2,5/12),(3,1/6),(4,1/24)$? $\endgroup$ – user342209 Jun 10 '16 at 15:09
  • $\begingroup$ @user342209: From $0\le (9-2n)/(6n)\lt 1/2$, we have $n=2,3,4$. For each $n$, you can get $\alpha$ from $\alpha=(9-2n)/(6n)$. $\endgroup$ – mathlove Jun 10 '16 at 15:11
  • $\begingroup$ Thank you Sir. Actually I had meant to ask how you got the values of $n$. $\endgroup$ – user342209 Jun 10 '16 at 15:12
  • $\begingroup$ Did you perhaps use the Wavy Curve method, and then just consider the Integral values in the intervals for the values of $n$? $\endgroup$ – user342209 Jun 10 '16 at 15:13
  • $\begingroup$ @user342209: Since $n\not=0$, you can multiply it by $6n^2\ (\gt 0)$ to get $0\le n(9-2n)\lt 3n^2$, i.e. $n(2n-9)\le 0$ and $n(5n-9)\gt 0$. I think that you can continue from here. $\endgroup$ – mathlove Jun 10 '16 at 16:05
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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.

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    $\begingroup$ If $d\ge0.5,2x\ge2k+1\implies[2x]=?$ $\endgroup$ – lab bhattacharjee Nov 18 '16 at 17:37
  • $\begingroup$ @labbhattacharjee very well pointed out, will edit the answer now $\endgroup$ – RGS Nov 18 '16 at 17:38
  • $\begingroup$ @RSerrao-I don't think the second case, i.e., $d \in [0.5, 1)$ is just mimicking the case $d \in [0, 0.5)$. What I was left after writing $k$ in terms of $d$, we get $$k=\dfrac{(8-3d)\pm\sqrt{9d^2+168d+136}}{4(3d+1)}$$ Now, as $k\in\Bbb{Z}$, hence the precondition is that the discriminant be a perfect square. And on finding the value of $D$ at $d=0.5$ and $d=1$, we get $222.25\le D\lt313$. So for $D$ to be a perfect square $D=15,16,17$, which occurs at the values $d\approx0.515,0.688,0.87$ respectively. As all the values of $d$ are irrational hence we arrive at the conclusion that $k$ (Cont.) $\endgroup$ – user350331 Nov 19 '16 at 4:59
  • $\begingroup$ (Cont.)cannot be an integer hence the condition $d \in [0.5, 1)$ is not satisfied by the equation $$\dfrac{1}{\left\lfloor{x}\right\rfloor}+\dfrac{1}{\left\lfloor{2x}\right\rfloor}=\{x\}+\dfrac{1}{3}$$ It took me a long time to determine the values of $d$ for which the discriminant is a perfect square. If you caould provide me with a better way to analyse the conditions I would be very thankful. $\endgroup$ – user350331 Nov 19 '16 at 5:02
  • $\begingroup$ @RSerrao-Is my analysis regarding the case $d \in [0.5, 1)$ correct and if possible can you suggest a shorter method if there is one. $\endgroup$ – user350331 Nov 19 '16 at 13:39
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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.

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  • $\begingroup$ Sir, how did you know which respective intervals to check for? For example, how did you know about the values of the LHS and RHS for $0\le x<2$ and then for 2\le x\<2.5$? It doesn't seem intuitive for me to break down the number line into such intervals $\endgroup$ – user342209 Jun 10 '16 at 15:05
  • $\begingroup$ The RHS always lies between $\frac{1}{3}$ and $1\frac{1}{3}$, whereas the LHS gets smaller as $x$ increases. For $x<2$ it is obviously at least $1+\frac{1}{3}$. For larger $x$, the LHS moves down in jumps at $x=2,2.5,3,3.5,\dots$, so it is natural to consider those intervals. $\endgroup$ – almagest Jun 10 '16 at 15:13
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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

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