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This question already has an answer here:

Solve for real $x$: $$\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = x - \lfloor x \rfloor + \frac{1}{3}$$

Hello! I hope everybody is doing well. I was not able to solve the above problem. And this problem becomes even more difficult with the fact that $x $ is not necessarily a positive real.

Here is what I did:

Let $x=p+r, 0 \leq r <1$ Now we divide into cases whether $r $ is less than $0.5 $ or not and I reach here(1st part): $0 \frac{3}{2p} - \frac13 < \frac12$. Now I could have done that $\frac95<2\le [x]\le4<\frac92 $ but since $x$ is not necessarily positive, I guess this is not always true.

Any help would be appreciated. Thanks.

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marked as duplicate by mathlove algebra-precalculus Oct 18 at 16:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Are you going to sign off your earlier question before asking another? $\endgroup$ – S. Dolan Oct 18 at 14:06
  • $\begingroup$ Oh, I am sorry I almost forgot that you can mark an answer correct too. $\endgroup$ – Vasu090 Oct 18 at 14:10
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Your method looks fine. So, for the first case:-

$$\frac{3}{2p}=r+\frac{1}{3}$$

and therefore $$3r+1=\frac{9}{2p}.$$

Now $0\le r <\frac{1}{2}$ and so $p$ is positive. The possibilities for $p$ are $2,3,4$.

These give $x$ as $2\frac{5}{12}, 3\frac{1}{6}, 4\frac{1}{24}.$

Now, if $r\ge \frac{1}{2}$, then $$\frac{1}{p}+\frac{1}{2p+1}=r+\frac{1}{3}.$$

Again $p$ must be positive. For $p=1$, the LHS is too large. For $p\ge2$, the LHS is too small.

Therefore there are no further solutions.

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I would approach the problem in this way.

Let's put $$ {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = x - \left\lfloor x \right\rfloor + {1 \over 3} = \left\{ x \right\} + {1 \over 3} = t $$

Since the fractional part of $x$ has values in the range $[0, \, 1)$ we have the the RHS is $$ 0 \le \left\{ x \right\} = t - {1 \over 3} < 1\quad \Rightarrow \quad {1 \over 3} \le t < {4 \over 3} $$

At the same time, we know that $$ x - 1 < \left\lfloor x \right\rfloor \le x $$ thus for the LHS we get this set of inequalities $$ \left\{ \matrix{ {1 \over 3} \le {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = t < {4 \over 3} \hfill \cr {1 \over x} + {1 \over {2x}} \le {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} < {1 \over {x - 1}} + {1 \over {2x - 1}} \hfill \cr} \right. $$

Translating this into inequalities for $x$, we shall have that $$ \left\{ \matrix{ {1 \over 3} < {1 \over {x - 1}} + {1 \over {2x - 1}} \hfill \cr {1 \over x} + {1 \over {2x}} < {4 \over 3} \hfill \cr} \right. $$ Solving the quadrics give $$ \left\{ \matrix{ x \in \left( {1/2,\;3 - \sqrt {22} /2} \right) \cup \left( {1,\;3 + \sqrt {22} /2} \right) \hfill \cr x \in \left( { - \infty ,\;0} \right) \cup \left( {9/8,\;\infty } \right) \hfill \cr} \right.\quad \Rightarrow \quad x \in \left( {9/8,\;3 + \sqrt {22} /2} \right) \approx \left( {1.1,\;5.34} \right) $$

So we have just to find the possible solutions for $$ \left\{ x \right\} = {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} - {1 \over 3} \quad \left| {\;x \in \left[ {1,\;3/2} \right)\,\; \cup \;\left[ {3/2,\;2} \right)\; \cup \; \cdots \,\; \cup \;\left[ {5,\;11/2} \right)} \right. $$

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