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I am talking about this question. This seems to be OP's first question, and OP didn't show any of their work. This is why the question got closed. However, I find the question itself very interesting and would like to know the answer. I have done some work of my own and I'm wondering if I can ask this as a seperate question. I just think that the probability of OP editing the question to include their work so that the question may be reopened is low. Should I wait or can I ask this as a question of my own?

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  • $\begingroup$ Probably it might be worth having a look at this post: Interested in a question which is closed because of "no effort". What to do? Other questions linked there might be of interest, too. (With the usual caveat that the linked question is already ten years old; the site might have changed since then a bit.) $\endgroup$ May 29, 2023 at 13:33
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    $\begingroup$ Personally, I don't see a problem with asking it again. Cite the original, of course, and supply the context/effort that the OP didn't. If the OP ever returns to improve their post, you can delete yours. Good mathematics should transcend weak questioners. $\endgroup$
    – lulu
    May 29, 2023 at 13:38
  • $\begingroup$ Another reason for the closure is like the +3 comment of Blue, asking for the definition of $\{x\}$ (I also do not know what $\{x\}$ means in this question). If you do ask this question again, then be sure to address this comment somehow. $\endgroup$
    – user1729
    May 29, 2023 at 14:09
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    $\begingroup$ Regarding waiting - I suggest waiting a day or two. There is no rush, and it gives the OP a chance to improve their question. $\endgroup$
    – user1729
    May 29, 2023 at 14:11
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    $\begingroup$ It's an intriguing question, isn't it? I had some thoughts that I'll probably flesh out more if you do repost it. Also, I think that's the notation for "fractional part of". $\endgroup$
    – JonathanZ
    May 29, 2023 at 14:46
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    $\begingroup$ Rescuing questions is a noble goal, but be warned that there are haters out there – I've done it a few times, and my efforts have attracted some downvotes. $\endgroup$ May 29, 2023 at 21:24
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    $\begingroup$ Just to say, $\{x\}$ is standard notation for the fractional part of $x$. It's just $x-\lfloor x \rfloor$, which is to say it is $x$ minus the greatest integer which fails to exceed $x$. This is very standard notation. $\endgroup$
    – lulu
    May 29, 2023 at 23:06
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    $\begingroup$ Horrible luck : I didn't check the edits to the question which had taken out some brackets. With those in place, here's an answer to the question on AoPS. This post on MSE does the job for a specific exponent but gives the general idea away, somewhat. I also hope that you were not solving the incorrect question. $\endgroup$ May 30, 2023 at 6:45
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    $\begingroup$ @SarveshRavichandranIyer, unfortunately I was solving the incorrect question. I found two solutions to the incorrect question: $x=1$ and $x=1+\phi.$ Interestingly, they both seem to satisfy the condition of the incorrect question. $\endgroup$
    – aqualubix
    May 30, 2023 at 12:30
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    $\begingroup$ @aqualubix The next solution to $\{x\}+\frac 1x = 1$ is $2+\sqrt{3} = 3.73205\ldots$, for which the condition of the incorrect question doesn't seem to hold from my check. $\endgroup$ May 30, 2023 at 13:22
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    $\begingroup$ @SarveshRavichandranIyer, but it should, shouldn't it? I just realised that for $x>1,$ we can swap $\frac{1}{x}$ and $\frac{1}{x^n}$ with $\{\frac{1}{x}\}$ and $\{\frac{1}{x^n}\}$ respectively. $\endgroup$
    – aqualubix
    May 31, 2023 at 1:34
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    $\begingroup$ It should, right? Yeah, I might have made a mistake somewhere. Anyway, keeping things on topic, that question hasn't been edited and is not going to be reopened in its current form (it has two reopen votes but I would vote to close in a heartbeat if it reopened right now). I think you can ask whatever question you want to at this point of time. Use the links I provided as context if you want. The other question will go with time. $\endgroup$ May 31, 2023 at 4:43
  • $\begingroup$ @SarveshRavichandranIyer, actually, the links you provided completely answered the question. The case of $n=3$ gives away the idea, like you said. For the general case, one needs to use the binomial theorem. $\endgroup$
    – aqualubix
    May 31, 2023 at 15:21
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    $\begingroup$ @aqualubix It's good that we've solved this eventually. $\endgroup$ May 31, 2023 at 16:03

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