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The golden ratio satisfies the property that $$\{\phi^{-1}\}=\{\phi\}=\{\phi^2\} = 0.618\cdots$$ where $\{x\}$ is the fractional part of $x$, equal to $x-\lfloor x\rfloor$. Inspired by that, I was wondering for what subsets $S$ of $\mathbb{Z}\setminus\{0\}$ (e.g. $S=\{-1,1,2\}$ as with the golden ratio), there exist $x\in\mathbb{R}$ such that for all $n\in S$, $\{x^n\}$ is equal (and not equal to $0$, because otherwise, there would just be trivial integer/integer root solutions).

If $|S|=2$, then I think there must exist a solution. Write $S=\{m,n\}$. If both elements are positive, on $[0,2^{mn})$, $\{x^m\}$ and $\{x^n\}$ have differing number of discontinuities (I think $2^n-1$ and $2^m-1$), but they're both increasing from $0$ to $1$ except at those discontinuities. So there must exist some intersection point. If both elements are negative, instead of a solution to $\{x^m\}=\{x^n\}$, you can just consider the solution to $\{x^{-m}\}=\{x^{-n}\}$ and just take the inverse of that. Finally, if one element's positive and one's negative, the graph for the negative one would be monotonically decreasing to $0$ after $x=1$, while the graph for the positive one would be increasing (except at those discontinuities) from $0$ to $1$, so there would be some intersection in the graphs.

Obviously, what's a lot more tricky is when $|S|\ge 3$. I'm not even sure if there's any solution with $|S|=3$ other than when $S=\{-1k,1k,2k\}$ with $k\in\mathbb{Z}$. I did get that if $S=\{1,2\}$, the set of solutions for $x$ is given by $$\left\{-\sqrt{m+\frac{3-\sqrt{5+4m}}{2}}:m\in\mathbb{Z}_{\ge 0}\right\}\bigcup\left\{\sqrt{m+\frac{1+\sqrt{1+4m}}{2}}:m\in\mathbb{Z}_{\ge 0}\right\}\bigcup\{0\}$$

Also, if $S$ works, then $kS=\{ks:s\in S\}$ works, where $k$ is an integer. Is there a simple way to characterize the sets $S$ that work? More specifically, is there any way to find out what sets with only three elements work?

Edit: This is a small comment but might motivate looking at it through the lens of algebra. If $\{x^a\}=\{x^b\}=\{x^c\}$ with $a>b>c\ge 1$, then there should exist an integer $m$ such that $x^a-x^b-m$ is reducible over $\mathbb{Z}[x]$. In fact, we would need integers $m,n$ such that $\deg(\gcd(x^a-x^b-m,x^b-x^c-n))\ge 1$.

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  • $\begingroup$ I think you want to exclude, not just integers, but square roots of integers, cube roots of integers, etc. E.g., for $x=\sqrt2$, we have $\{x^n\}=0$ for all $n$ in $S=\{2,4,6,\dotsc\}$. $\endgroup$ Commented Jul 17, 2023 at 6:35
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    $\begingroup$ When $|S|=3$ is a Solution , then every SubSet of $S$ with 2 elements will be a Solution. Eg $\{-1,+1,+2\}$ works with $\phi$ , thus $\{1,2\}$ & $\{-1,1\}$ will work with $\phi$. What that implies : We must try to figure out all Solutions to $|S|=2$ , then check whether it allows adding a new element. When it allows (we know it allows , for the known Cases) , then that will give a Solution to $|S|=3$. When there are other Solutions , that will be obtained via this Method ! $\endgroup$
    – Prem
    Commented Jul 17, 2023 at 6:55
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    $\begingroup$ Recording the vote tallies as the bounty starts: Gerry 7, Eric 5, the rest at 0. Whichever answers gets the most upvotes during the bounty period will have it. $\endgroup$ Commented Aug 25, 2023 at 19:02
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    $\begingroup$ If anyone sees a cool question /answer, do bring it to the attention of users visiting the Pearl Dive. $\endgroup$ Commented Aug 25, 2023 at 19:03
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    $\begingroup$ During the bounty period Gerry got 4 votes, Eric 3 and Cactus 2. The voters have spoken. Thanks all the voters (as well as other visitors) for checking this question out. Too bad we didn't get much new material. May be the question is too tough? See ya! $\endgroup$ Commented Sep 1, 2023 at 20:06

3 Answers 3

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$S=\{-4,3,1\}$ works. Let $x$ be the real root of $x^3-x-1$, then the fractional parts of $x^{-4}$, $x$, and $x^3$ are all equal. We have $x^3-x=1$, so the fractional parts of $x$ and $x^3$ are equal, and $x^4=x^2+x$, $x^5=x^3+x^2=x^2+x+1$, so $x^5-x^4=1$, so $x-x^{-4}=1$, so $x$ and $x^{-4}$ have the same fractional part.

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    $\begingroup$ Nice find! I wonder if I can use this to come up with other working sets too. $\endgroup$ Commented Jul 17, 2023 at 19:12
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    $\begingroup$ I made some unsuccessful efforts to find similar examples, but maybe I didn't try hard enough, or was looking in the wrong places. $\endgroup$ Commented Jul 18, 2023 at 1:10
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    $\begingroup$ As it turns out, this value of $x$ also works for $S=\{5,4,-9\}$. Also, if $y$ is root of $y^3-y^2-1$, it works for $S=\{3,2,-5\}$. $\endgroup$ Commented Jul 21, 2023 at 23:56
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This just a partial solution - I show that arithmetic progressions of 3 numbers don’t work. There seem to be sporadic solutions like $\{-1,1,2\}$ and $\{-4,1,3\}$ (as noticed by Gerry) and $\{3,2,-5\}$ and $\{5,4,-9\}$ (as noted by Peter).

Note that if any set of numbers has a common divisor, you can just take $x$ to that power, so it suffices to just consider the cases where the numbers of $S$ are relatively prime.

Separately, if all of $S$ share a common divisor (say $d$) then you can let $x=2^{1/d}\not \in \mathbb Z$ and get that for all $a\in S$ that $\{x^a\}=\{2^{a/d}\}=0$. This is kind of cheating however, so I’m going to require that the fractional parts are nonzero going forward.

Any two element set works. Note that for all integer $a<b$ with $b$ positive, we have that $x^b-x^a-1$ has no integer solutions (since they would have to divide the constant and so be $\pm 1$ which doesn’t work), and it’s negative at $1$ and positive for sufficiently large $x$, so it has some real root. Using that real root $x$ gives $\{x^a\}=\{x^b\}$. Any integers which are rational exponents of each other are just the same integer to various powers, and no powers of the same integer are adjacent, so the fractional part is nonzero. Using $1/x$ covers the case when they’re both negative.

Consider any arithmetic progression $\{a, a+d, a+2d\}$. This would give that $x^d=(x^{a+2d}-x^{a+d})/(x^{a+d}-x^{a})\in \mathbb Q $. Since $x$ is an algebraic integer (being a solution to a monic polynomial) it’s an algebraic integer, so $x^d$ is a rational algebraic integer, so an integer. Let $m=x^d$. Let $n=x^{a+d}-x^{a}$. Then, $n=(m-1)x^{a}$, so $x^a$ is a rational algebraic integer, so an integer. Thus, $\{x^a\}=0$, no nontrivial solution exists.

$\{-1,1,2\}$ is particularly nice since the terms of the trinomials which $x$ is a root of have degrees $1,-1,0$ and $2,1,0$. These are just offset from each other, so appropriate coefficient choice allows them to perfectly overlap. For arbitrary $m<n<p$, WLOG at most one of which is negative, we get that that only occurs when $0,m,n,p$ is an arithmetic progression meaning either having a ratio of $0:1:2:3$ which is disallowed from the above or $m:0:n:p$ is in a progression which gives $-1:0:1:2$ which gives $\phi$. I’d guess this is true for all 3 element sets. It’s not clear which (if any) 3 element sets might lead to trinomials with overlapping roots.

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  • $\begingroup$ Note that if we let $x^r=\phi$, we can take $S=\{-r,r,2r\}$. A trivial variation on $\{-1,1,2\}$, but, still.... $\endgroup$ Commented Jul 17, 2023 at 6:43
  • $\begingroup$ Based on Gerry's comment and your answer, I updated my question to say that the fractional parts are nonzero. Also, I think your same approach applies (with some modification) to other sets (like $\{1,2,4\}$). $\endgroup$ Commented Jul 17, 2023 at 6:51
  • $\begingroup$ Agreed. Note that finding a 3 element solution (other than $-1,1,2$) is equivalent to finding the polynomials of the form $x^a-x^b-c$ with a common root. If you restrict to $c=\pm 1$, you get that that’s either irreducible or has roots of unity (I.e. $x^{2n}-x^n+1$) which helps get you $\{-4n,3n,n\}$. I can’t find any other examples. $\endgroup$
    – Eric
    Commented Jul 17, 2023 at 20:52
  • $\begingroup$ You wrote $x^a=x^b$ rather than $\{x^a\}=\{x^b\}$, and $x^a=0$ rather than $\{x^a\}=0$. $\endgroup$
    – mr_e_man
    Commented Aug 28, 2023 at 3:18
  • $\begingroup$ Thanks, fixed those (I forgot to backslash the curly brackets in the latex). $\endgroup$
    – Eric
    Commented Aug 28, 2023 at 4:11
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It doesn't work for $S = \{1,2,3\}$ for example. Indeed, assume that $\{x\} = \{x^2\} = \{x^3\}$ and $x \notin \mathbb{Z}$, then there exists integers $(n,m)$ such that $x^3 = x + n$ and $x^2 = x + m$.

The second equations implies that $x$ is integral of degree at most two over $\mathbb{Z}$, and of degree exactly two because it is not an integer. Therefore, its minimal polynomial is $X^2 - X - m$. $x^3 - x - n = 0$ is thus equivalent to $(X^2 - X - m)|(X^3 - X - n)$ and, doing the Euclidean division of the one by the other, $$ X^3 - X - n = (X^2 - X - m)(X + 1) + mX + m - n. $$ We deduce that $(X^2 - X - m)|(X^3 - X - n)$ if and only if $mX + m - n = 0$, or in other words, $m = n = 0$. In this case, we have $x^3 = x^2 = x$, which is impossible unless $x \in \{0,1\}$ and they are integers. Notice that I didn't even need the stronger hypothesis that $\{x^s\} \neq 0$ when $s \in S$.

I think that for most $S$ of cardinality $3$, the same reasoning holds and the ones for which there is a solution are probably pretty rare.

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