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 $\deg(\gcd(x^a-x^b-m,x^b-x^c-n))\ge 1$.
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.