For which positive integers $m$ and $n$ do $x^m-x$ and $x^n-x$ being integers imply that $x$ is an integer

Asked Aug 12 '23 at 18:42

Active Aug 13 '23 at 23:32

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This is inspired by Prove that $x$ is an integer if $x^4-x$ and $x^3-x$ are integers.

For which positive integers $m$ and $n$ do $x^m-x$ and $x^n-x$ being integers imply that $x$ is an integer.

$x$ is assumed to be real.

I don't really have an idea of what to do except for the idea that this probably holds if $m$ and $n$ are relatively prime, in which case there are integers $a$ and $b$ such that $am-bn=1$.

edited Aug 13 '23 at 22:44

asked Aug 12 '23 at 18:42

marty cohen

110,450

2

FYI, using an Approach0 search, there's the AoPS threads x^2-x,x^n-x integers and If a polynomial is integer for a x show that x is integer, which indicate it's true for $m = 2$ and any $n \ge 3$. – John Omielan Aug 12 '23 at 19:01
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$x \in \mathbb{R}$ or $x \in \mathbb{C} $? If $\gcd(m - 1, n - 1) \ge 3$ then the system $x^m - x = x^n - x = 0$ has complex solutions. – Denis Shatrov Aug 12 '23 at 19:40
$x$ is a real number. – marty cohen Aug 13 '23 at 22:43
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There is some resemblance to https://math.stackexchange.com/questions/4737448/real-number-x-such-that-xn-is-constant-for-all-n-in-s/ – Gerry Myerson Aug 14 '23 at 03:53
For $(1, n), (n, 1), (n, n)$ this boils down to $x^n-x\in\mathbb{Z}$. If $n = 1$ then take $x$ to be any non-integer, if $n\geq 2$ then $x^n-x-1 = 0$ has a solution in $(1, 2)$. So those three cases don't imply that $x$ is an integer, we can safely assume that $n\neq m$ and $n, m\geq 2$. – Jakobian Aug 14 '23 at 13:17

For which positive integers $m$ and $n$ do $x^m-x$ and $x^n-x$ being integers imply that $x$ is an integer

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