5

It is known that, if a positive integer $m$ is squarefree, then the following properties hold:

  1. If $m \mid n^2$ holds, then $m \mid n$ is true (where $n$ is a positive integer).
  1. The equation $m = \operatorname{rad}(m)$ holds, where $\operatorname{rad}(m)$ is the radical or squarefree kernel of $m$.

Here is my question:

Are there other properties, not mentioned above, that characterize squarefree integers?

One property per answer only, please.

Jose Arnaldo Bebita Dris
  • 8,471

6 Answers6

4

A natural number $n$ is squarefree iff $\gcd(a,b)=1$ whenever $n=ab$.

Hagen von Eitzen
  • 382,479
4

Using the usual notation $\omega(n)$ for the number of distinct prime factors of $n$ and $\Omega(n)$ for the number of prime factors of $n$ counted with multiplicity: $n$ is squarefree if and only if $\Omega(n)-\omega(n)=0$.

In isolation this might seem a little underwhelming; however, it is known that for every nonnegative integer $k$, the density of the set of integers $n$ such that $\Omega(n)-\omega(n)=k$ exists (that is, $\Omega(n)-\omega(n)$ has a limiting density function, which is a "limiting histogram"). Squarefree numbers are the case $k=0$ of this general fact (and their density is well known to equal $6/\pi^2$).

Greg Martin
  • 92,241
3

$m$ is square free if and only if for all integers $n$, $n^{\phi(m) + 1} \equiv n \mod m$. Here, $\phi$ is the totient function.

Mark Saving
  • 33,541
2

Using $\mu$ (the möbius function), $m$ is a square-free positive integer if and only if $\mu(m) = -1$ (when $m$ has an odd number of prime factors) or $\mu(m) = 1$ (when $m$ has an even number of prime factors).

Note Greg Martin's comment states

Perhaps a flashier version of this answer would be "$m$ is squarefree if and only if the sum of the primitive $m$th roots of unity is nonzero".

John Omielan
  • 52,653
  • 2
    I was about to complain that this is circular per the usual definition of $\mu$, but the linked WP article actually defines $\mu$ as sum of primitive roots of unity … – Hagen von Eitzen Dec 16 '22 at 06:42
  • 1
    Perhaps a flashier version of this answer would be "$m$ is squarefree if and only if the sum of the primitive $m$th roots of unity is nonzero". – Greg Martin Dec 16 '22 at 18:32
  • @GregMartin Thanks for your excellent suggestion. I've added it to my answer. – John Omielan Dec 16 '22 at 19:12
2

If m is square free any abelian group of order m is cyclic. Also see the fundamental theorem of finite abelian groups for a better understanding.

Baris Aytekin
  • 66
1

A positive integer $m$ is square-free if and only if, for every prime $p$, we have that $p \nmid m \, \lor \, p^2 \nmid m$. Basically equivalently, $m$ is square-free iff there's no prime $p$ where $p^2 \mid m$.

John Omielan
  • 52,653
  • Why $p \nmid m , \lor , \dots$? – Solomon Ucko Dec 16 '22 at 14:09
  • 1
    @SolomonUcko I'm not exactly sure what you're asking. If it's why it works then, for all but a finite # of primes, the first condition holds, i.e., $p\nmid m$. For primes which divide $m$, as the first condition is false, the second condition must be true, i.e., $p^2\nmid m$. However, if you're asking why I stated this then, although it's basically the same as my simpler $2$nd sentence statement, I thought it was an interesting, & to me at least non-obvious, necessary & sufficient condition for $m$ to be a square-free positive integer. If you're asking about something else, please clarify. – John Omielan Dec 16 '22 at 19:09
  • 1
    Thank you for clarifying Solomon's point, @JohnOmielan! No worries, I perfectly understood. – Jose Arnaldo Bebita Dris May 14 '23 at 10:56