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I'm looking for a set of conditions and maybe a proof of said conditions for the thought proposed in the title. It seems to me that what was stated in the title always is true when N is not a perfect square and N < m, but I can find instances that work for perfect squares N, e.g. m = 12, N = 4.

If it's not clear what I'm asking, I'm looking for a general case of what's asked in this question.

Any help is appreciated! Thanks.

Edit: m and N are positive integers.

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ifangy
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3 Answers3

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It is $(3)$ in the chracterizations of a squarefree integer $q\,$ below.

Theorem $\ $ Let $\rm\ 0 \ne q\in \mathbb Z\:.\ \ $ The following are equivalent.

$(1)\rm\quad\ \ \ \, n^2\,|\ q\ \ \Rightarrow\ \ n\ |\ 1\qquad\ $ for all $\rm\:\ n\in \mathbb Z $

$(2)\rm\quad\ \ \ \, n^2\, |\, qm^2 \!\Rightarrow n\ |\ m\qquad\! $ for all $\rm\: \ n,m\in \mathbb Z$

$(3)\rm\qquad\ q\ |\ n^2\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z $

$(4)\rm\qquad\ q\ |\ n^k\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z,\ k\in \mathbb N $

$(5)\rm\quad\:\ \: q^q\ |\ n^n\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb N,\ $ for $\rm\ q > 0 $

See this answer for proofs.

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Bill Dubuque
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(I assume $m$ integer and $N$ non-zero integer.)

The proposed statement is true for all $m\in\mathbb{Z}$ if and only if $N$ is square-free (this is obvious). If $N=k^2l$ with $l$ square-free and $k\neq\pm1$ then $m=kl$ is a counterexample.

More precisely: We have that $N\mid m^2\implies N\mid m$ for all $m\in\mathbb{Z}$ if and only if $N$ is a square-free integer.

Example for $N$ not a perfect square: $N=12$, $m=6$.

If you wanted $m>N$: $N=12$, $m=18$.

A general statement is not really interesting: We have that $N\mid m^2\implies N\mid m$ if and only if $N\mid m$ or $N\nmid m^2$. This is not quite useful, is it?

yo'
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  • What about m = 8 and N = 4? N is not square-free, but N is a multiple of 8 and 64. – ifangy Aug 31 '16 at 20:20
  • @ifangy I start with "for all $m\in\mathbb{Z}$" for a reason. A general statement would be useless. I'll expand on this. – yo' Aug 31 '16 at 20:22
  • @user236182 See my edit, I assume $N\neq0$ as for $N$ zero things are pretty weird. – yo' Aug 31 '16 at 20:27
  • What about N|m if and only if all of the squares of the prime factors of N are divisors of $m^2$? Makes sense, although now that I think about that it's trivial as well. Thank you for all your help! – ifangy Aug 31 '16 at 20:33
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As phrased not true.

$36$ is a multiple of $18, 6$ is not a multiple of $18$

or $100$ is a multiple of $4, 10$ is not a multiple of $4$

Doug M
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