When is this true: If $m^2$ is a multiple of N then m is a multiple of N 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.
 A: 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.
A: (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?
A: 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 $4$
