Divisibility crieteria

Question

This is a follow-up question.

The problem is:

Given two natural numbers, $m$ and $n$, and $n \vert m^2$.

Find necessary and sufficient conditions for $n \vert m$.

Here are what I find:

• Necessity

  1. $m \geq n$ (trivial)
  2. ?

• Sufficiency

  1. $n$ is prime - follows directly from Matthias's answer
  2. $\color{#c00}{n \text{ is} ~square\!-\!free,}$ $\color{#c00}{\text{i.e., has no} ~repeated~prime~factor}$ (stated in JHance's comment to my answer) - as pointed out by gammatester, it is wrong.
  3. ?

Help me to complete this list, folks.

Answer 1

Your statement is wrong. For every $n$ set $m=n$ and you have $n|m^2$ and $n|m.\;$ But $n$ is not necessarily square-free! Ex. $4|16$ and $4|4$ but $4$ is not square-free.

Answer 2

I think I found the solution. Do correct me if I am wrong.

$n|m$ if and only if there is no such prime $p$ such that $p^2|n$.

If part then it is easy, for example $4|14^2$ but $4\nmid 14$.

For the only if part, we can show a contradiction.