I am trying to show that if $m \in \mathbb{Z}$ and $m \ne 0$, then there exists an integer $k > 1$ such that $m \mid k$. However, I seem to be going in circles and arriving at incorrect conclusions.
If $m \ne 0$, then there exists some integer, $n$ such that $1 \le m \le n$ and $m \mid n$.
If $m \mid n$, then there exists some integer $k$ such that $n=km$.
From here I am stuck. I am not sure if I can use the fact that $k=\frac{n}{m}$?
Or, I've considered the division algorithm ($m,n \in \mathbb{Z}$ and $m >0$ then there exists some $q,r \in \mathbb{Z}$ such that $n=qm+r$ where $0 \le r < m$), but again, I don't get anywhere with it.
It seems like this should be quite simple to show since I know that if $m \ne 0$ then there must be some integer, $k$, that it divides and $k = nm$ for some integer $n$. But, as above, I seem to get to a fraction instead.
Edit to Add: For the suggestion of simply picking any value of $k$ and showing it works - Is there any way to solve for $k$ so that it works for all values of $m$?
For example, can we say $k = j\cdot m$ and show $j$ is negative when $m$ is negative; and, that $j$ is positive when $m$ is positive (in order to ensure $>1$)?
