Zero divisors of ${\Bbb Z}_n = $ integers $\!\bmod n$ Consider the following proposition:

A nonzero element $m\in{\bf Z}_n$ is a zero divisor if and only if $m$ and $n$ are not relatively prime.

I don't know if this is a classical textbook result. (I didn't find it in Gallian's book). 
For the "only if" part, one may like to use the Euclid's lemma. But I cannot see how can one prove the "if" part:
If $m_1>0$, $(m_1,n)=d>1$, and $n|m_1m_2$, then $n\nmid m_2$. 
Edit:
The "if" part, should be:
If $m_1>0$ and $(m_1,n)=d>1$, then there exists $m_2$ such that $n|m_1m_2$, and $n\nmid m_2$.
Does one need any other techniques other than "divisibility"?
Questions:


*

*How to prove the proposition above?  

*How many different proofs can one have?

 A: Let $d=GCD(m,n)>1$ and write $n=dk$. Thus $1<k<n$ so that $k\not\equiv0\bmod n$.
Yet $km\equiv0\bmod n$ since $d\mid m$.
A: Okay so it is given that $(m,n) = d > 1$ and we wish to prove that $m$ is a zero divisor. Note that both $\frac{m}{d}>1$ and $\frac{n}{d}>1$ are in $Z_n$ and as $(\frac{m}{d})n=0$ in $Z_n$(being a multiple of n) so we have $0=(\frac{m}{d})n=m(\frac{n}{d})$, i.e. m is indeed a zero divisor.
A: You have been trying to prove that
"If $m_1>0$, $(m_1,n)=d>1$, and $n|m_1m_2$, then $n\nmid m_2$."
But this is general false.  Counterexamples are easy to find.  Take for example $m_1=2$, $n=2$, $m_2=2$.
A good approach to solving the original problem is to look at what happens in  few concrete cases.  Take for example $m=2$, $n=6$.  You want to show $m$ is a zero divisor in $\mathbb{Z}_n$.  Can you come up with a number $x$ not congruent to $0$ modulo $6$ such that $2x \equiv 0$? Sure, easy, $x=3$ will do.
What about $m=12$, $n=14$?  Sure, we want multiply $12$ by some $x$ not divisible by $14$ but such that the result is divisible by $14$.  Well, the $2$ in $12$ helps, we only need to multiply $12$ by $7$.  
Look at a few more examples.  We can after a while see that the "$d$" that is common to $m$ and $n$ can be used to come up with an $x$ not divisible by $n$ such that $mx$ is divisible by $n$. Then the formal proof writes itself.
A: Hint $\rm\,\ d\mid n,m\,\Rightarrow\,\ mod\ n\!:$ $\rm\displaystyle\:\ 0\equiv n\:\frac{m}d\ =\ \frac{n}d\: m\ $ and $\rm\, \dfrac{n}d\not\equiv 0\,$ if $\rm\,d>1$
