Show the equivalence: $ab|c \iff a|c$ and $b|c$ Let $a,b \in \mathbb{Z}$ \ {$0$} with $gcd(a, b) = 1$ and let $c \in \mathbb{Z}$. Show the equivalence: 
$ab|c \iff a|c$ and $b|c$ 
Also give an example of numbers  $a,b \in \mathbb{Z}$ \ {$0$}  and $c \in \mathbb{Z}$ with $gcd \not= 1$ where the equivalence above does not hold
My work(for the first part)
Write $ax+by=1$, $c=aa'$, $c=bb'$. Let $t=b'x+a'y$.
Then $abt=abb'x+baa'y=c(ax+by)=c$ and so $ab \mid c$.
an also I am stuck on the example. Any help is appreciated

How will anyone find a counter-example ? First of all we need to find $a,b$ such that gcd$(a,b)\neq 1$...so lets start with $2$ and $4$ thus $ab=8$...now you need to find $c$ such that $2|c$, $4|c$ but $8\not| \ c$...can you find such $c$ ?
