I know this to be true but I do not know why I can't prove it. Here is my proof so far:
$a$ doesn't divide $b$ means $b = a k_1 + r_1$ for some $0 < r_1 < a$.
$a$ doesn't divide $c$ means $c = a k_2 + r_2$ for some $0 < r_2 < a$.
$a$ divides $bc$ means $bc = a k_3 + r_3$ with $r_3 = 0$.
So I can write $bc = (a k_1 + r_1)(a k_2 +r_2)$ which expands to
$bc = a(a k_1 k_2 + k_1 r_2 + k_2 r_1) + r_1 r_2$, where $r_1 r_2$ need to be $0$, but I started off saying that $r_1$ and $r_2$ were strictly positive.
So, I have proven by contradiciton that there are no integers $a,b,c$ that satisfy the property, yet $a=6$, $b=3$, and $c=8$ is an example. Where did I go wrong in my proof?