Prove that $GCD(a,b)=1$ if for all natural numbers $c, a|bc $ then $a|c$.

Question

I'm trying to prove a theorem out of my text:

Theorem: Let $a$ and $b$ be natural numbers. Then $GCD(a,b)=1$ if and only if for all natural numbers $c$, if $a|bc$ then $a|c$.

I did come across this proof, which is slightly similar, but not exactly the same.

Answer 1

If $(a,b)=1$ then there are $s, t$ so $1=sa+tb$ iff $c=sac+tbc$. If $a$ divides $bc$ then $a$ divides the entire right-hand side. So $a$ divides $c$.

In the other direction (via contrapositive), if $(a,b)=d\neq 1$ then for some $n, m$ we have $a=nd$, $b=md$. Take $c=n$. So $a\mid mdn=bc$ but $a\not\mid n=c$ (or else $a$ is a proper divisor of itself).