If $\gcd( a, b ) = 1$, then is it true to say $\gcd( ac, bc ) = c$? I tried to prove this theorem, but I'm really confused about its correctness.
Let's say 

$$\gcd( a, b ) = 1 \Rightarrow am + bn = 1 \text{ where } m, n\in\mathbb Z$$ 

Then by multiplying both side by $c$, we have: $amc + bnc = c$. 
From here, is it sufficient to say $c = \gcd( ac, bc )$?
I feel like this proof needs some more explanation, but I can't think of one. Any idea?

Thanks,
What you have shown is that there is a linear combination of $ac$ and $bc$ that equals $c$. Now, for any $x$, $y$, the gcd of $x$ and $y$ divides any linear combination of $x$ and $y$. In this case, this means that the gcd of $ac$ and $bc$ divides $c$.

You are almost done, but you also need to argue that $c$ divides the gcd of $ac$ and $bc$, to conclude the equality (if $z$ and $w$ are positive, and $z$ divides $w$, and $w$ divides $z$, then they are equal).

But, any linear combination of $ac$ and $bc$ is (obviously) a multiple of $c$ since $c$ divides both $ac$ and $bc$. So we are in fact done.
Since $\gcd(a,b)=1$, there exist two integer $x$ and $y$ such that $ax+by=1$.

Now, $acx+bcy=c$.

It follows that $\gcd(ac,bc)=c.$