Suppose $(a,b)=1$. If $a$ divides $c$ and $b$ divides $c$ prove that $ab$ divides $c$.
Use theorem $1.4$:
If $a$ divides $bc$ and $(a,b)=1$, then $a$ divides $c$.
$a$ divides $c \implies c=ak$
$b$ divides $c \implies bt=c=ak$
$(a,b)=1$, $au+bw=1$
I tried moving things around and plugging things in but I can't figure out how to finish the proof from here.
I suppose you mean that the gcd (greatest common divisor) of $a$ and $b$ is 1? you are almost finished: since $ak = bt$, we have that $a$ divides $bt$ and since $\text{gcd}(a,b) = 1$, the theorem 1.4 shows that $a$ divides $t$, therefore we can write $as = t$ for some $s$. This implies that $c = bt = b(as) = (ab)s$, so $ab$ divides $c$.
just multiply the last equation by c, i.e. multiply au + bw = 1 through by c, getting auc + bwc = c. now can you finish?
If $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $(ab)|c$. If so, there is a proof here.
– projectilemotion Jan 29 '17 at 21:43