How do I prove that if $a|c$ and $b|c$ then $ab | c$ if $\gcd(a,b) = 1$?

Question

How do I prove that if $a|c$ and $b|c$ then $ab | c$ if $\gcd(a,b) = 1$?

I got stuck with the following steps:

1. Express $c = ma = nb$ for some $m, n$ integers.
2. Multiply together the $2$ expressions to get $c^2 = mn(ab)$

But I cannot show that $c = \frac{mn}{c}(ab)$ because $mn/c$ might not be an integer.

How do I use the result that $\gcd(a,b) = 1$?

Answer 1

HINT: As usual, start with the fact that there are integers $k,\ell$ so that $ka+\ell b=1$. Multiply by $c$ to get $kac+\ell bc=c$. Can you see that $ab$ divides each term on the left?

Answer 2

This is one of these results that is easier to think in terms of factors.

By the fundamental theorem of arithmetic, a, b and c factor uniquely. Since their GCD is 1, a nd b do not share any factors.

Can you see how the joke ends?

Answer 3

One way to prove $ab|c$ is to observe that $ma = n b$ and $\gcd(a,b)=1$ imply that $b|m$. That is, $m = k b$. Therefore, $c = ma = k ab$.