Prove for integers a, b, and c, if gcd(a, b) = 1, a|c, and b|c then ab|c

Question

Prove for integers $a$, $b$, and $c$, if $\gcd(a, b) = 1$, $a|c$, and $b|c$ then $ab|c$.

Part b of this question is: "Is the converse true? Prove or disprove accordingly?"

Hey, so I've been drawing a blank for at least an hour now. I played around with the definition of divisibility and the gcd of one but couldn't get anywhere. Could someone help out?

Answer 1

Extended Euclidean algorithm allows one to write the gcd of $a$ and $b$ in the form $xa+yb$, and many results can be proved from there.

Answer 2

converse is obvious since $a|ab$ and $b|ab$ then given $ab|c $ we conclude that $a|c$ and $b|c$

Answer 3

Theorem. For $a,b,c\in\mathbb Z$, if $a\mid bc$ and $\gcd(a,b)=1$, then $a\mid c$.

Proof. From Bézout's identity we know that there exist $u,v\in\mathbb Z$ such that $au+bv=1$. This gives us $$c=(au+bv)c = a\cdot uc+bc\cdot v.$$ The number $a$ divides both summands, hence $a\mid c$.

The above result is called Euclid's lemma.

Corollary. For $a,b,c\in\mathbb Z$, if $a\mid bc$ and $\gcd(a,b)=1$, then $a\mid c$.

We have $c=ka$ for some $k\in\mathbb Z$. Since $b\mid ka$ and $\gcd(a,b)=1$, we get from Euclid's lemma that $b\mid k$. This implies $ab\mid ka=c$.