Show: $\gcd(a,b)=\gcd(a,c)=1\implies\gcd(a,bc)=1$

Question

Let $a,b,c\in\mathbb Z$. Then $$ \gcd(a,b)=\gcd(a,c)=1\implies\gcd(a,bc)=1. $$

I tried using Bezout's Lemma, that states: if $a$ and $b$ are relatively prime, then $a\mid bc\implies a\mid c$.

This means in our case that $a\mid bc\implies a\mid c\ \land\ a\mid b$.

Is this the right approach? How to continue from here?

EDIT

Can I conclude from Bezout's lemma, that in our case, if a number divides $bc$, that is divides both $a$ and $b$? The reverse holds also. So the common divisors of $a$ and $bc$ are the same are the common divisors of $a$ and $b$, and $a$ and $c$?

Answer 1

There exist $m,n,j,k$ with $$am + bn = 1$$ and $$aj + ck = 1$$ so that (after multiplying) $$a^2jm + abjn + ackm + bckn = 1.$$ Thus $$a(ajm + bjn + ckm) + bc(kn) = 1.$$