If $(b,c) = 1$, then $(a,bc) = (a,b)(a,c)$

Question

I do not know how to solve the following:

Show that If $(b,c) = 1$, then $(a,bc) = (a,b)(a,c)$

Thanks.

Answer 1

Distributing: $\ (a,b)(a,c) = ((a,b)a,(a,b)c) = (aa,ab,ac,bc) = (a\!\!\!\!\!\overbrace{(a,\color{#c00}{b,c})}^{\large =1\ {\rm by}\ \color{#c00}{(b,c)=1}}\!\!\!\!\!,bc) = (a,bc)$

Remark $ $ We used only the GCD Distributive Law $\ x(y,z) = (xy,xz)\ $ and other universal gcd laws (associative, commutative) therefore the result holds true in any gcd domain (vs. proofs using Bezout or primes, which are not as general).

Answer 2

Hint

Assume that $p$ is a prime number and that $p|a$ and $p|bc.$ That is $p|a$ and $p|b$ or $p|c.$ So $p|\gcd(a,b)$ or $p|\gcd(a,c).$ In any case $$p|\gcd(a,b)\gcd(a,c).$$

Can you finish now?