GCD proof? $(a,bc) = (a,b)(a,c)$ if $(b,c)=1$

Question

This seems like it should be an easy problem, but I just can't seem to get anywhere.

What I've tried:

trying to show that $(a,bc)$ and $(a,b)(a,c)$ divide each other by writing out equations (like $a=dk$ for $d= (a,bc)$)
trying to expand using Bezout's Identity

It's true more generally if $\ (a,b,c)=1,\ $ see my answer. – Bill Dubuque Apr 29 '15 at 21:08 — Bill Dubuque, Apr 29 '15 at 21:08

score 3 · Answer 1 · answered Apr 29 '15 at 21:05

3

$(a,b)(a,c) = ((a,b)a,(a,b)c) = (a^2,ab,ac,bc) = (a(a,\color{#c00}{b,c}),bc) = (a,bc)\,$ by $\,\color{#c00}{(b,c)}=1$

answered Apr 29 '15 at 21:05

Bill Dubuque

Can you explain the first equality? – futuremathteacher Apr 29 '15 at 21:09
It's the gcd distributive law $\ d(a,c) = (da,dc),$ then applied again in the second equality, etc. See here for a handful of proofs of the fundamental GCD distributive law. – Bill Dubuque Apr 29 '15 at 21:10

1 Answers1