How to show: if $b \mid a$ and $c \mid a$ and $\mathrm{gcd}(b,c) = 1$, then $bc \mid a$?

Question

A little stumped on this problem, any help would be greatly appreciated.

Show that for all $a,b,c \in \mathbb{Z}$, if $b \mid a$ and $c \mid a$ and $\mathrm{gcd}(b,c) = 1$, then $bc \mid a$.

Try to prove this: if $b\mid a$ and $c\mid a$, then $\text{lcm}(b,c)\mid a$. You can probably work it out (though it might be messy) using the prime factorisations of $b$ and $c$. — Ian Coley, May 14 '13 at 20:19

score 3 · Answer 1 · answered May 14 '13 at 20:20

3

We are given $a=nb$, $a=mc$, $1=ub+vc$. Then $n=nub+nvc=ua+nvc=(um+nv)c$, hence $a=(um+nv)bc$.

answered May 14 '13 at 20:20

Hagen von Eitzen

score 0 · Answer 2 · answered May 15 '13 at 03:15

0

If the highest power of prime $p$ in $a,b,c$ are $r_a,r_b,r_c$ respectively,

$r_a$ must be $\ge$ max$(r_b,r_c)$

$\implies $lcm $(b,c)$ divides $a$

If gcd$(b,c)=1,$lcm $(b,c)=b\cdot c$

answered May 15 '13 at 03:15

lab bhattacharjee

2 Answers2