abstract algebra - Let $amid bc$ then prove or disprove $amid (a,b)c$

Prove or disprove:

Let $a\mid bc$ then $a\mid (a,b)c$

Here is my approach, but I am not sure if I am doing this correctly or efficiently.

Let $a\mid bc$. It follows that either

1. $a\mid b$ Proof: $b=ar, a\mid bc => (ar)c = a \rightarrow a(rc)=a \rightarrow a|a(rc)$




2. $a\mid c$.

Since $rc$ is an integer. $a\mid bc$. Similar for $(2). a|c$

Let $a\mid (a,b)c$.

Using: the definition of $\gcd(a,b)=1=ax+by$ if $\exists x,y \in Z$

then we can rewrite it as $a\mid dc$. This is as far as I go. I can't manipulate it so that I show that $a\mid (a,b)c$. Does this mean that I would have to disprove $a\mid bc$ then $a\mid (a,b)c$?

Any help would be appreciated.

Answer

I'm really confused about your solution. $a\mid bc$ doesn't imply that either $a\mid b$ or $a\mid c$ unless $a$ is a prime number. Anyway, the statement you try to prove/disprove is true. You can write $(a,b)=ka+lb$ when $k,l\in\mathbb{Z}$. Then $(a,b)c=kac+lbc$. $a$ divides $a$ and so $a\mid$. Also $a\mid bc$ which implies $a\mid lbc$. And hence $a$ divides the sum $kac+lbc=(a,b)c$.