If $a\mid b+c$ and $\gcd(b,c)=1$, prove $\gcd(a,b)=\gcd(a,c)=1$

Question

I have the following:

$b+c=av$ for some integer $v$, and $a=dm$ and $b=dn$ for $d=\gcd(a,b)$ and some integers $m,n$.

Then, $c=av-b=dmv-dn=d(mv-n)$.

So, $d|c$, and we know that $d|a$ and $d|b$.

I don't know where to go from here.

Answer 1

You almost got it. Since $d\mid b$ and $d\mid c$, $d=1$.

Then, $b=av-c$. So $a$ and $c$ are coprime, too.

Answer 2

More conceptually, $ $ let $\,A = b\!+\!c\,$ below $ $ (use $\,b\leftrightarrow c\,$ symmetry to get $\,(a,c) = 1).$

Lemma $\ \ (a,b) = 1\iff (A,b) = 1\,$ for some multiple $\,A\,$ of $\,a$

Proof $\ \ (\Rightarrow)\ \ $ Let $\,A = a.\ $ $\ (\Leftarrow)\ \ $ $\,a\mid A\,\Rightarrow\,(a,b) = (a,b,A) = 1\,$ by $\,(b,A)=1.$