prove by contradiction:gcd and divisibility

Question

Let's say there are $3$ natural numbers $a,b$ and $c$. $a|(b+c)$ and $\text{gcd}(b,c) = 1$,prove that $\text{gcd}(a,b)=1$

I know that I can prove this statement by contradiction.

Let's suppose $\text{gcd}(a,b) = d$ with $d>1$, and $d|a$, $d|b$. But how can I get $\text{gcd}(b,c)>1$? And how to approach $\text{gcd}(a,d)$ can not be greater than $1$ therefore it must be $1$?

Thanks so much for the help!

Answer 1

You don't need to use contradiction:

Let $d:=\gcd(a,b).$ Then $d\mid b$ and $d\mid a$ and since $a\mid b+c$ then $d\mid b+c$ and so $d\mid(b+c)-b=c,$ which implies that $d$ is a common divisor of $b$ and $c$ and thus $d\mid\gcd(b,c)=1,$ which implies that $d=1.$

Answer 2

We have: $b+c = ma, pb+qc = 1 \Rightarrow pb + q(ma-b) = 1 \Rightarrow mqa + (p-q)b = 1 \Rightarrow \text{gcd}(a,b) = 1$