# Suppose $b,c \in \textbf Z^+$ are relatively prime (i.e., $\gcd(b,c) = 1$), and $a \,|\, (b+c)$. Prove that $\gcd(a,b) = 1$ and $\gcd(a,c) = 1$

Suppose $b,c \in \textbf Z^+$ are relatively prime (i.e., $\gcd(b,c) = 1$), and $a \,|\, (b+c)$. Prove that $\gcd(a,b) = 1$ and $\gcd(a,c) = 1$.

I've been trying to brainstorm how to prove this. I have determined that $\gcd(b, b + c) = 1$, but I am not sure if this fact will aid in proving this statement at all.

You can prove that if $\gcd(a,b)>1$, then $\gcd(b,c)>1$:
Indeed, let $d>1$ a divisor of both $a$ and $b$. As $a$ divides $b+c$, so does $d$. This implies $d$ divides $b+c-b=c$, hence $\gcd(b,c)\ge d>1$. The argument is symmetrical if $\gcd(a,c)>1$.
since $a \,|\, (b+c)$ thus there is $n$ such that $an=b+c$ suppose $\gcd(a,b) = d$ we show that $d=1$.$d \,|\, b$ , $d \,|\, a$ thus $d \,|\, an$ thus $d \,|\, c=an-b$ hence $d \,|\, \gcd(b,c)=1$ so $d=1$
We can write $bx+cy=1$ for some integers $x, y$. Also $b+c=aq$ for some integer $q$.
Then $c=aq-b$. And so $1=bx+cy=bx+(aq-b)y=a(qy)+b(x-y)$. So $\gcd(a,b)=1$.
A similar argument works for $\gcd(a,c)$.