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. 

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 $\gcd(a,c)$.
