# gcd

• Oct 5th 2008, 07:14 PM
bigb
gcd
(a,b)=1 and c divides a+b prove (a,c)=1 and (b,c)=1
• Oct 5th 2008, 08:47 PM
o_O
Let $d_{1} = (a,c) \geq 1$ and $d_{2} = (b,c) \geq 1$.

Since $c \mid (a+b)$, then that means $d_{1} \mid (a+b)$ and $d_{2} \mid (a+b)$ as both $d_{1}, d_{2}$ are factors of $c$.

However, since $d_{1} \mid (a+b)$ and $d_{1} \mid a$ & $d_{2} \mid (a+b)$ and $d_{2} \mid b$, then $d_{1} \mid b$ and $d_{2} \mid a$.

This means $d_{1} \mid (a,b)$ and $d_{2} \mid (a,b)$. But $(a,b) = 1$. Conclude?

Can you conclude?