(a,b)=1 and c divides a+b prove (a,c)=1 and (b,c)=1

Results 1 to 2 of 2

- Oct 5th 2008, 06:14 PM #1

- Joined
- Sep 2008
- Posts
- 98

- Oct 5th 2008, 07:47 PM #2
Let $\displaystyle d_{1} = (a,c) \geq 1$ and $\displaystyle d_{2} = (b,c) \geq 1$.

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

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

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

Can you conclude?