GCD Direct Proof

Question

I need to show that if $a,b,c$ are ints such that $\gcd(a,b) = 1$ and $c|(a+b)$, then $\gcd(c,a) = \gcd(c,b) = 1$

I want to try and prove this directly because I think it will be more straightforward then a indirect. Also, I believe this has something to do with relatively prime numbers.

The help is appreciated!$$$$

Answer 1

If $k=\gcd(c,a)$ then $k|a$ and $k|c$ and so $k|(a+b)$, meaning $k|b$ and $k|\gcd(a,b)=1$, so $\gcd(c,a)=k=1$.

Similarly $\gcd(c,b)=1$.

Answer 2

Hint: let $a+b = kc$. Let $x,y \in \mathbb{Z}$ such that $xa+yb=1$. Then:

$$kc = a+b \implies ykc = ya+yb$$

So

$$ykc + (x-y)a = xa+yb = 1$$

Hence, $\gcd(c,a)=1$. Can you show a similar proof for $\gcd(c,b) = 1$?