<s>[INST] The greatest common divisor Problem : Confirm the following properties of the greatest common divisor: 
If $\gcd(a,b) = 1$, and $c \mid (a+b)$, then $\gcd(a,c) = \gcd(b,c) = 1$.
Is this right?  This is my answer:


*

*$\gcd(a,b) = 1$ => There exist integers $x$ and $y$ such that: $ax + by = 1$.

*$c \mid (a+b)$ => There exist an integer k such that: $ck = a + b$ => $a = ck - b$ and $b = ck - a$.

*Because $ax + by = 1$ and $a = ck - b$ => $(ck - b)x + by = 1$ <=> $c(xk) + b(y - x) = 1$ ( $xk$ and $(y - x)$ are integers ) => $\gcd(c, b) = 1$.

*Because $ax + by = 1$ and $b = ck - a$ => $ax + (ck - a)y = 1$ <=> $a(x - y) + c(ky) = 1$ ( $(x - y)$ and $ky$ are integers ) => $\gcd(a, c) = 1$.
=> $\gcd(a,c) = \gcd(b,c) = 1$.
 [/INST] I think your reasoning is right, and I can provide a shorter proof: $\gcd(a,a+b)=\gcd(a,b)=1$ and similarly  $\gcd(b,a+b)=\gcd(b,a)=1$. Then $c\mid a+b$ implies $\gcd(a,c) = \gcd(b,c) = 1$
</s>