Number theory: prove that if $a,b,c$ odd then $2\gcd(a,b,c) = \gcd(a+b,b+c, c+a)$ Please help! Am lost with the following:
Prove that if $a,b,c$ are odd integers, then $2 \gcd(a,b,c) = \gcd( a+b, b+c, c+a)$
Thanks a lot!!
 A: Let $d=\gcd(a+b,b+c,c+a)$.
$$d\mid a+b,b+c\implies d\mid (a+b)-(b+c)\implies d\mid a-c$$
$$\begin{cases}d\mid a-c\\d\mid a+c\end{cases}\implies d\mid (a+c)-(a-c)\implies d\mid 2c$$
Similarly, we have $d\mid 2a$, $d\mid 2b$.
$$\begin{cases}d\mid 2a\\d\mid 2b\\d\mid 2c\end{cases}\implies d\mid \gcd(2a,2b,2c)\implies d\mid 2\gcd(a,b,c)$$
So we've proved that $\gcd(a+b,b+c,c+a)\mid 2\gcd(a,b,c)$. What is left to prove is that $2\gcd(a,b,c)\mid \gcd(a+b,b+c,c+a)$.
$$\begin{cases}\gcd(a,b,c)\mid a\\\gcd(a,b,c)\mid b\\\gcd(a,b,c)\mid c\end{cases}\implies \begin{cases}\gcd(a,b,c)\mid a+b\\\gcd(a,b,c)\mid b+c\\\gcd(a,b,c)\mid c+a\end{cases}\implies \gcd(a,b,c)\mid \gcd(a+b,b+c,c+a)\implies $$
$$\implies 2\gcd(a,b,c)\mid \gcd(a+b,b+c,c+a)$$
(since $\gcd(a+b,b+c,c+a)$ is even and $\gcd(a,b,c)$ is odd). $\ \ \ \square$
A: Let $d=\gcd(a,b,c)$. Then solve $d=ax+by+cz$. 
Use that $2a=(a+b)+(a+c)-(b+c)$, $2b=(a+b)+(b+c)-(a+c)$ and $2c=(a+c)+(b+c)-(a+b)$. Then 
$$2d=2ax+2by+2cz = (a+b)(x+y-z) + (a+c)(x-y+z) + (b+c)(y+z-x)$$
So we have a solution to:
$$2d = (a+b)X+(a+c)Y + (b+c)Z$$
So we know $\gcd(a+b,a+c,b+c)\mid 2\gcd(a,b,c)$. The other direction is easier (and there, you use that $a,b,c$ are odd.)
A: Hint $\,(a,b,c)\to (a\!+\!b,b\!+\!c,c\!+\!a)$ is linear with det $= \color{#c00}2\,$ so by the simple method in this answer 
$$ \gcd(a,b,c)\mid  \gcd(a\!+\!b,b\!+\!c,c\!+\!a)\mid \color{#c00} 2 \gcd(a,b,c)$$
