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!!

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.)
