Greatest common divisor proof attempt I am trying to prove the following assertion: 

"If $a$ and $b$ are odd, then $(2a,2b)=(a+b, a-b)$". 

$(x,y)$ denotes the greatest common divisor of $x$ and $y$. I am trying to prove it by showing that $(a,b) \vert (a+b, a-b)$ and $(a+b, a-b) \vert (a,b)$.  
Denote $(a+b, a-b)$ by $d.$ Since $d \vert (a+b)$ and $d \vert (a-b)$ then $d \vert (a+b) + (a-b) = 2a$ and also $d \vert (a+b)-(a-b) = 2b$. This implies that $d \vert (2a,2b) = 2(a,b)$. 
The other way of the proof is where I am struggling. Suppose $(a,b) = y$. Since $a$ and $b$ are odd we have that $a=2n+1$ and $b=2m+1$ for some integers $n,m.$ The sum $a+b$ and the difference $a-b$ is therefore even, and also we have that
$a=yr$ and $b=ys$ for some integers $r,s$ so $a+b = y(r+s)$ and $a-b = y(r-s).$ This implies that $y \vert (a+b)$ and $y \vert (a-b)$ which implies that $y \vert (a+b, a-b)$. However, I have not been able to show that $2y \vert (a+b, a-b)$. Any pointers in the right direction would be much appreciated! 

$ D=(a,b)$ so that $D|a,\ D|b,\ 2D=(2a,2b)$ and $D$ is odd
Hence $D|a+b,\ D|a-b$ so that $D|(a+b,a-b)$
Since $a+b,\ a-b$ are even, $2D|(a+b,a-b)$
When $D'=2kD|(a+b,a-b),\ k>1$, then
$$2kD=D'|(a+b)-(a-b)=2b
,\ kD|b$$
Similarly, $kD|a$ so that it is a contradiction to $D=(a,b)$
