First, we set d=gcd(x,y) and e=gcd(x+y,x-y)

Now we will show that 2d<=e and that e<=2d. I would say that this is the most common way of proving equalities involving gcd.

First, observe that since d|x and d|y, we have d|(x+y) and d|(x-y). Since x and y are both odd we also have 2|(x+y) and 2|(x-y). Once again, because x and y are both odd, we know that d is also odd and so it follows that 2d|x+y and 2d|x-y. Hence, 2d<=e.

For the other direction, observe that since e|(x+y) and e|(x-y), we have e|(x+y+x-y) and e|(x+y-(x-y)). So, e|2x and e|2y. Since x and y are both odd, this implies that e<=2d.

Therefore, 2d=e as desired.

Hope this helps!

Optikal