
GCD Proof
if I have x and y be odd integers. Then 2gcd(x,y) = gcd(x+y,xy). I am completely lost for what to do. I may be over think this but help would greatly be appreciated. This is what I have so far:
Let d = gcd(x,y) and e = gcd(x+y, xy). To prove that e = 2d it suffices to show that 2de and e2d. To show that 2de we need to show that 2d(xy) and 2d(x+y). Thus e = (x+y)s + (xy)t and d = xk + yl and 2d = 2xk + 2yl. (xy) with x and y being always odd will always return even so xy = 2n and (x+y) with x and y being always odd will always return even so xy = 2m for some s,t,k,l,n and m in Z. Thus,
2n = (2xk + 2yl)z
2n = 2xkz +2ylz
n = xkz + ylz
2m = (2xk + 2yl)z
2m = 2xkz + 2ylz
m = xkz+ylz
But honestly I dont even know if this is right. I need help!! I have 2 others like this and have no clue where to go!!

Re: GCD Proof
First, we set d=gcd(x,y) and e=gcd(x+y,xy)
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 dx and dy, we have d(x+y) and d(xy). Since x and y are both odd we also have 2(x+y) and 2(xy). Once again, because x and y are both odd, we know that d is also odd and so it follows that 2dx+y and 2dxy. Hence, 2d<=e.
For the other direction, observe that since e(x+y) and e(xy), we have e(x+y+xy) and e(x+y(xy)). So, e2x and e2y. Since x and y are both odd, this implies that e<=2d.
Therefore, 2d=e as desired.
Hope this helps!
Optikal