# Prove that if x is odd and y is even, then gcd(x+y,x-y)=gcd(x,y)

It is trivial to prove that gcd(x,y) divides gcd(x+y,x-y). How is it possible to prove gcd(x+y,x-y) divides gcd(x,y)? I don´t know how to use the fact that x is odd and y is even. Can anybody help me prove the statement?

Suppose $$d\,|\,\gcd (x+y,x-y)$$. Then $$d\,|\,(x+y+x-y)=2x$$. Now, the parity assumptions tell us that both $$x\pm y$$ are odd so $$d$$ must be odd. Hence $$d\,|\,2x\implies d\,|\,x$$.
• Well, $\gcd(x+y,x-y)$ is a divisor of $\gcd(x+y,x-y)$. – lulu Apr 4 at 21:01
• I can't understand your confusion. In my post, I proved that any $d$ which divides $\gcd(x+y,x-y)$ also must divide $\gcd(x,y)$. So, just take $d=\gcd(x+y,x-y)$ and you are done. – lulu Apr 4 at 21:16