# 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?

## 1 Answer

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 $$d\,|\,2x\implies d\,|\,x$$.

Can you finish from here?

• What I see is that we could do an analogous reasoning for y and say that d divides y and, together with d divides x, we could say that d divides gcd(x,y). How can I conclude? I am stuck. – Luis Gimeno Sotelo Apr 4 at 21:00
• 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
• Oh, now I see it. Sorry a lot! I read your comment wrong! Now I understand! Thank you very much !! – Luis Gimeno Sotelo Apr 4 at 21:18
• No problem, glad to help. – lulu Apr 4 at 21:19