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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?

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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?

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  • $\begingroup$ 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. $\endgroup$ – Luis Gimeno Sotelo Apr 4 at 21:00
  • $\begingroup$ Well, $\gcd(x+y,x-y)$ is a divisor of $\gcd(x+y,x-y)$. $\endgroup$ – lulu Apr 4 at 21:01
  • $\begingroup$ 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. $\endgroup$ – lulu Apr 4 at 21:16
  • $\begingroup$ Oh, now I see it. Sorry a lot! I read your comment wrong! Now I understand! Thank you very much !! $\endgroup$ – Luis Gimeno Sotelo Apr 4 at 21:18
  • $\begingroup$ No problem, glad to help. $\endgroup$ – lulu Apr 4 at 21:19

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