We have $m,n \in \mathbb{N}$ and $gcd(m,n) = 1.$
How to prove that $gcd(m+n,m-n) = 1$ or $2$?
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Hint: $\gcd(m+n,m-n) \mid \gcd\big((m+n)+(m-n), (m+n)-(m-n)\big)=\gcd(2m,2n)\,$.
dxiv
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It would be more helpful if You could write solution whit steps. – Zvnoky Brown Mar 10 '17 at 08:51
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@ZvnokyBrown For any integers $u,v,$, $d=\gcd(u,v) \mid u$ and $d \mid v,$ by the definition of $\gcd,$. It follows that $d \mid u+v$ and $d \mid u-v,$, therefore $d \mid \gcd(u+v, u-v),$. The above is just using this property for $u=m+n$ and $v=m-n,$. – dxiv Mar 10 '17 at 16:57