GCD proof of (m,2n)=(m+2n,n)

Question

(m,2n)=(m+2n,n). I was testing cases before believing that this was true, and to me it only seems to be true for odd m, n, is this correct?

Answer 1

Say $d$ is the GCD of $m+2n$ and $n$; i.e., $d$ divides both $n$ and $m + 2n$. Then it must divide their difference, $d \mid m + 2n - n = m+n$. Again by the same logic, $d$ must divide $m + n - n = m$. So the GCD of $m+2n$ and $n$ is equal to the GCD of $m$ and $n$.

But if that is true, then there are obvious cases where the GCD of $m$ and $n$ is not the same as the GCD of $m$ and $2n$; specifically, if $m$ is even and $n$ is odd, then the GCD of $m$ and $n$ cannot be even, because $n$ is not divisible by $2$; but the GCD of $m$ and $2n$ will be even, since $2$ divides both.