Prove gcd(m,n)=1 implies gcd(2m+n,2n)=1 (n odd)

[Skolem]

I've been reading through the new book "Algebra: Chapter 0" by Aluffi in my spare time, but I can't seem to get this one: Prove gcd(m,n)=1 implies gcd(2m+n,2n)=1 where n is odd.

I know the basic properties of gcd, and also about min{am + bn as a,b in Z} = gcd(m,n) and all that, but I think I'm just missing something fundamental.


Skolem

 

[hamster143]

gcd(2m+n,2n) must divide 2n. And it must be odd, because 2m+n is odd. Therefore it must divide n. Since it divides 2m+n, it must divide m. Therefore it must divide gcd(m,n).