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

  1. Sep 6, 2010 #1
    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.

  2. jcsd
  3. Sep 6, 2010 #2
    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).
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