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Homework Help: Disibility proof

  1. Sep 29, 2011 #1
    I'm at a loss here. I have no idea how to prove this.

    For all positive integers, k, m, n if k|mn then k|4m or k|4n.

    2. Relevant equations
    An integer r is divisible by an integer d if and only iff r=ds where s is some integer and d != 0.

    3. The attempt at a solution
    I tried rewriting the divisibilities.

    k|4m
    4m = ks

    k|4n
    4n = kq

    k|mn
    mn = ky

    but I don't know where to go from here.
     
  2. jcsd
  3. Sep 29, 2011 #2
    Are you sure you wrote it right? Take k = 21, m = 3, and n = 7 - doesn't work.
     
  4. Sep 29, 2011 #3
    Now I look foolish. I tried a couple of arrangements of numbers and it worked out, so I assumed it to be true. Thanks.
     
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