1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Disibility proof
  1. A proof (Replies: 3)

  2. Trig Proof (Replies: 4)

  3. Geometrical Proof (Replies: 6)

  4. "Arithmetic?" proof (Replies: 3)

Loading...