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Number theory. modulu question.

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    x cong 1(mod m^k) implies x^m cong 1(mod m^(k+1))

    2. Relevant equations
    x cong 1(mod m^k) <=> m^k|x-1 <=> ym^k=x-1


    3. The attempt at a solution

    starting with ym^k=x-1 add one to both sides
    ym^k+1=x now rise to the power m.

    (ym^k+1)^m=x^m <=> subtract the 1^m from the end of the expantion to get
    x^m-1^m-(...)=(y^m)(m^(km)) where (...) is the rest of the binomial expansion.

    I am stuck here. somehow I need to get a m^(k+1) on the LHS and say it divides everything on the RHS. I don't think my approach is the best.
     
  2. jcsd
  3. Sep 29, 2009 #2

    Hurkyl

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    Wouldn't it be much easier to just reduce everything modulo mk+1?
     
  4. Sep 30, 2009 #3
    you mean reduce everything in the binomial expansion mod m^k+1?
     
  5. Sep 30, 2009 #4

    Hurkyl

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    Yep.
     
  6. Sep 30, 2009 #5
    so i couldn't figure out what you meant or maybe i did... but instead of doing the binomial expansion i factored the x^m-1.

    so I have m^(k+1)|x^m-1 and I expanded x^m-1 to give me m^(k+1)|(x-1)(x^m+x^(m-1)+...+x+1).

    now (x-1) is m^k so I have m^(k+1)|m^k(x^m+...+x+1) I need to somehow pull an m out of (x^m+...+x+1) to make this work but I don't know how.
     
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