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!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

Similar Discussions: Number theory. modulu question.