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How to prove this?

  1. Feb 23, 2006 #1
    My teacher posed this as an extra credit problem on an exam, i couldn't quite prove it. i'm wondering if you guys could help me, i would really like to see the proof even though i already took the exam:

    Prove that if n and m are positive integers, then n^m - n is divisible by m.

    I took a few approaches:

    I tried to pose the problem in mod m. so if n^m - n is divisible by m, then it'll be congruent to 0 mod m. so then i tried to induction on n, but i got stuck on the (k+1)^m expansion. I don't believe this is the correct way.

    Another way I tried was to break it up into cases:
    Case #1: m|n, then it must divide n^m - n
    Case #2: m does not divide n. so then n = mq + r, where r is greater than zero and less than q. then i'm still stuck with a binomial expansion.

    any help guys???
     
  2. jcsd
  3. Feb 23, 2006 #2

    StatusX

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    That's wrong as you've stated it. For example, 6 does not divide 2^6-2=62. But it looks similar to Fermat's little theorem, and you might get something valid if you changed some of the conditions on n and m.
     
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