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Euler's function

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Let phi denote Euler's totient function, ie for k in N, phi(k) is the number of natural numbers less than k which are relatively prime to k.
    Prove that for every n in N we have n| phi((2^n) -1).
    (Hint: Compute o[2] in U_(2^n)-1, the group of units in Z mod (2^n)-1)

    3. The attempt at a solution
    I understand that in the hint, o[2]=n...but I don't know how to relate it to the problem.

    I also found that the value of phi(k) for any k is an even integer.
  2. jcsd
  3. Sep 26, 2008 #2


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    How does phi((2^n)-1) relate to U_((2^n)-1)? I.e. what sort of numbers in {0, ..., 2^n - 1} appear in U_((2^n)-1)? Next, if g is an element of a finite group G, what can you say about the relationship between o[g] and |G|?
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