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A number theory question

  1. Aug 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that the equation [itex]x^k \equiv 1 (mod p)[/itex] has exactly k solutions if [itex] k|p-1[/itex].

    I'm also curious to know if it's possible to generalize this theorem this way:
    Prove that the equation [itex]x^k \equiv 1 (mod n)[/itex] has exactly k solutions if [itex]k|\varphi(n)[/itex] where [itex]\varphi(n)[/itex] indicates the Euler's totient function.
    2. Relevant equations



    3. The attempt at a solution
    Well, this is what I've done so far, but at some point I couldn't progressed any futher:

    Consider the polynomial xk-1. By a theorem we know that this polynomial would have k solutions at the maximum because its degree is k. Now, since k|p-1, there exists an integer number t such that: kt=p-1. we'll have:

    [itex] x^{p-1} - 1 = x^{kt} -1 = (x^k - 1)(x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1)[/itex]

    well, since p is a prime number, by Euler's theorem (or Fermat's little theorem to be more precise) we know that [itex] x^{p-1} - 1 \equiv 0 (mod p)[/itex] would have exactly p-1 solutions (since all the numbers from 1 to p-1 would be prime relative to p).

    now we'll have:

    [itex] x^{p-1} - 1 \equiv 0 (mod p)[/itex]
    [itex] x^{kt} -1 = (x^k - 1)(x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1) \equiv 0 (mod p)[/itex]
    [itex] p | (x^k - 1)(x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1) [/itex]
    since p is prime, p divides [itex]x^k -1[/itex] or [itex]x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1[/itex], therefore:

    [itex] x^k - 1 \equiv 0 (mod p)[/itex]
    [itex] x^{k(t-1)} + x^{k(t-2)} + ... + x^k + 1 \equiv 0 (mod p)[/itex]

    Now this is where I'm stuck. I don't know how to go further. I'm not sure if what I've done is useful to conclude anything either.
    Any helps would be appreciated.
     
  2. jcsd
  3. Aug 24, 2012 #2
    By the way, I found a counter example to that general statement.
    x2=1 (mod 8) has 4 solutions even though that 2 divides 4.
     
  4. Aug 25, 2012 #3
    I think p is prime in this question.
     
  5. Aug 25, 2012 #4
    Yes, it is.
    I'm sorry if having not said that caused confusion because I thought that was obvious from my solution to the problem.
     
  6. Aug 27, 2012 #5
    No ones knows how to prove it???
     
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