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Binomial Coefficient of a Prime Power

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Let p be a prime, k be positive integer, and m ∈ {1, 2, 3, ..., pk-1}. Without using Lucas' theorem, prove that p divides [itex]\binom{p^k}{m}[/itex].
    2. Relevant equations
    The definition of the binomial coefficients: [itex]\binom{a}{b} = \frac{a!}{b! (a-b)!}[/itex]

    3. The attempt at a solution
    I've managed to prove the statement when k=1 by arguing that p must divide p! but cannot possible divide m!(p-m)!. I figure I should be able to use a similar counting argument for this proof, likely by counting the number of factors of p that appear in the numerator and in the denominator of [itex]\binom{p^k}{m}[/itex]. It looks to me as though the number of factors of p in [itex]p^k ![/itex] is [itex]\sum_{i=1}^{k} p^{k-i}[/itex]. However, I'm having applying this to the problem. If anyone could give some advice on how to do so, or perhaps nudge me in the direction of a different proof it would be much appreciated.

    Thanks!
     
    Last edited: Nov 23, 2014
  2. jcsd
  3. Nov 24, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That counting argument should work. You can can find a similar formula for m! and (pk-m)! and try to evaluate the difference.

    Or directly consider the fraction as
    $$\frac{p^k \cdot (p^k-1) \cdot \dots}{m! \cdot (m-1)! \cdot \dots}$$ and try to find matching factors there. Hmm... induction over m?
     
  4. Nov 24, 2014 #3
    Thanks for the help. I followed your advice and worked out similar formulas for m! and for (pk-m)!, and managed to prove the result by bounding the difference using geometric series. I appreciate your comment, it gave me the little push in the right direction that I needed.
     
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