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Congruence help

  1. Apr 6, 2005 #1
    Anybody know how to do this one:

    For each odd prime p and for each k such that 0<= k <=(p-1), prove that:

    the combination (p-1, k) is congruent to (-1)^k (mod p)

    Any help would be great.
  2. jcsd
  3. Apr 6, 2005 #2


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    What is "the combination"?
  4. Apr 6, 2005 #3


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    I'm guessing the 'combination' is refering to the binomial coefficient.

    In that case consider (x+y)^p mod p. Expand/simplify it in a couple ways, using the fact that p is prime (Fermat's little theorem) and then using the binomial theorem on (x+y)(x+y)^(p-1). Then compare coefficients.
  5. Apr 7, 2005 #4
    combination(p-1, k) = [(p-1)!]/ [k!*(p-1-k)!]

    sorry if i didn't make sense :)
  6. Apr 7, 2005 #5


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    In other words...

    \binom{p-1}{k} = \frac{p-1}{1} \frac{p-2}{2} \cdots \frac{p-k}{k}

  7. Apr 8, 2005 #6
    Going back to the originial question, Does anybody know how to do this one? The answer is that you use induction. The basis will work for 0 giving:

    [tex]\frac{(p-1)!}{0!(p-1)!} =1=(-1)^0 [/tex]
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