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Math Help - Summation up to p-1 modulo p, p prime

  1. #16
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    Re: Summation up to p-1 modulo p, p prime

    the original question asked for what is: \sum_{n=1}^{p-1} n^k\ (mod\ p)? perhaps the difficuly is in what you meant by "it" in your post...your "it" and Pinkk's "it" may be 2 different things.

    your orignal post (#11) was clearer.

    i think this is an interesting problem, which "as-stated" doesn't have a definitive answer. i haven't looked very far into it, but my feeling is, that if p is a divisor of the "denominator" for a closed form expression of \sum_{n=1}^{p-1} n^k, each power will resolve to 1, and we get p-1. the "denominators" in question are the even-numbered terms in this sequence:

    A064538 - OEIS

    which are also the denominators of the even Bournoulli numbers, expressed in "lowest terms". it seems to me that this question is a "hard" question, which is belied by its simple statement.

    the complete answer of this question, then, rests on proving the following:

    let k be n even integer, and let B_k denote the k-th Bernoulli number. if p divides the denominator of B_k written in lowest terms, then \sum_{n=1}^{p-1} n^k = p-1\ (mod\ p)

    conjecture: the statement "p divides the denominator..." can be replaced by p-1 divides k.

    any takers?
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  2. #17
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    Re: Summation up to p-1 modulo p, p prime

    From my previous post here we know that S_p^n \equiv 0 \pmod{p} when 0\le n \le p-2.
    If n=k(p-1) for some k\in\mathbb{N} then S_p^n \equiv p-1 \pmod{p} by Fermat's Little Theorem. Otherwise, for n \ge p-2 which is not divisible by p-1 we get S_p^n \equiv S_p^r \pmod{p} where 1 \le r \le p-2 is the remainder of n upon division by p-1. So again we obtain S_p^n \equiv 0 \pmod{p}. In conclusion:

    For n=k(p-1), k\in\mathbb{N} we have S_p^n \equiv p-1 \pmod{p}, and otherwise S_p^n \equiv 0 \pmod{p}.
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  3. #18
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    Re: Summation up to p-1 modulo p, p prime

    Quote Originally Posted by Unbeatable0 View Post
    From my previous post here we know that S_p^n \equiv 0 \pmod{p} when 0\le n \le p-2.
    If n=k(p-1) for some k\in\mathbb{N} then S_p^n \equiv p-1 \pmod{p} by Fermat's Little Theorem. Otherwise, for n \ge p-2 which is not divisible by p-1 we get S_p^n \equiv S_p^r \pmod{p} where 1 \le r \le p-2 is the remainder of n upon division by p-1. So again we obtain S_p^n \equiv 0 \pmod{p}. In conclusion:

    For n=k(p-1), k\in\mathbb{N} we have S_p^n \equiv p-1 \pmod{p}, and otherwise S_p^n \equiv 0 \pmod{p}.
    excellent! right, by FLT, each summand is congruent to 1 mod (p-1) (because p-1 is the order of the multiplicative group, and p-1 divides the power we're raising to). i should have thought of that myself. and every other case, we can reduce mod p to knock the exponents down to less than p-1.
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