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Math Help - legendre symbol proof

  1. #1
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    legendre symbol proof

    Hi,

    Need help with this proof. (a/p) is legendre symbol not to be confused with 1 divided by p;

    Prove

    (1.2/p) + (2.3/p) + (3.4/p) + ..... + ((p-2)(p-1)/p) = -1

    Thanks
    Last edited by htata123; April 18th 2009 at 07:35 PM.
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  2. #2
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    no one?? this is must be a really tough one.
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  3. #3
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    Quote Originally Posted by htata123 View Post

    (a/p) is legendre symbol not to be confused with 1 divided by p; Prove \sum_{k=1}^{p-2}\left ( \frac{k(k+1)}{p} \right)= 0, \ \ p > 2.
    Quote Originally Posted by htata123 View Post

    no one?? this is must be a really tough one.
    be patient! also your question is not hard, it's just false! each summand of your sum is either 1 or -1 and we have an odd number of them. so they can never add up to 0.
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  4. #4
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    but then why is it a question on the text book? I'm sure there is a proof because, my professor told us it we had to use the fact (a.b/p) = (a/p).(b/p) and the fact that (a/p) = (a*/p) where a* is the multiplicative inverse of a.
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  5. #5
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    Quote Originally Posted by htata123 View Post
    but then why is it a question on the text book? I'm sure there is a proof because, my professor told us it we had to use the fact (a.b/p) = (a/p).(b/p) and the fact that (a/p) = (a*/p) where a* is the multiplicative inverse of a.
    i don't know why it's in your text book, but you can see it for yourself that the sum is not 0. for example what do you get for p = 3?

    also, are you sure you've posted the question as it is in your text book?
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  6. #6
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    I'm so sorry. It's actually supposed to equal -1. Apologies.
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  7. #7
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    Quote Originally Posted by htata123 View Post

    I'm so sorry. It's actually supposed to equal -1. Apologies.
    ok! first note that \sum_{k=1}^{p-2}\left ( \frac{k(k+1)}{p} \right)=\sum_{k=1}^{p-1}\left ( \frac{k(k+1)}{p} \right)=S_1. for any 0 \leq j \leq p-1, let S_j=\sum_{k=1}^{p-1}\left ( \frac{k(k+j)}{p} \right). then S_0=p-1 and for any 1 \leq j \leq p-1: \ S_j=S_1. why?

    also note that for any integer k: \ \ \sum_{j=0}^{p-1}\left ( \frac{k+j}{p} \right)= \sum_{j=1}^{p-1}\left ( \frac{j}{p} \right)=0. therefore: (p \ - \ 1)(S_1+1)=\sum_{j=0}^{p-1}S_j=\sum_{k=1}^{p-1} \sum_{j=0}^{p-1}\left ( \frac{k(k+j)}{p} \right)=\sum_{k=1}^{p-1}\left ( \frac{k}{p} \right)\sum_{j=0}^{p-1}\left ( \frac{k+j}{p} \right)=0. thus S_1=-1.
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