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Math Help - Quadratic reciprocity

  1. #1
    MHF Contributor alexmahone's Avatar
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    Quadratic reciprocity

    Prove that \displaystyle\sum\limits_{j=1}^{p-1} \left(\frac{j}{p}\right)=0, p an odd prime.

    My attempt:

    If p \equiv 3\ (mod\ 4),

    \left(\frac{-j}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{j}{p}\right)=(-1)^{(p-1)/2}\left(\frac{j}{p}\right)=-\left(\frac{j}{p}\right) and the result follows immediately.

    But what if p \equiv 1\ (mod\ 4)?
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  2. #2
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    Re: Quadratic reciprocity

    Quote Originally Posted by alexmahone View Post
    Prove that \displaystyle\sum\limits_{j=1}^{p-1} \left(\frac{j}{p}\right)=0, p an odd prime.

    My attempt:

    If p \equiv 3\ (mod\ 4),

    \left(\frac{-j}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{j}{p}\right)=(-1)^{(p-1)/2}\left(\frac{j}{p}\right)=-\left(\frac{j}{p}\right) and the result follows immediately.

    But what if p \equiv 1\ (mod\ 4)?
    Theorem: For an odd prime p there are exactly (p-1)/2 quadratic residues and exactly (p-1)/2 quadratic nonresidues.*

    So restating the theorem, half of the terms (1/p), (2/p), ..., ((p-1)/p) are -1 and half are 1.
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