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Thread: Sum of Legendre Symbols

  1. #1
    Member Haven's Avatar
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    Sum of Legendre Symbols

    Let $\displaystyle f(x) = a*x^2 + b*x + c$ and $\displaystyle d = b^2 - 4*a*c $
    Let p be prime, such that p doesn't divide a.
    Show if p doesn't divide d then

    $\displaystyle \sum_{x=1}^{p}\ (\frac{f(x)}{p}) = (-1)*(\frac{a}{p}) $

    We're given that the sum as $\displaystyle \sum_{x=1}^{p}\\x^k\\\equiv $ 0 (mod p) if p-1 doesn't divide k and is = p-1 if p-1 divides k.

    My prof also gave the hint that using euler's criterion will let us use the identity given, but i don't see it

    Again sorry about the bad notation, i'm still figuring out LaTeX
    Last edited by Haven; Jul 30th 2009 at 08:51 PM.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that: $\displaystyle 4a\cdot{f(x)}=(2a\cdot x +b)^2-(b^2-4ac)$

    So we can write: $\displaystyle \left( {\tfrac{a}
    {p}} \right)_L \cdot \left( {\tfrac{{f\left( x \right)}}
    {p}} \right)_L = \left( {\tfrac{{4af\left( x \right)}}
    {p}} \right)_L = \left( {\tfrac{{\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)}}
    {p}} \right)_{_L } $

    That is: $\displaystyle
    \left( {\tfrac{{f\left( x \right)}}
    {p}} \right)_L = \left( {\tfrac{a}
    {p}} \right)_L \cdot \left( {\tfrac{{\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)}}
    {p}} \right)_{_L }
    $ and therefore: $\displaystyle \sum_{x=1}^{p}{
    \left( {\tfrac{{f\left( x \right)}}
    {p}} \right)_L }= \left( {\tfrac{a}
    {p}} \right)_L \cdot \sum_{x=1}^{p}{\left( {\tfrac{{\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)}}
    {p}} \right)_{_L } }
    $ $\displaystyle (*)$

    By Euler's Criterion: $\displaystyle
    \sum\limits_{x = 1}^p {\left( {\tfrac{{\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)}}
    {p}} \right)_{_L } } \equiv \sum\limits_{x = 1}^p {\left[ {\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)} \right]^{\tfrac{{p - 1}}
    {2}} } (\bmod. p)
    $

    Apply the Binomial Theorem to each term: $\displaystyle
    \sum\limits_{x = 1}^p {\left[ {\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)} \right]^{\tfrac{{p - 1}}
    {2}} } = \sum\limits_{x = 1}^p {\sum\limits_{k = 0}^{\tfrac{{p - 1}}
    {2}} {\binom{\tfrac{p-1}{2}}{k}\left( {2ax + b} \right)^{2k} \cdot \left( { - d} \right)^{\tfrac{{p - 1}}
    {2} - k} } }
    $ $\displaystyle =
    \sum\limits_{k = 0}^{\tfrac{{p - 1}}
    {2}} {\binom{\tfrac{p-1}{2}}{k}\left( {\sum\limits_{x = 1}^p {\left( {2ax + b} \right)^{2k} } } \right) \cdot \left( { - d} \right)^{\tfrac{{p - 1}}
    {2} - k} } $

    Apply The binomial Theorem to: $\displaystyle
    \sum\limits_{x = 1}^p {\left( {2ax + b} \right)^{2k} }
    $ getting, by the sum you gave down there: $\displaystyle
    \sum\limits_{x = 1}^p {\left( {2ax + b} \right)^{2k} } \equiv _p \left\{ \begin{gathered}
    0{\text{ if }}k < \tfrac{{p - 1}}
    {2} \hfill \\
    \sum\limits_{x = 1}^p {\left( {2ax} \right)^{p - 1} } \equiv - 1{\text{ if }}k = \tfrac{{p - 1}}
    {2} \hfill \\
    \end{gathered} \right.
    $

    Thus: $\displaystyle
    \sum\limits_{x = 1}^p {\left[ {\left( {2ax + b} \right)^2 - \left( {b^2 - 4ac} \right)} \right]^{\tfrac{{p - 1}}
    {2}} } \equiv{-1} (\bmod.p)
    $ and by $\displaystyle (*)$ we get the result
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