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Math Help - Sum of Legendre Symbols

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

    Let  f(x) = a*x^2 + b*x + c and 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

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

    We're given that the sum as \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; July 30th 2009 at 08:51 PM.
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  2. #2
    Super Member PaulRS's Avatar
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    Note that: 4a\cdot{f(x)}=(2a\cdot x +b)^2-(b^2-4ac)

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

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

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

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

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

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