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Thread: Legendre symbol

  1. #1
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    Legendre symbol

    Hey guys, I need a bit of help here.

    1. Show that the equation $\displaystyle x^2-71y^2=19$ has no integer solution using Legendre symbol.

    So $\displaystyle \left(\frac{a}{p}\right) \equiv a^{(p-1)/2}\ \pmod{p}$

    $\displaystyle \left(\frac{71}{19}\right) \equiv 71^{(19-1)/2}\ \pmod{19} $

    $\displaystyle 71^9\ \pmod{19} = 18 = -1$

    So because it equals $\displaystyle -1$, $\displaystyle 71$ is a quadratic non-residue modulo $\displaystyle 19$ and so there is no integer solution to $\displaystyle x^2-71y^2=19$.

    Is this enough to show no integer solutions?


    2. Evaluate the Legendre symbol $\displaystyle \left(\frac{p^2-1}{p}\right)$ for odd prime $\displaystyle p$.

    So $\displaystyle \left(\frac{p^2-1}{p}\right) \equiv (p^2-1)^{(p-1)/2}\ \pmod{p}$

    What is the formal way to go from here?
    I have evaluated this for several cases of odd prime p and kind of see a pattern.

    Also, this value alternates between -1 and 1 for odd numbers.

    Any help guys?

    Thanks for your help.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Legendre symbol

    1. $\displaystyle x^2-71y^2=19 \implies x^2\equiv 19 \pmod{71}$

    $\displaystyle \left(\frac{19}{71}\right)=$$\displaystyle \left(\frac{71}{19}\right)=-1$

    So, 19 is a quadratic non-residue modulo 71 and therefore the equation has no integer solutions.

    2. $\displaystyle \left(\frac{p^2-1}{p}\right)=\left(\frac{(p+1)(p-1)}{p}\right)$

    $\displaystyle =\left(\frac{1}{p}\right)\left(\frac{-1}{p}\right)$

    $\displaystyle =(-1)^{(p-1)/2}$
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  3. #3
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    Re: Legendre symbol

    Oh cool, thanks alex, I understand it!
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