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Thread: 2 Quadratic Reciprocity Proofs

  1. #1
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    Quadratic Reciprocity Proof (working now)

    Hi all, messed up using Latex at first. Okay, got two problems on quadratic reciprocity. I've worked up something of a proof for each, but I feel very uncertain. Thought maybe someone could give me some pointers.

    Okay, simple one first:
    Suppose p is a prime such that $\displaystyle p\equiv 3 (mod 4)$, and let a be a QR modulo p.
    Show that $\displaystyle x=a^\frac{p+1}{4}$ is a solution to
    $\displaystyle x^2\equiv a(mod\, p)$

    My proof:
    a - QR mod p $\displaystyle \Rightarrow $ the Legendre symbol
    $\displaystyle (\frac{a}{p})= 1 \equiv a^{\frac{p-1}{2}}$,
    so clearly $\displaystyle 1\equiv a^{\frac{p-1}{2}} (mod\, p)$.
    Multiplying both sides by a yields
    $\displaystyle a\equiv a^{\frac{p-1}{2}}*a\equiv a^{\frac{p+1}{2}}$.
    But if $\displaystyle x=a^{\frac{p+1}{4}}$,
    we have that $\displaystyle x^{2}=a^{\frac{p+1}{2}}$.
    We know that $\displaystyle a^{\frac{p+1}{2}}$ is congruent to a modulo p, so
    $\displaystyle x=a^{\frac{p+1}{4}} $ is a solution to the congruence.

    My trouble is that this seems too easy, and not once did I apply $\displaystyle p\equiv 3(mod\, 4)$.

    Any pointers? Thanks in advance.
    Last edited by swashbucklord; Jun 19th 2011 at 03:29 PM. Reason: Used latex wrong
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Quadratic Reciprocity Proof (working now)

    The 'solution' $\displaystyle x= a^{\frac{p+1}{4}}$ must of course be an integer and that happens only if $\displaystyle p \equiv 3\ (\text{mod}\ 4)$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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