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Math Help - Quadratic congruence - proof

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    Quadratic congruence - proof

    Show that the quadratic reciprocity theorem could also be written (p/q)(q/p) = (-1)^[(p-1)(q-1)/4] for odd primes p and q.
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    MHF Contributor Bruno J.'s Avatar
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    That's usually how it's written - which form do you have on hand?
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    My text uses the following:

    If p and q are odd primes and p==q==3(mod 4), then (p/q) = -(q/p). Otherwise, (p/q) = (q/p).
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by tarheelborn View Post
    My text uses the following:

    If p and q are odd primes and p==q==3(mod 4), then (p/q) = -(q/p). Otherwise, (p/q) = (q/p).
    Well, there's four cases to verify, so a quick case by case proof should do the trick.
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  5. #5
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    p==q==3(mod 4), then (p/q) = -(q/p).

    Meaning p==3(mod 4), q==3(mod 4), p==q(mod 4) and?
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Here are the four cases:

     p\equiv1\mod{4} and q\equiv1\mod{4}
     p\equiv1\mod{4} and q\equiv3\mod{4}
     p\equiv3\mod{4} and q\equiv1\mod{4}
     p\equiv3\mod{4} and q\equiv3\mod{4}

    I'll verify one case for you:
     p\equiv1\mod{4} and q\equiv3\mod{4}

    We know  \left(\frac{p}{q}\right) = \left(\frac{q}{p}\right) .
    Therefore  \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = \left(\frac{p}{q}\right)^2 = (\pm1)^2 = 1

    We're also given that  p=4k+1 and  q=4n+3 .
    So  \tfrac14(p-1)(q-1) = 2k(2n+1) = 2x .
    Thus  (-1)^{\frac{(p-1)(q-1)}{4}} = (-1)^{2x} = 1 .

    (Notice that is proves the case  p\equiv3\mod{4} and q\equiv1\mod{4} too since  p and  q are arbitrary.
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