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Math Help - proof with quadratic residue

  1. #1
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    proof with quadratic residue

    Prove that if p \equiv 3 (\mod{8}) and (p-1)/2 is prime then (p-1)/2 is a quadratic residue (\mod{p}).

    I believe I need to show that (((p-1)/2)/p) = 1. Will someone show me where to start? I have been thinking about this problem all weekend.

    Thanks in advance,
    Last edited by Zalren; July 17th 2011 at 02:03 PM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Re: proof with quadratic residue

    Well, if p \equiv 3 \mod p, that doesn't leave you much choice for the value of p, does it?
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  3. #3
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    Re: proof with quadratic residue

    Oops, I typed it wrong. I corrected it!
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  4. #4
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    Re: proof with quadratic residue

    What if I started like this:

    Using Quadratic Reciprocity Theorem: (((p-1)/2)/p)(p/((p-1)/2)) = (-1)^{((((p-1)/2)-1)/2)((p-1)/2)} = 1 since (p-1)/2 is an even exponent. That tells me that (((p-1)/2)/p)=(p/((p-1)/2))

    p \equiv 1 (\mod{(p-1)/2}), I believe this is a true statement, how do I back it up?

    So (((p-1)/2)/p)=(p/((p-1)/2))=(1/((p-1)/2))=1
    Last edited by Zalren; July 17th 2011 at 02:44 PM.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Re: proof with quadratic residue

    p=2\times ((p-1)/2)+1.

    And (p-1)/2 is not even if p \equiv 3 \mod 8...
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  6. #6
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    Re: proof with quadratic residue

    So can I assume p>3, for example p = 11, 19, ..., and just show the specific case where p=3 and show that (((p-1)/2)/p) = (((3-1)/2)/3) = (1/3) = 1?
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