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Math Help - Legendre symbols

  1. #1
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    Legendre symbols

    hi, can somebody please help me with the following proof:

    Let r be an odd prime. Show that there is an integer a such that

    <br />
(\frac{a}{r})=+1<br />

    and

    [(\frac{a+1}{r})=-1


    I know i need to consider the smallest positive quadratic nonresidue
    mod r.
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  2. #2
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    The first condition by itself is easy - try a=1.

    [Something seems wrong if both conditions must hold at the same time - try r=7. The numbers mod 7 are 1,2,3,4,5,6 whose quadratic residues are:
    1,4,2,2,4,1 and it doesn't look like any of them satisfy your 2 conditions.]

    Ignore the above - misinterpreted the Legendre symbol.
    Last edited by qmech; March 10th 2010 at 01:17 PM. Reason: Error!
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  3. #3
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    Quote Originally Posted by qmech View Post
    the first condition by itself is easy - try a=1.

    Something seems wrong if both conditions must hold at the same time - try r=7. The numbers mod 7 are 1,2,3,4,5,6 whose quadratic residues are:
    1,4,2,2,4,1 and it doesn't look like any of them satisfy your 2 conditions.

    both of the condtions have to hold at the same time, since the definition states that:

    Legendre symbols-untitled.jpg
    Last edited by Maths; March 10th 2010 at 10:43 AM. Reason: typo
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by qmech View Post
    The first condition by itself is easy - try a=1.

    Something seems wrong if both conditions must hold at the same time - try r=7. The numbers mod 7 are 1,2,3,4,5,6 whose quadratic residues are:
    1,4,2,2,4,1 and it doesn't look like any of them satisfy your 2 conditions.
     \left(\frac{2}{7}\right)=1

     \left(\frac{2+1}{7}\right)=\left(\frac{3}{7}\right  )=-1

    so our hypothesis is satisfied in this case.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    If our hypothesis was false, then that would mean  \left(\frac{a}{r}\right)=1 \implies \left(\frac{a+1}{r}\right)=1 .

    We know  \left(\frac{1}{r}\right)=1 , so that would mean  \left(\frac{2}{r}\right)=1 \implies \left(\frac{3}{r}\right)=1 ... an so on.

    So we've just shown that every number is a square modulo  r which is a contradiction.
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