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Math Help - LQR problem

  1. #1
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    LQR problem

    Can someone check my argument on this problem? If p is prime and p\equiv 1 ( mod 4) and m^2+n^2=p and m>0 odd. Then (\frac{m}{p})=1
    By LQR,
    (\frac{m}{p})=(\frac{p}{m})=(\frac{m^2+n^2}{m})=(\  frac{n^2}{m}). Here, do I need to check the cases when m is prime and when m is composite to figure out if (\frac{n^2}{m})=1?
    In class, I think we have a property that (\frac{n^2}{p})=1 with p prime.
    Last edited by namelessguy; October 21st 2008 at 11:40 PM.
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    (\frac{m}{p})=(\frac{p}{m})?
    I'm not sure I get that line
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by SimonM View Post
    I'm not sure I get that line
    He's using the Legendre symbol, if that's what you don't get.
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    Quote Originally Posted by alexmahone View Post
    He's using the Legendre symbol, if that's what you don't get.
    I get that, I just don't quite get what he has done which allows that, it looks like quadratic reciprocity, but not quite
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    Quote Originally Posted by SimonM View Post
    I get that, I just don't quite get what he has done which allows that, it looks like quadratic reciprocity, but not quite
    Hi SimonM
    (\frac{m}{p})=(\frac{p}{m}) since p\equiv ( 1 mod 4 )
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    Quote Originally Posted by alexmahone View Post
    He's using the Legendre symbol, if that's what you don't get.
    Quote Originally Posted by SimonM View Post
    I get that, I just don't quite get what he has done which allows that, it looks like quadratic reciprocity, but not quite
    This is all okay. Because namelessguy is using Quadradic Reciprocity generalized to Jacobi symbols.
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  7. #7
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    Quote Originally Posted by namelessguy View Post
    Here, do I need to check the cases when m is prime and when m is composite to figure out if (\frac{n^2}{m})=1?
    In class, I think we have a property that (\frac{n^2}{p})=1 with p prime.
    To prove that, all you need to show is that (n^2,m)=1. Can you do this?

    Assume the opposite
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  8. #8
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    Quote Originally Posted by SimonM View Post
    To prove that, all you need to show is that (n^2,m)=1. Can you do this?

    Assume the opposite
    I only need to know that (n,m)=1, which is obviously true from the hypothesis, we know that n, m, p are relatively prime. Then I use a theorem for Jacobi symbol to conclude that (\frac{n^2}{m})=1
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