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Math Help - Legendre Symbol (5/p)

  1. #1
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    Legendre Symbol (5/p)

    Hi, I have this problem.

    Let p be an odd prime. Show that the legendre symbol (5/p)= 1 if and only if p\equiv{-1}(mod 5).

    I am really struggling to understand this legendre symbol stuff so if anyone could give me a clue I would be very grateful.

    Thanks,
    Katy
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  2. #2
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    Quote Originally Posted by harkapobi View Post
    Hi, I have this problem.






    Let p be an odd prime. Show that the legendre symbol (5/p)= 1 if and only if p\equiv{-1}(mod 5).

    I am really struggling to understand this legendre symbol stuff so if anyone could give me a clue I would be very grateful.

    Thanks,

    Katy

    By quadratic reciprocity, \binom {5}{p}=\binom{p}{5} , since 5=1\!\!\!\pmod 4, and \binom{p}{5}=1\Longleftrightarrow p=\pm 1\!\!\!\pmod 5 ...and thus the claim is false, and indeed:we've, for example, \binom{5}{11}=1 since 4^2=5\!\!\!\pmod{11}, but nevertheless 11\ne -1\!\!\!\pmod 5 .

    Tonio
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  3. #3
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    The Legendre symbol (5/p) =1 means that 5 is a quadratic residue mod p. The number 4^2 = 16 = 5 (11) shows that 5 is a quadratic residue mod 11, however 11 = 1(5), not -1.
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  4. #4
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    Oh, yes your absolutely right, the question said p\equiv{\pm1}(mod 5)

    So because and 5\equiv1(mod 4)
    \binom {5}{p}=1\equiv{p^2}(mod 5)

    \Rightarrow {p^2}\equiv1(mod 5) \Rightarrow p\equiv{\pm1}(mod 5)

    Is this right? How do you know that \binom {p}{5} = 1?
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  5. #5
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    Quote Originally Posted by harkapobi View Post
    Oh, yes your absolutely right, the question said p\equiv{\pm1}(mod 5)

    So because and 5\equiv1(mod 4)
    \binom {5}{p}=1\equiv{p^2}(mod 5)

    \Rightarrow {p^2}\equiv1(mod 5) \Rightarrow p\equiv{\pm1}(mod 5)


    What is this?? What does p^2 have to do in all this?

    Is this right? How do you know that \binom {p}{5} = 1?

    Because the only quadratic residues modulo 5 are 0, 1, -1. As p is a prime number it can't be 0 mod 5, so...

    Tonio
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  6. #6
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    The p^2 comes from:

    \binom {n}{p}\equiv{n^{0.5(p-1)}}(mod p)
    is this not right?

    Then:
    \binom {p}{5}\equiv{p^{0.5(5-1)}} = {p^2}(mod p)

    If this is not right, I dont understand how you get p\equiv{\pm1}(mod 5)
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  7. #7
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    You've already done the hard work. Once you know that
    <br />
\binom {5}{p} = \binom {p}{5}<br />
    you just have to look at the quadratic residues mod 5. Now any number, not just a prime p, is equivalent to 1,2,3 or 4 mod 5 (we ignore numbers divisible by 5). So we test them 1 by 1.

    <br />
1^2 = 1 \equiv 1 mod 5<br />

    <br />
2^2 = 4 \equiv -1 mod 5<br />

    <br />
3^2 = 9 \equiv 4 \equiv -1 mod 5<br />

    <br />
4^2 = 16 \equiv 1 mod 5<br />

    So we've tested all possible numbers and found that 1 and -1 are the only quadratic residues, and that 2 and 3 are not quadratic residues for any number. It's as simple as that.
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  8. #8
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    Quote Originally Posted by harkapobi View Post
    The p^2 comes from:

    \binom {n}{p}\equiv{n^{0.5(p-1)}}(mod p)
    is this not right?

    Then:
    \binom {p}{5}\equiv{p^{0.5(5-1)}} = {p^2}\pmod p

    But then here it is \binom {p}{5}\equiv p^2\!\!\!\!\pmod 5 , ain't it?

    Tonio


    If this is not right, I dont understand how you get p\equiv{\pm1}(mod 5)
    .
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  9. #9
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    Yes your right I meant to write mod 5. Sorry.

    So what do you do when you cannot know if \binom {p}{q} = \binom {q}{p} or \binom {p}{q} = -\binom {q}{p}?

    For example \binom {7}{p}. You dont know if \binom {7}{p} \binom {p}{7} is +1 or -1.

    Do you just look at both cases? Because I am suppose to prove that \binom {7}{p} = 1\Longleftrightarrow p\equiv{\pm1}, {\pm3} or {\pm9}(mod 28)

    I dont understand where the (mod 28) comes from here. I really dont understand this stuff very well.

    Thank you for all your help
    Katy
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