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Math Help - Legendre Problem

  1. #1
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    Legendre Problem

    Let p be an odd prime. Show that the Diophantine equation x^2+py+a=0; gcd(a, p)=1 has an integral solution if and only if (-a/p)=1.
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  2. #2
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    Is correct this equation?

    Hi,
    I do not know if this helpful to you but:
    Solving to y we have:
    y=-x^2/p-a/p
    Now if (-a/p) is not 1, there is still solution. Check for a=-15, p=11 (x=2, y=1).
    If this is what you mean...
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  3. #3
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    Quote Originally Posted by gdmath View Post
    Hi,
    I do not know if this helpful to you but:
    Solving to y we have:
    y=-x^2/p-a/p
    Now if (-a/p) is not 1, there is still solution. Check for a=-15, p=11 (x=2, y=1).
    If this is what you mean...
    I am not sure what you are doing. The meaning of (-a/p) is not a fraction, it means this.

    Quote Originally Posted by cathwelch View Post
    Let p be an odd prime. Show that the Diophantine equation x^2+py+a=0; gcd(a, p)=1 has an integral solution if and only if (-a/p)=1.
    There exists an x and y that solve this equation if and only if x^2 + a = p(-y) if and only if p|(x^2+a) if and only if x^2\equiv -a(\bmod p), and this has a solution if and only if (-a/p)=1.
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  4. #4
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    Sorry i just missunderstood the question.
    I am not doing something suspicious, if this you mean.
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