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Sums of Legendre Symbols Question

  1. Jul 1, 2012 #1

    [tex]\sum_{i=0}^{p-1} (\frac{i^2+a}{p})=-1[/tex] for any odd prime p and any integer a. (I am referring to the Legendre Symbol).

    I was reading a paper where they claimed it was true for the a=1 case and referred to a source that I don't have immediate access to. So I was wondering if anyone knows if this is true or a source that talks about this? I know it doesn't mean it's necessarily true, but this proposition has been true with all the examples I've looked at with Mathematica. Thanks!
  2. jcsd
  3. Jul 2, 2012 #2
    Nevermind, I found this in the exercises of Ireland and Rosen on page 63. Just thought I would post where I found it in case anyone else needs to know.
  4. Jul 2, 2012 #3
    In your formula, the integer a should not be zero, because

    [tex]\sum_{i=0}^{p-1} (\frac{i^2}{p})=p-1[/tex]

    (Note: [tex](\frac{0}{p}) = 0[/tex] for p > 2 and
    [tex](\frac{x^2}{p})= 1[/tex] for gcd(x,p)=1)
    Last edited: Jul 2, 2012
  5. Jul 2, 2012 #4

    Even if the integer is zero modulo p the formula works, since [itex]\,p-1=-1\pmod p[/itex]

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