## Sums of Legendre Symbols Question

Proposition:

$$\sum_{i=0}^{p-1} (\frac{i^2+a}{p})=-1$$ 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!
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 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.
 Recognitions: Gold Member In your formula, the integer a should not be zero, because $$\sum_{i=0}^{p-1} (\frac{i^2}{p})=p-1$$ (Note: $$(\frac{0}{p}) = 0$$ for p > 2 and $$(\frac{x^2}{p})= 1$$ for gcd(x,p)=1)

## Sums of Legendre Symbols Question

 Quote by RamaWolf In your formula, the integer a should not be zero, because $$\sum_{i=0}^{p-1} (\frac{i^2}{p})=p-1$$ (Note: $$(\frac{0}{p}) = 0$$ for p > 2 and $$(\frac{x^2}{p})= 1$$ for gcd(x,p)=1)

Even if the integer is zero modulo p the formula works, since $\,p-1=-1\pmod p$

DonAntonio

 Tags legendre symbol, number theory

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