Sums of Legendre Symbols Question

[doubleaxel195]

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!

 

[doubleaxel195]

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.

 

[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)

 

[DonAntonio]

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