$\sum_{y=0}^{p-1 } (\frac{y^2 - b^{-1}}{p}) =-1$

Question

Given Legendre symbol: $(\frac{y^2 - b^{-1}}{p})$. Please show a detailed proof of $\sum_{y=0}^{p-1 } (\frac{y^2 - b^{-1}}{p}) =-1$

Answer 1

This question shows that for⁽¹⁾⁽²⁾ $c\neq0$ we have $$\sum_{y}\left(\frac{y(y+c)}{p}\right)=-1$$ This is done by replacing⁽³⁾ $y(y+c)$ by $(y+c)/y$ which parametrizes⁽⁴⁾ $\mathbb F_p\setminus\{1\}$, which reduces the problem to the well known⁽⁵⁾ $$\sum_{y}\left(\frac{y}{p}\right)=0$$

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In this case the question is different in that $y^2-b$ only factors as above if $b$ is a square mod $p$.⁽⁶⁾ What we can do is invert the order of summation: $$\sum_{y}\left(\frac{y^2-b}{p}\right)=\sum_{x}\left(\frac{x}{p}\right) \left(1+\left(\frac{x+b}{p}\right)\right)$$ because $1+\left(\frac{x+b}{p}\right)$ is the number of solutions to $y^2-b\equiv x$.

This reduces to problem to the identity above.⁽⁷⁾

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More details

1. Strictly speaking only for $c=1$; the general case is proved analoguously.
2. For $c=0$ all terms (except for $y=0$) are $1$, and the sum equals $p-1$
3. Note that $\left(\frac{ab}p\right) = \left(\frac{ab/a^2}p\right) = \left(\frac{b/a}p\right)$ for $a\neq 0$.
4. When $y$ runs through $1, \ldots, p-1$, $1+c/y$ runs through $2, 3, \ldots, p$: it cannot equal $1$ mod $p$, and $a = 1+c/y$ iff $y=(a-1)/c$ shows that $a$ can take all values from $2$ to $p$ (mod $p$), the only restriction being that $y \neq0$.
5. Because there are as many QR's as non QR's mod $p$ (not counting $0$)
6. It factors as $(y-b)(y+b)$, but making the change of variables $z=y-b$ gives the problem from the beginning, with $c=2b$.
7. We are left with $\sum_{x}\left(\frac{x}{p}\right)=0$ and $\sum_{x}\left(\frac{x}{p}\right)\left(\frac{x+b}{p}\right)=0$, note that $\left(\frac{ab}p\right) = \left(\frac{a}p\right)\left(\frac{b}p\right)$.