This question is from David burton section: Quadratic reciprocity law, page 184.
Question: If p is an odd prime, show that $\sum_{a=1}^{p-2} (a(a+1) /p) =-1$ .
I can only say that a, a+1 are always co-prime but could not think of using it.
Can you please tell how to solve this particular problem.
There are some questions where one can't provide much as attempt.
$(\frac{x}{p})=1$ iff the $\it{residue}$ $\it{class}$ $\bar{x}:=x+p\Bbb Z$ is a non-zero square in the finite field $$\Bbb F_p:=\{x+p\Bbb Z:x\in\Bbb Z\}$$ $\bar{x}=\bar{y}$ simply means $p\;\Big\vert\;(x-y)$ or, equivalently, $x\equiv y\mod{p}$ and so addition/multiplication of residue classes is understandably well-defined as $$\bar{x}+\bar{y}=\overline{x+y}\;\;\;\;\;\;\;\;\;\;\;\;\bar{x}\cdot\bar{y}=\overline{x\cdot y}$$ where $\bar{x}^{-1}$, denoted again $\frac{\bar{1}}{\bar{x}}$, is that residue class satisfying $\bar{x}\cdot\bar{x}^{-1}=\bar{1}$.
$$\chi(\bar{x}):=(\frac{x}{p})=\chi(\bar{x})=\Bigg\{\begin{matrix}0\;\text{if}\;\bar{x}=\bar{0}\\1\;\text{if}\;\bar{x}\;\text{is}\;\text{a}\;\text{non-zero}\;\text{square}\\-1\;\text{otherwise}\end{matrix}$$ defines a multiplicative function $\chi:\Bbb F_p\to\{\pm 1\}$. Note $\bar{x}$ is a non-zero square in $\Bbb F_p$ iff $\bar{x}^{-1}$ is as well. Note that given $\chi(\bar{x_0})=-1$ then because $\bar{x}\mapsto\bar{x_0}\cdot\bar{x}$ defines a permutation of $\Bbb F_p$ we see $$\sum_{\bar{x}\in\Bbb F_p}\chi(\bar{x})=\sum_{\bar{x}\in\Bbb F_p}\chi(\bar{x_0}\cdot\bar{x})=\sum_{\bar{x}\in\Bbb F_p}-\chi(\bar{x})=-\sum_{\bar{x}\in\Bbb F_p}\chi(\bar{x})\;\;\;\;\;\therefore\;\sum_{\bar{x}\in\Bbb F_p}\chi(\bar{x})=0$$ $$\sum_{a=1}^{p-2}(\frac{a(a+1)}{p})=\sum_{\bar{0},\bar{-1}\neq\bar{x}\in\Bbb F_p}\chi(\bar{x}\cdot(\bar{x}+\bar{1}))=\sum_{\bar{0},\bar{-1}\neq\bar{x}\in\Bbb F_p}\chi(\frac{\bar{x}+\bar{1}}{\bar{x}})=\sum_{\bar{0},\bar{-1}\neq\bar{x}\in\Bbb F_p}\chi(1+\bar{x}^{-1})=\sum_{\bar{0},\bar{1}\neq\bar{x}\in\Bbb F_p}\chi(\bar{x})$$ $$=\sum_{\bar{x}\in\Bbb F_p}\chi(\bar{x})-\chi(\bar{0})-\chi(\bar{1})=0-1=-1$$