Prove $\bigg(\frac{1(1-m)}{p}\bigg) + \bigg(\frac{2(2-m)}{p}\bigg) + \dots + \bigg(\frac{(p-1)(p-1-m)}{p}\bigg) = -1$.

Question

How would I go about proving this? Note that I'm not allowed to use quadratic reciprocity since my textbook hasn't covered it.

Answer 1

Define $\eta(x)=\left(\frac{x}{p}\right)$ (in fact, $\eta$ is the quadratic character of $\mathbb{F}_p$). Note that $\eta\left(x^{-1}\right)=\eta(x)$ for all $x\in\mathbb{F}_p^*$. Then the desired sum is

$$S=\sum_{x\in\mathbb{F}_p^*}\eta(x(x-m))=\sum_{x\in\mathbb{F}_p^*}\eta\left(x^{-1}(x-m)\right)=\sum_{x\in\mathbb{F}_p^*}\eta\left(1-mx^{-1}\right).$$

Write $y=1-mx^{-1}$. As $x$ runs through $\mathbb{F}_p^*$, $y$ runs through $\mathbb{F}_p\setminus\{1\}$. Then $$S=\sum_{y\in\mathbb{F}_p}\eta\left(y\right)-\eta(1)=0-1=-1.$$