Prove that $\sum_{a\in\mathbb{Z}/p\mathbb{Z}} (a/p)a^i$ is $0$ if $i\neq (p-1)/2$ and $p-1$ if $i= (p-1)/2$.

Question

Let $0\leq i\leq p-1$. I want to prove that $$\sum_{a\in\mathbb{Z}/p\mathbb{Z}} \left(\frac{a}{p}\right) a^i$$ is $0$ is $i\neq (p-1)/2$ and $p-1$ if $i= (p-1)/2$.

I know that this holds for $i=0$ but I don't know how to approach it in general.

Answer 1

Suppose $p$ is fixed (odd prime), and denote $(\mathbb{Z}/p\mathbb{Z})^\times:=(\mathbb{Z}/p\mathbb{Z})\setminus\{0\}$ simply by $G$. Then $$\sum_{a\in G}f(a^2)=\sum_{a\in G}f(a)\left[1+\left(\frac{a}{p}\right)\right]\qquad\forall f:G\to G,$$ since each square $a\in G$ contributes $2f(a)$ to both sums, and each nonsquare contributes nothing.

It remains to substitute $f(a)=a^i$ and use the known formula $$\sum_{a\in G}a^k=\begin{cases}0,&p-1\not\mid k\\p-1,&p-1\mid k\end{cases}.$$

Answer 2

You can also use $$\left(\frac{a}{p}\right)=a^{\frac{p-1}{2}}$$ (the equality is considered in $\mathbb{F}_p$). Therefore, for an integer $i$, $0\leq i\leq p-1$, we get $$\sum_{a\in\mathbb{F}_p}\,\left(\frac{a}{p}\right)\,a^i=\sum_{a\in\mathbb{F}_p}\,a^{i+\frac{p-1}{2}}\,.$$ This equals $0$, unless $p-1\mid i+\dfrac{p-1}{2}$, which happens only if $i=\dfrac{p-1}{2}$. The claim follows immediately.