Prove $\sum\limits_{j=1}^{p-1} j\left(\frac{j}{p}\right) = 0 $ for an odd prime $p$ with $p\equiv 1\text{ mod } 4$

Question

I want to show for an odd prime $p$ with $p\equiv 1\text{ mod } 4$, that

$$\sum\limits_{j=1}^{p-1} j\left(\frac{j}{p}\right) = 0 $$

where $\left(\frac{j}{p}\right) $ is the Jacobi symbol. I got so far:

Let $w$ be a primitive root modulo $p$, then we get: $\sum\limits_{j=1}^{p-1} j\left(\frac{j}{p}\right) \equiv \sum\limits_{r=1}^{p-1} w^r\left(\frac{w^r}{p}\right) = \sum\limits_{j=1}^{p-1} w^r\cdot(-1)^r = \sum\limits_{j=1}^{p-1} (-w)^r = \sum\limits_{j=0}^{p-2} (-w)^r= \frac{(-w)^{p-1} -1}{-w-1} \equiv 0 \text{ mod } p $, where I used that $-w\neq1$ because then $p=3$, but $p\equiv 1\text{ mod } 4$ and in the last equation I used Fermat's Little Theorem. So how do I get $"=0"$ and not $"\equiv 0 \text{ mod } p"$?

Answer 1

1. $\left(\frac{j}{p}\right)=\left(\frac{p-j}{p}\right)$ since $p\equiv1\bmod4$.

2. $\sum_1^{(p-1)/2}\left(\frac{j}{p}\right)=(1/2)\sum_1^{(p-1)/2}\left[\left(\frac{j}{p}\right)+\left(\frac{p-j}{p}\right)\right]=(1/2)\sum_1^{p-1}\left(\frac{j}{p}\right)=0$ since there are exactly as many quadratic residues as nonresidues modulo $p$.

3. $\sum_1^{p-1}j\left(\frac{j}{p}\right)=p\sum_1^{(p-1)/2}\left(\frac{j}{p}\right)=0$, where the first equation follows from the calculations in the comments on the question.