Find the sum of the value $\left|\sum_{j=1}^{\frac{p-1}{2}}w^{j^2}\right|$

Question

Let $p$ be prime number,and $p\equiv 3\pmod 4$, and $w=e^{\frac{2\pi}{p}\cdot i}$. Find the value $$\left|\sum_{j=1}^{\frac{p-1}{2}}w^{j^2}\right|$$

it seem use Quadratic Residues?

Answer 1

First you can observe that $$\sum_{j=1}^{\frac{p-1}{2}} w^{j^2} = \sum_{j=\frac{p-1}{2}+1}^{p-1} w^{(p-j)^2} = \sum_{j=\frac{p-1}{2}+1}^{p-1} w^{j^2}$$

So $$\sum_{j=1}^{\frac{p-1}{2}} w^{j^2} = \frac12 \sum_{j=1}^{p-1} w^{j^2} = \frac12 \left(\sum_{j=0}^{p-1} w^{j^2} - 1\right) = \frac12 (g(1;p) - 1) = \frac12 (-1 +i\sqrt p)$$