sums of legendre symbol I am studying the sums of Legendre symbol. It's easy to prove that for $p\equiv 1 \pmod 4$,
$$\sum_{0<n<p/2}\left(\frac{n}{p}\right)=0.$$
by writing the sum over all residue classes and then using the fact that sum over $0$ to $p/2$ is same as $p/2$ to $p$ (by a change of variable). But the same idea is not working with the following two equalities:
(1) If $p\equiv 3 \pmod 8$, then
$$\sum_{0<n<p/4} \left(\frac{n}{p}\right) =0.$$
(2) If $p\equiv 7 \pmod 8$, then
$$\sum_{p/4<n<p/2} \left(\frac{n}{p}\right) =0.$$
Thanks in advance for any help.

(1) If $p \equiv 3 \pmod{8}$ then $(\frac{-1}{p})=(\frac{2}{p})=-1$. Let $p=8k+3$. Observe that 
\begin{align}
\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})}& =\sum_{1 \leq n \leq 4k+1}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k}{(\frac{2n}{p})}+\sum_{1 \leq n \leq 2k+1}{(\frac{2n-1}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{p+1-2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{2k+1 \leq n \leq 4k+1}{(\frac{2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+(\frac{-1}{p})(\frac{2}{p})\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})} \\
& =-\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})}
\end{align}
Thus
\begin{equation}
\sum_{0<n<\frac{p}{4}}{(\frac{n}{p})}=\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}=0
\end{equation}
(2) If $p \equiv 7 \pmod{8}$ then $(\frac{-1}{p})=-1, (\frac{2}{p})=1$. Let $p=8k+7$. Observe that
\begin{align}
\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})}& =\sum_{1 \leq n \leq 4k+3}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k+1}{(\frac{2n}{p})}+\sum_{1 \leq n \leq 2k+2}{(\frac{2n-1}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{1 \leq n \leq 2k+2}{(\frac{p+1-2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{2k+2 \leq n \leq 4k+3}{(\frac{2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+(\frac{-1}{p})(\frac{2}{p})\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}-\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})}
\end{align}
Thus
\begin{equation}
\sum_{\frac{p}{4}<n<\frac{p}{2}}{(\frac{n}{p})}=\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})}=0
\end{equation}
