Proving that $F(x)=\sum\limits_{k=1}^{p-1}\left(\frac{k}{p}\right)x^k$ has at least $\frac{p-1}{2}$ different complex roots 
Let $p$ be odd prime number. Show that
$$F(x)=\sum_{k=1}^{p-1}\left(\frac{k}{p}\right)x^k$$
has at least $\dfrac{p-1}{2}$ different complex roots $z$ with $|z|=1$,
where $\left(\dfrac{k}{p}\right)$ is Legendre's symbol.

I try: since $F(1)=\sum\limits_{k=1}^{p-1}\left(\dfrac{k}{p}\right)=0$,so $1$ is a one complex root in $M:=\{z\mid z\in \mathbb C,\ |z|=1\}$. If $p\equiv 1\pmod 4$, then we have
$$F'(1)=\sum_{k=1}^{p-1}k\left(\frac{k}{p}\right)=\sum_{k=1}^{p-1}(p-k)\left(\frac{p-k}{p}\right)\\
=(-1)^{\frac{p-1}{2}}\sum_{k=1}^{p-1}(p-k)\left(\frac{k}{p}\right)=-(-1)^{\frac{p-1}{2}}F'(1).$$
Since $p\equiv 1\pmod 4$, so we have $F'(1)=0$. But I can't continue solving this problem.
 A: $$F(e^{i t})= \sum_{k=1}^{p-1} \left(\frac{k}p\right) e^{ikt}=
\sum_{k=1}^{p-1} \left(\frac{p-k}p\right) e^{i(p-k)t}\\=
e^{ipt}\left(\frac{-1}p\right)\sum_{k=1}^{p-1} \left(\frac{k}p\right) e^{-ikt}=
e^{ipt}i^{p-1} \overline{F(e^{it})}
 $$
So $e^{-ipt/2}i^{-(p-1)/2} F(e^{it})$ is real valued.
For $p\nmid a$ with $ab\equiv 1\bmod p$
$$F(e^{2i\pi a/p}) =
\sum_{k=0}^{p-1} \left(\frac{k}p\right) e^{2i\pi ak/p}=
\sum_{k=0}^{p-1} \left(\frac{bk}p\right) e^{2i\pi 
abk/p}\\=\left(\frac{b}p\right) F(e^{2i\pi /p})= \left(\frac{a}p\right)F(e^{2i\pi/p}) $$
Next $$\sum_{k=0}^{p-1} \left(\frac{k}p\right)\left(\frac{k+1}p\right)=
\left(\frac{a^2}p\right)\sum_{k=0}^{p-1} \left(\frac{bk}p\right)\left(\frac{bk+1}p\right)$$ $$=
\sum_{k=0}^{p-1} \left(\frac{abk}p\right)\left(\frac{abk+a}p\right)=
\sum_{k=0}^{p-1} \left(\frac{k}p\right)\left(\frac{k+a}p\right)$$
so that
$$\sum_{k=0}^{p-1} \left(\frac{k}p\right)\left(\frac{k+1}p\right)=\sum_{k=0}^{p-1} \left(\frac{k}p\right)\frac1{p-1}\sum_{a=1}^{p-1}\left(\frac{k+a}p\right)= 
\sum_{k=1}^{p-1} \left(\frac{k}p\right)\frac{-1}{p-1}\left(\frac{k}p\right) = -1 $$
This implies that there are $\frac{p-1}{2}$ values of $k\in 1\ldots p-2$ such that $\left(\frac{k}p\right)\left(\frac{k+1}p\right)=-1$

For the other $\frac{p-3}{2}$ values it will be $\left(\frac{k}p\right)\left(\frac{k+1}p\right)=1$. For each of those $k$ we'll get that $e^{-ipt/2}i^{-(p-1)/2} F(e^{it})$ changes of sign on $t\in (\frac{2\pi k}p,\frac{2\pi (k+1)}p)$.

Whence we get at least $\frac{p-3}{2}$ zeros of $F(e^{it})$ on $[\frac{2\pi}p,2\pi-\frac{2\pi}p]$. Together with $F(1)=0$ it gives at least $\frac{p-1}{2}$ distinct zeros of $F(x)$ on the unit circle.
