Proving that $F(x)=sumlimits_{k=1}^{p-1}left(frac{k}{p}right)x^k$ 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 $F(1)=sumlimits_{k=1}^{p-1}left(dfrac{k}{p}right)=0$,so $1$ is a one complex root in $M:={zmid zin mathbb C, |z|=1}$. If $pequiv 1pmod 4$, then we have
$$F'(1)=sum_{k=1}^{p-1}kleft(frac{k}{p}right)=sum_{k=1}^{p-1}(p-k)left(frac{p-k}{p}right)\
=(-1)^{frac{p-1}{2}}sum_{k=1}^{p-1}(p-k)left(frac{k}{p}right)=-(-1)^{frac{p-1}{2}}F'(1).$$
Since $pequiv 1pmod 4$, so we have $F'(1)=0$. But I can’t continue solving this problem.