Silverman Rational Points on Elliptic Curves Exercise 4.3c: Counting points on the Fermat curve over mod p Is there any somewhat elementary way to prove the Hasse-Weil bound for the Fermat curve? i.e.
Let $p$ be a prime and $m$ be some factor of $p-1$. Let $C$ be the projective curve $x^m+y^m+z^m=0$. Then
$$\left|\#C\left(\mathbb F_p\right)-p-1\right|\leq(m-1)(m-2)\sqrt p$$
Silverman also hinted that this is easier when $m$ is a prime.
I've tried a few approached but none seems to reach the answer. My closest attempt was:
Let $\chi$ be a character of $\mathbb F_p$ that has values in the $m$th roots of unity, so $\chi\left(x\right)=1$ iff $x\equiv y^m\pmod p$ for some $y$. Then with $0^0=1$
\begin{align*}
\#C\left(\mathbb F_p\right)&=\underbrace{m}_{(x:y:0)}+\sum_{x=0}^{p-1}\underbrace{\sum_{i=0}^{m-1}\chi\left(-x^m-1\right)^i}_{\text{Number of points on }x^m+y^m+1=0}\\
&=m+p+\sum_{i=1}^{m-1}\underbrace{\sum_{x=0}^{p-1}\chi\left(-x^m-1\right)^i}_{\gamma_i}\\
&=1+p+\sum_{i=1}^{m-1}\Re\left(\gamma_i+1\right)
\end{align*}
however I can't seem to find anyway to prove that $\left|\Re\left(\gamma_i+1\right)\right|\leq(m-2)\sqrt p$, which I've tested to be true numerically for the first 50 primes.
Alternatively I've considered the sum ($\zeta_p=e^{\frac{2\pi i}p}$)
\begin{align*}
\frac1{p-1}\left(\sum_{x=0}^{p-1}\sum_{y=0}^{p-1}\sum_{z=0}^{p-1}\frac1p\sum_{k=0}^{p-1}\zeta_p^{k\left(x^m+y^m+z^m\right)}-1\right)&=\frac1{p^2-p}\left(\sum_{x=1}^{p-1}\sum_{y=1}^{p-1}\sum_{z=1}^{p-1}\sum_{k=0}^{p-1}\zeta_p^{k\left(x^m+y^m+z^m\right)}\right)+3m\\
&=\frac1{p^2-p}\left(\sum_{k=1}^{p-1}\sum_{x=1}^{p-1}\zeta_p^{kx^m}\sum_{y=1}^{p-1}\zeta_p^{ky^m}\sum_{z=1}^{p-1}\zeta_p^{kz^m}+(p-1)^3\right)+3m
\end{align*}
For this I'm looking to have some bound on the power exponential sum via Fourier transforms(similar to quadratic gauss sums):
Let $f(x)=\zeta_p^{kx^m}$, since $\frac{f(x)+f(x+1)}2=\sum c_n$ where $c_n$ are the Fourier coefficients for the function $f(u+x)$ defined on $u\in[0,1]$, we can express the sum $\sum_{x=0}^{p-1}f(x)$ as
\begin{align*}
\sum_{x=0}^{p-1}f(x)&=\sum_{n=-\infty}^\infty\sum_{x=0}^{p-1}\int_0^1f(u+x)e^{-2\pi inu}du\\
&=\sum_{n=-\infty}^\infty\int_0^pf(x)e^{-2\pi inx}dx\\
&=\sum_{n=-\infty}^\infty\int_0^pe^{\frac{2\pi ikx^m}p}e^{-2\pi inx}dx\\
&=\sum_{n=-\infty}^\infty\int_0^p\zeta_p^{kx^m-pnx}dx\\
&=\sum_{n=-\infty}^\infty\int_0^p\zeta_p^{k\left(x^m-\frac{pn}kx\right)}dx\\
&=\sum_{n=-\infty}^\infty\int_0^p\zeta_p^{k\left(\left(x-\frac1m\sqrt[m-1]{\frac{pn}k}\right)^m+\dots+\frac1{m^m}\sqrt[m-1]{\frac{pn}k}^m\right)}dx\\
\end{align*}
However it doesnt look like the sum can be split up into smaller parts and the whole integral looks rather hard to bound.
As Silverman suggested trying the case that $m$ is prime first, it seems quite likely that there is a somewhat elementary solution, but I'm kinda lost at what to do currently, any suggestions for how should I continue this problem or a different method?
 A: Turns out I'm seriously overcomplicating this problem:
Let $\chi$ be a character with values in the $m$th roots of unity
Let $N(P)$ be the number of roots over $\mathbb F_p$ for some algebraic curve $P$
\begin{align*}
\#C\left(\mathbb F_p\right)&=\underbrace{N\left(x^m+1\right)}_{z=0}+N\left(x^m+y^m+1\right)\\
&=N\left(x^m+1\right)+\sum_{k=0}^{p-1}N(x^m=k)N(y^m=1-k)\\
&=N\left(x^m+1\right)+\sum_{k=0}^{p-1}\sum_{i=0}^{m-1}\chi(k)^i\sum_{j=0}^{m-1}\chi(1-k)^j\\
&=N\left(x^m+1\right)+\sum_{i=0}^{m-1}\sum_{j=0}^{m-1}\sum_{k=0}^{p-1}\chi(k)^i\chi(1-k)^j\\
\end{align*}
When $(i,j)=(0,0)$, the sum over $k$ is $p$.
When $i+j=m$,
\begin{align*}
\sum_{k=0}^{p-1}\chi(k)^i\chi(1-k)^j&=\sum_{k=0}^{p-1}\chi(k)^{m-j}\chi(1-k)^j\\
&=\sum_{k=1}^{p-1}\chi(1-\frac1k)^j\\
&=-\chi(-1)^j\\
\end{align*}
Importantly, $\sum_{j=0}^{m-1}\chi(-1)^j=N\left(x^m+1\right)$, hence combining all of these, we get
$$\#C\left(\mathbb F_p\right)=p+1+\sum_{i=1}^{m-1}\sum_{j=1,i+j\neq p}^{m-1}\sum_{k=0}^{p-1}\chi(k)^i\chi(1-k)^j$$
Importantly, notice that $\sum_{k=0}^{p-1}\chi(k)^i\chi(1-k)^j$ is a Jacobi sum and $i+j\neq m$, hence it can be factored into $\frac{g\left(\chi^i\right)g\left(\chi^j\right)}{g\left(\chi^{i+j}\right)}$ where $g$ is the Gauss sum, hence its norm is $\sqrt p$ and since we have $\sum_{i=1}^{m-1}\sum_{j=1,i+j\neq m}^{m-1}$ summing over $(m-1)(m-2)$ numbers whose norm is $\sqrt p$, giving us the result
$$\left|\#C\left(\mathbb F_p\right)-p-1\right|\leq(m-1)(m-2)\sqrt p$$
As a final note this doesn't quite show $\Re\left(\gamma_i+1\right)\leq(m-2)\sqrt p$ but the $\Re$ is quite awkward to handle anyways
