Consider the Fermat cubic
$$
X = \{x\_0^3+\dots +x\_n^3 = 0\}\subset\mathbb{P}^n\_{\mathbb{F}\_{q}}
$$
over a finite field $\mathbb{F}\_{q}$ with $q$ elements.


If $q \equiv 2 \mod 3$ then the projection $\pi:X\rightarrow \mathbb{P}^{n-1}\_{\mathbb{F}\_{q}}$ defined by $\pi(x\_0,\dots,x\_n) = (x\_0,\dots,x\_{n-1})$ yields a bijection $X(\mathbb{F}\_{q})\rightarrow \mathbb{P}^{n-1}\_{\mathbb{F}\_{q}}(\mathbb{F}\_{q})$ so that the number of $\mathbb{F}\_{q}$-points of $X$ is $\frac{q^n-1}{q-1}$.


Does there exist a formula for the number of $\mathbb{F}\_{q}$-points of $X$ when $q \not\equiv 2 \mod 3$?



Fermat cubic hypersurfaces over finite fields
A. Weil, in "Numbers of solutions of equations in finite fields" (Bull. Am. Math. Soc. 55, 497-508 (1949)), proved that the number of $\mathbb{F}\_q$-points when $q\equiv 1 \bmod 3$ is
$$\frac{q^n-1}{q-1} +\frac{1}{q-1} \sum\_{\chi\_0,\ldots,\chi\_n} J\_0(\chi\_0,\ldots,\chi\_n)$$
where the sum is over $(n+1)$-tuples of non-trivial characters $\chi\_i \colon \mathbb{F}\_q^{\times} \to \mathbb{C}^{\times}$ of order $3$ such that $\prod\_i \chi\_i$ is trivial. Here $J\_0$ is the Jacobi sum
$$J\_0(\chi\_0,\ldots,\chi\_n):=\sum\_{t\_0+\ldots+t\_n=0} \prod\_i \chi\_i(t\_i).$$
($\chi\_i(0)$ is defined to be $0$.) The Jacobi sum is expressible as
$$\frac{q-1}{q}\prod\_i g(\chi\_i)$$
where $g$ is the Gauss sum $g(\chi\_i)=\sum\_{t \in \mathbb{F}\_q^{\times}}\chi\_i(t)e^{\frac{2\pi i}{p}\mathrm{Tr}(t)}$. In particular,
$$|J\_0(\chi\_0,\ldots,\chi\_n)|=(q-1)q^{\frac{n+1}{2}-1}.$$
In fact Weil studied more general diagonal hypersurfaces and his results inspired him to formulate his famous conjectures. See also Chapters 8 and 10 in Ireland and Rosen's book "A Classical Introduction to Modern Number Theory" (particularly Theorem 2 in Chapter 10).


Finally, if $q \equiv 0 \bmod 3$ then the number of $\mathbb{F}\_q$-points is like in the $q \equiv 2 \bmod 3$ case, since the equation reads $(x\_0+\ldots+x\_n)^3=0$.


*Addendum:* When $q \equiv 1 \bmod 3$, the total contribution of the Jacobi sums is $O(q^{\frac{n-1}{2}}q^n)$ which is exponentially smaller then the main term $(q^n-1)/(q-1)$ as long as $q \neq 4$. But for $q=4$ this is quite useless. However, for $q=4$, $x\_i^3$ is either $1$ (if $x\_i \neq 0$) or $0$ (if $x\_i=0$) so one can parametrize the points as tuples $(x\_0,\ldots,x\_n)$ (up to a multiplicative scalar) where the number of non-zero entries is even. This leads by the binomial theorem to the formula $$\frac{4^{n+1}+(-2)^{n+1}-2}{6}.$$
It can also be derived by computing the Gauss sums explicitly.


