I am interested in the hypersurface $X\subset\mathbb{A}^4\_{\mathbb{F}\_{5^n}}$ defined by
$$
X = \{x^3 + 3xy^2 + z^3 + 3zw^2 + 1 = 0\}
$$
over a finite field $\mathbb{F}\_{5^n}$ with $5^n$ elements. Via some computer experiment I have noticed that when $n$ is odd the number of points of $X$ is equal to the number of points of $\mathbb{A}^3\_{\mathbb{F}\_{5^n}}$.


Is this just a coincidence or is there a theoretical reason for this?


Thank you.



Points on affine hypersurface over finite field
Note that the defining equation for $X$ can be rewritten as
$$(x+y)^3+(x-y)^3+(z+w)^3+(z-w)^3=-2.$$


As the linear transformation $(x,y,z,w)\mapsto(x+y,x-y,z+w,z-w)$ is invertible over any field of characteristic not equal to $2$, the number of $\mathbb{F}\_{5^n}$-points on $X$ is equal to the number of $\mathbb{F}\_{5^n}$-points on the hypersurface $Y$ with defining equation
$$a^3+b^3+c^3+d^3=-2.$$


As $3$ does not divide $5^n-1$, every element of $\mathbb{F}\_{5^n}$ is a cube in a unique way, so the number of points on $Y$ are the same as the number of points on the hypersurface defined by
$$e+f+g+h=-2,$$
but this hypersurface is isomorphic to $\mathbb{A}^3,$ as desired.


