Let $k = \mathbb{F}\_q$ be a finite field with $q$ elements and $Q\subset\mathbb{P}^n\_k$ a quadric hypersurface defined over $k$.


By the Chevalley-Warning theorem if $n\geq 2$ then $Q$ has a point. Is there a formula for the number of points of $Q$ or at least a bound of the type $\#Q(k)\geq f(q,n)$ where $f$ is a function of $q$ and $n$?



Number of points of a quadric hypersurface over a finite field
Yes. The number of points of a smooth quadric hypersurface is $\frac{q^n-1}{q-1}$ if $n$ is even or $\frac{q^n-1}{q-1} \pm q^{ \frac{n-1}{2}}$ if $n$ is odd.


This can be proven in a number of ways. Here is a (perhaps extravagantly) geometric approach:


Fix a point $x$. A line through $x$ contains one other points of $Q$, unless that line is tangent to $Q$ at $x$ - in other words, is contained in the tangent plane of $Q$ at $x$. If so, then the line contains no other points of $Q$, except if the line is contained in $Q$, in which case it contains $q$ other points. So the number of points is


$$1 + \frac{ q^{n}-1}{q-1} - \frac{ q^{n-1}-1}{q-1} + qN = q^{n-1} +1 + qN$$


where $N$ is the number of lines through $x$ that lie in $Q$, since I have taken the point $x$, added the number of lines, subtracted the number of tangent lines, and added $qN$.


Now the lines through $x$ in the tangent plane are parameterized by an $n-2$-dimensional projective space, and the lines contained in $Q$ form a quadric surface in that projective space. So the number of points on a quadric hypersurface in $\mathbb P^n$ for $n\geq 2$ is $q^{n-1} +1$ plus $q$ times the number of points on a quadric hypersurface in $\mathbb P^{n-2}$. The formulas in above then follow by induction from the easy base cases $n=0, n=1$.


