Quadratic forms over $F_p$ in more than $2$ variables are isotropic

Question

I'm trying to find all anisotropic quadratic forms over $F_p=Z/pZ$. I have found that "If $F$ is a finite field and $(V, q)$ is a quadratic space of dimension at least three, then it is isotropic" (here https://en.wikipedia.org/wiki/Isotropic_quadratic_form). So I wonder if the fact is true and if it is, why. Thank you

Answer 1

First, an observation: if the number of variables of $q$ is less than the dimension, then we can get an isotropic vector by assigning $0$ to the appearing variables and $1$ to the non appearing ones. So we now assume the number of variables $n$ to be the dimension.

Then your result is an specific instance of the Chevalley-Warning theorem: If a set of polynomials $\{f_i\}$ over a finite field satisfies $n>\sum_i d_i$, where $n$ is the number of variables and $d_i$ the degree of $f_i$, then the system of equations $\{f_i=0\}$ has a number of solutions which is a multiple of the characteristic of the finite field.

In your case, you have just one polynomial $q$ of degree $2$ and $n\geq 3>2$, so the number of solutions is a multiple of $p$, hence can be $0,p,2p\ldots$. But we already know that $(0,\ldots,0)$ is a solution, so there must be at least other $p-1$ nonzero solutions $v$ which give $q(v)=0$.