Conditions for quadratic form to have non zero solutions in a finite field

Question

Does $x^TAx=0$ always have a non trivial solution when vectors are taken over $GF(k)^n$? I haven't found any counterexample yet (I was randomly trying, maybe the counterexample needs to be contructed specifically). Any thoughts?

Answer 1

A non degenerate quadratic form of rang $n\geq 3$ has always a zero over a finite field (it is a standard counting argument, see below)

For rank $1$, it does not, obviously.

For rank $2$, it will have a zero if and only if it is hyperbolic (isotropic+rank 2 implies hyperbolic) if and only if its determinant is $-u^2$ for some non zero $u$.

Proof of isotropy in rank $n\geq 3$. Clearly, it is enough to consider the case of rank $3$ quadratic forms . It is enough to prove that, given nonzero $a,b$, any $c$ maybe written as $c=ax^2+by^2$. Note that the sets $\{c-ax^2, x\in GF(q)\}$ and $\{ by^2, y\in GF(q)\}$ both have cardinality $(q-1)/2+1$ because $a,b$ are non zero (don't forget the zero square!), so they cannot be disjoint. QED