Quadratic forms of dimension at least three over odd finite fields are isotropic

By [L1], after choosing a basis we may write $q(x_1,\dots ,x_n)=a_1{x}_1^2+\cdots +a_n{x}_n^2$. If some $a_i=0$, then the corresponding basis vector is a nonzero isotropic vector. So we may assume $a_1,a_2,a_3\mathrm{\ne }0$ and restrict to the ternary subform $a_1{x}^2+a_2{y}^2+a_3{z}^2$.

Let $Q=\{u^2:u\in F\}$ be the set of square classes including $0$. By [L2] and [L3], $F^{\times }$ has even order, so the nonzero squares form an index-two subgroup and $\mathrm{\mid }Q\mathrm{\mid }=(\mathrm{\mid }F\mathrm{\mid }+1)\mathrm{/}2$. The sets $a_{1}Q$ and $-a_3-a_{2}Q$ therefore each have more than half the elements of $F$, so they intersect. Hence there exist $x,y\in F$ with $a_1{x}^2=-a_3-a_2{y}^2$, and then $(x,y,1)$ is a nonzero isotropic vector for the ternary subform and therefore for $q$.