Almost all local completions are isotropic in dimension at least three

By [L1], after scaling we may write $q=a_1{X}_1^2+\cdots +a_n{X}_n^2$ with integers $a_i$. If some $a_i=0$, then $q$ already has the rational isotropic vector with $X_i=1$ and every other coordinate $0$, hence it is isotropic over every ${\mathbb{Q}}_p$ and there is nothing more to prove. So assume from now on that every $a_i$ is nonzero. Exclude the finite set of primes dividing $2a_1\cdots a_n$. For any remaining odd prime $p$, all $a_i$ are units modulo $p$, so the reduction $\bar{q}$ over ${\mathbb{F}}_p$ still has dimension $n\ge 3$.

By [L2], the reduced form $\bar{q}$ has a nonzero isotropic vector $\bar{x}$. Since some coordinate ${\bar{x}}_j$ is nonzero and $a_j\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}p)$, the partial derivative $\mathrm{\partial }q\mathrm{/}\mathrm{\partial }{X}_j=2a_j{X}_j$ is nonzero at $\bar{x}$ modulo $p$. Fix lifts of the other coordinates and view $q$ as a polynomial in $X_j$ alone; then [L3] lifts the simple root ${\bar{x}}_j$ to a $p$-adic root. Thus $q$ is isotropic over ${\mathbb{Q}}_p$. Since only finitely many primes were excluded in step 1.1, the theorem follows.