Let $A$ be a symmetric matrix with even diagonal and elements in $\mathbb{Z}_p$ and nonzero discriminant. (Yes, $p$ can be two) and $a$ an nonzero integer. Suppose there exists a solution to $x^{\top}Ax/2=a$ where $x$ is a primitive vector. Is there an explicit $\delta>0$ depending only on $A$ and $a$ such that if $\Vert a_{ij}c_{ij} \Vert<\delta$ and $C$ has the same determinant as $A$ then $x^{\top}Cx/2=a$ will have a primitive solution?
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1$\begingroup$ No. Take $a=0$ and $A$ the zero matrix. Then you're asking if a quadratic form $q$ in $n$ variables over $\mathbf{Q}_p$ has a nontrivial solution to $q(x)=0$ when the coefficients of $q$ are all close to 0, but such closeness to zero is irrelevant since anything can be scaled to satisfy that condition. And there are many nondegenerate quadratic forms in 2, 3, or 4 variables over a local field with no nontrivial rational zero. Please give the actual motivation so it can be determined if you meant to ask something else. $\endgroup$ – nfdc23 Oct 23 '17 at 21:13

$\begingroup$ The actual motivation is that this is required for an algorithm for computing integral quadratic forms from genus symbols. Thanks for pointing out I had some trivial counterexamples to remove. Anyway, it looks like Lemma 5.1 of Cassel's can be adapted to prove this. $\endgroup$ – Watson Ladd Oct 23 '17 at 21:43
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