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Can anyone either direct me to an relatively elementary proof in the literature--or show me why this (Conjecture 1 stated below) is true--or if I am mistaken and it is not true:

  1. For any 4-tuple $\xi = (x_0,x_1,x_2,x_3) \in \mathbb{Z}^4$, let us define the quadratic form $Q(\xi) \doteq x_0^2+x_0x_3+2x^2_1 + x_1x_3+13x^2_2+2x^2_3$.

  2. Now let $q \in \mathbb{N}$ be a prime power such that $(26,q)=1$, and let $\cal{B}$ be the set $\{B=(b_0,b_1,b_2,b_3) \in \mathbb{Z}^4$; $Q(B) \equiv_q 1\}$.

  3. Now let $\cal{A}$ be the set $\{A = (a_0,a_1,a_2,a_3) \in \mathbb{Z}^4; Q(A) = 2^k$ for some nonnegative integer $k\}$.

Conjecture 1: Then using the notation as above, then for any $B =(b_0,b_1,b_2,b_3) \in \cal{B}$ there is an $A =(a_0,a_1,a_2,a_3) \in \cal{A}$ that satisfies $a_i \equiv_q b_i$ for each $i \in \{0,1,2,3\}$.

I do need this result for a paper that I am writing. But I must admit to not understanding the Strong Approximation Theorem well at all, nor the associated math around this--which is what I think this looks like. [I am in graph theory].

Now I believe that if the above holds for all $q$ as prescribed in 2 then it holds for all integers $q$ such that $(q,26) = 1$.

If it would further help things let $q$ as in 2. be a power of 3, at least to start with.

Many Thanks!

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    $\begingroup$ Your "quadratic form" has an $2x_2^3$ term. Probably it should be $2x_2^2$? $\endgroup$ – Joe Silverman May 12 '19 at 20:17
  • $\begingroup$ @JoeSilverman good catch! Yes that was a typo. I meant $2x^2_3$ [instead of $2x^3_2$ ]. I just fixed that $\endgroup$ – Mike May 12 '19 at 20:19
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The conjecture follows from Theorem 2.1 in Hsia-Jöchner: Almost strong approximations for definite quadratic spaces, Inventiones 129 (1997), 471-487. The paper is available here.

The details of this implication are nontrivial. Let $q=\ell^s$ be a prime power (so $\ell$ is the underlying prime), and let $B\in\mathcal{B}$. Let $k$ be a large positive integer divisible by $\varphi(q)$. Then the $2$-power $\alpha:=2^k$ is large, and $Q(B)\equiv 1\equiv\alpha\pmod{q}$. By a standard lifting argument, there exists a primitive vector $v_\ell\in B+q\mathbb{Z}_\ell^4$ such that $Q(v_\ell)=\alpha$. Let us also define $v_2\in\mathbb{Z}_2^4$ and $v_{13}\in\mathbb{Z}_{13}^4$ as the integer vector $(2^{k/2},0,0,0)\in\mathbb{Z}^4$. We observe that $Q(v_2)=Q(v_{13})=\alpha$, and $v_{13}\in\mathbb{Z}_{13}^4$ is primitive, but $v_2\in\mathbb{Z}_2^4$ is not. Now we apply the mentioned theorem of Hsia-Jöchner for the primes $T:=\{2,13,\ell\}$ and the above data. The conditions of the theorem are satisfied, because $Q$ is positive definite and isotropic over $\mathbb{Q}_2$ (cf. Section IV.2 in Serre: A course in arithmetic). We conclude that there exists $A\in\mathbb{Z}^4$ such that $Q(A)=\alpha$ and $A\equiv v_p\pmod{p^s}$ for all $p\in T$. In particular, $A\in\mathcal{A}$ and $A\equiv B\pmod{q}$. We are done.

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  • $\begingroup$ Thank you @GH from MO! Give me some time to digest this and then I will accept your answer. $\endgroup$ – Mike May 13 '19 at 22:23
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    $\begingroup$ @Mike: You are welcome! If you can use it in your paper, please thank me there as "GH from MO". $\endgroup$ – GH from MO May 13 '19 at 22:24
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    $\begingroup$ I will acknowledge you, I will send you a copy of the paper first and you can decide how you would like to be acknowledged.... $\endgroup$ – Mike May 13 '19 at 22:29
  • $\begingroup$ ...if that is possible to do that via DM here....otherwise my email is michael_capalbo AT hotmail $\endgroup$ – Mike May 13 '19 at 22:30
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    $\begingroup$ @Mike: No need to send me the paper, just share the link when it becomes public (e.g. on arXiv). Glad I could help! $\endgroup$ – GH from MO May 13 '19 at 22:32

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