# Strong Approximation for sol'ns to quadratic diophantine equations

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!

• Your "quadratic form" has an $2x_2^3$ term. Probably it should be $2x_2^2$? May 12, 2019 at 20:17
• @JoeSilverman good catch! Yes that was a typo. I meant $2x^2_3$ [instead of $2x^3_2$ ]. I just fixed that
– Mike
May 12, 2019 at 20:19

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.

• Thank you @GH from MO! Give me some time to digest this and then I will accept your answer.
– Mike
May 13, 2019 at 22:23
• @Mike: You are welcome! If you can use it in your paper, please thank me there as "GH from MO". May 13, 2019 at 22:24
• 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....
– Mike
May 13, 2019 at 22:29
• ...if that is possible to do that via DM here....otherwise my email is michael_capalbo AT hotmail
– Mike
May 13, 2019 at 22:30
• @Mike: No need to send me the paper, just share the link when it becomes public (e.g. on arXiv). Glad I could help! May 13, 2019 at 22:32