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:

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

Now let $q \in \mathbb{N}$ be a prime

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