Can anyone either direct me to an 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 $Q(\xi) = x_0^2+x_0x_3+2x^2_1 + x_1x_3+13x^2_2+2x^3_2$.
Now let $q$ 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 $\{A = (a_0,a_1,a_2,a_3); Q(A) = 2^k$ 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].
Many Thanks!