Which residue classes of $\Bbb Z$ sit in the index two subgroups of the quotient of $\Bbb Z_2^\times$ by $\langle\overline3\rangle$?
Let $\Bbb Z_2$ be the ring of 2 adic integers
Let $\Bbb Z\subset\Bbb Z_2$ by the obvious inclusion
Let $\langle3\rangle=\{3^i:i\in\Bbb Z\}$
Let $\langle\overline3\rangle$ be the closure of $\langle3\rangle$ under the 2-adic metric.
Then I'm informed unless I misunderstand this comment that $\langle\overline3\rangle$ is an index two subgroup of $\Bbb Z_2^\times$.
Let $X\subseteq \{\{a,b\}:\forall y\in\Bbb Z,ay+b\in\Bbb Z_2^\times/\langle\overline3\rangle\}$ be some subset of the set of ordered pairs $a,b$ such that the residue classes $x\equiv b\pmod a$ cover the odd numbers $\Bbb 2\Bbb Z-1$.
Which residue classes of $\Bbb Z$ sit in each of the two cosets $X$ and $\not X$? Or if they can't be defined by residue classes, which integers defined by some other means.
Given $a\ge 3$, $$(1+2^a(1+2b))^2 \equiv 1+2^{a+1}(1+2b)+2^{2a}(1+2b)^2\equiv 1+2^{a+1}\bmod 2^{a+2}$$ thus by induction on $a$ there is a power of $1+8$ which is equal to $1+8x \bmod 2^a $, ie. $(1+8)^{\Bbb{Z}}$ is dense in $1+8 \Bbb{Z}_2$.
The closure of $(1+2)^{\Bbb{Z}}$ is $(1+2)^{\Bbb{Z}_2}$, it contains $(1+2)^{2\Bbb{Z}_2}=(1+8)^{\Bbb{Z}_2}=1+8 \Bbb{Z}_2$.
Whence $\Bbb{Z}_2^\times / (1+2)^{\Bbb{Z}_2}$ is a quotient of $\Bbb{Z}_2^\times/(1+8\Bbb{Z}_2) = \Bbb{Z/8Z}^\times$ and we obtain
$$\Bbb{Z}_2^\times = (1+2)^{\Bbb{Z}_2}\cup 5(1+2)^{\Bbb{Z}_2}$$
If $n \equiv 1$ or $3\bmod 8$ then $n\in (1+2)^{\Bbb{Z}_2}$, if $n\equiv 5$ or $7\bmod 8$ then $n\in 5(1+2)^{\Bbb{Z}_2}$.