This is not a complete answer to all of your questions. This is to show you some things you need to investigate. The first question is answered. The second question has an example. I do not know complete answers to the third and fourth questions, but I give a try on explaining your plot of $m=61$.
From your last sentences, it looks like you are interested in the case when $m$ is a prime. Let $m=p$ be an odd prime. Then consider $p\equiv 1$ mod $4$, and $p\equiv 3$ mod $4$.
In the former case $p\equiv 1$ mod $4$, we see the symmetric black dots. This is because the Legendre symbol at $-1$ is $1$. That is
$$
\left( \frac{-1}p \right)=1.
$$
This means $-1$ is a square of something in $\mathbb{Z}/p\mathbb{Z}$. Suppose $x\equiv y^2$ mod $p$, then we have $-x \equiv z^2$ mod $p$ for some $z\in\mathbb{Z}/p\mathbb{Z}$.
Your example $m=61$ is a prime that is $1$ mod $4$. Thus, we have a symmetric black dots.
In general when $p$ is an odd prime, the image of the square mapping is
$$\{ x^2 \ \mathrm{mod} \ p| 0\leq x \leq \frac{p-1}2 \}.$$
Note that the black dots represent image of the square mapping.
Thus, the number of black dots is $\frac{p+1}2$. In your example of $m=61$, we have $31$ black dots.
Now we use a primitive root $g$ in $\mathbb{Z}/p\mathbb{Z}$. Then any element $x\in \mathbb{Z}/p\mathbb{Z} - \{0\}$, we have some integer $a$ such that $x\equiv g^a$ mod $p$. Then a cycle formed by square mapping which includes $x$ can be written as
$$
\{g^{a\cdot 2^k} \ \mathrm{mod} \ p| k=0, 1, 2, \ldots \}.
$$
To see if we have cycles, try solving
$$
a\cdot 2^k \equiv a \ \mathrm{mod} \ p-1.
$$
In your plot of $m=61$, we have a primitive root $g=10$ and the following are cycles of length greater than $1$. All of these should be considered with modulo $61$.
$$
(g^{20} g^{40}),
$$
$$
(g^4 g^8 g^{16} g^{32}),
$$
$$
(g^{12} g^{24} g^{48} g^{36}),
$$
$$
(g^{56} g^{52} g^{44} g^{28})
$$
I am not sure if you consider these as cycles, because there can be numbers in front of these, such as
$$
g^5 \mapsto g^{10} \mapsto g^{20},
$$
and comes in to the cycle $(g^{20} g^{40})$.
Best Answer
In the procedures I use A is the complete set of rational numbers.
$A^n$ is then rational. $(1-A^n)$ are also rational, but how to demonstrate that the roots are irrational, I.e., that no A expands to $(1-A^n)$.
The procedure of subtraction produces for one set +x, +y, for the other +x, -y.
The difference in orientation indicates that the two sets have no overlap, that A does not expand to the reflective set. That is an important conclusion.