Let $P = $ the prime numbers in $\Bbb{N}$. Define $P^i = \{ \pm q_1 \cdots q_i : q_j \in P\}$. These sets $\{ P^j\}_{j\geq 0}$ with $P^0 := \{\pm 1\}, P^{-1} := \{0\}$ form a basis for a topological space on $\Bbb{Z}$ turning it into a topological monoid.
The basis sets $P^i$ are both open and closed since each equals $\Bbb{Z} \setminus (\biguplus\limits_{i\neq j=-1}^{\infty} P^j)$.
Now let the open sets be unions of sets of the form $U_a = \{ a n^2 - 1 : n \in \Bbb{Z}\}$. If $a' = b^2 a$ then $U_{a'} \subset U_{a}$. Conversely if every $a'n^2 - 1$ is of the form $am^2 - 1$ then for $n = 1$ we have $a' = am^2$.
Thus I have proved that $U_b \subset U_a$ if and only if $b = am^2$ for some $m \in \Bbb{Z}$. I don't know if that helps, but my question is:
Can we write each $U_a$ as a closed set as well?
(Note. Thanks to commenter we can only ask the question on the space $\Bbb{Z}\setminus \{-1\}$. In other words consider the subspace topology on $X = \Bbb{Z}\setminus \{-1\}$ generated by $\{U_a \cap X\}_{a\in \Bbb{Z}}$.)
This is in the spirit of Furstenberg's use of the evenly spaced integer topology.
Since the $P^k, k \geq -1$ and the $U_a, a \in \Bbb{Z}$ are both bases for a topology on $\Bbb{Z}$ then the intersections of these basis elements, namely the collection of all $\{P^k \cap U_a:k \geq -1, a \in \Bbb{Z}\}$ is also a basis.