The Kirch topology is the same as the prime integer topology These topologies are defined on the set $X=\{1,2,...\}$ of positive integers and provide interesting examples of a countable space that is connected, locally connected and Hausdorff.  Notation: $\mathbb{N}=\{0,1,...\}$. The topologies are defined in terms of suitable (one-sided) arithmetic progressions
$$a+\alpha\mathbb{N} = \{a,a+\alpha,a+2\alpha,...\}.$$
The Kirch topology (Amer. Math. Monthly vol. 76 (1969), pp. 169-171) is defined by taking as a subbase all sets $a+p\mathbb{N}$ with $a>0$, $p$ prime and $p>a$.  (So for example $1+2\mathbb{N}$ and $10+17\mathbb{N}$ are in the subbase, but $10+3\mathbb{N}$ is not.)
On the other hand, Steen & Seebach, Counterexamples in Topology, example #61 defines the prime integer topology in the same way, except for the condition $p>a$.  Their topology admits as a subbase all sets of the form $a+p\mathbb{N}$ with $a>0$, $p$ prime, and $p$ not dividing $a$.  Equivalenty, a base for the topology is given by all sets of the form $a+\alpha\mathbb{N}$ with $a>0$, $\alpha>0$ and squarefree, and $a$ and $\alpha$ relatively prime.
Clearly, every set in the subbase for the Kirch topology is open in the prime integer topology, so the prime integer topology is a finer topology.  Are the two topologies in fact the same?

(Added Jun 6, 2022) I realized I did not properly reflect the definition from Steen & Seebach.  In their notation, they define the prime integer topology as being generated by the subbase consisting of all sets of the form
$$U_p(b)=(b+p\mathbb{Z})\cap X,$$
with $p$ prime and not dividing $b$.  That is, the subbasic open sets are arithmetic progressions extending as far as possible in both directions (while remaining a subset of $X$). If we take $b'\equiv b\pmod p$ with $0<b'<p$, then $U_p(b)=U_p(b')=b'+p\mathbb{N}$ with $b'<p$.  So this is the same as a subbasic open set in the Kirch article.
The question whether the topologies generated by the $a+p\mathbb{N}$
described at the top with and without the condition $a<p$ are the same or not is still valid, and the answer is they are the same.
For example, the article The Kirch space is topologically rigid by Banakh, Stelmakh, Turek takes the definition without the condition $a<p$.
Similarly the closely related Golomb topology on $X$ admits as a base the collection of all sets of the form $a+\alpha\mathbb{N}$ with $a>0$, $\alpha>0$, and $a$ and $\alpha$ relatively prime.  The subcollection of such sets with the extra condition $a<\alpha$ also forms a base for the topology.
 A: Yes, the two topologies are the same.
Let $\tau$ denote the Kirch topology.  It's enough to show that any set $a+p\mathbb{N}$ (with $p\nmid a$) in the subbase for the prime integer topology is open in $\tau$. Take an arbitrary element $a+pn\in a+p\mathbb{N}$.  Let $a'$ be the residue of $a$ modulo $p$, so $0<a'<p$ and $a\equiv a' \pmod{p}$.  Then $a'+p\mathbb{N}$ is open in $\tau$ by definition.  Choose a prime $q$ larger than $p$ and $a+pn$.  The set $(a+pn)+q\mathbb{N}$ is open in $\tau$.  So the intersection
$$((a+pn)+q\mathbb{N})\cap(a'+p\mathbb{N})=(a+pn)+pq\mathbb{N}$$
is open in $\tau$.  That's an open nbhd of $a+pn$ contained in $a+p\mathbb{N}$.  Since $a+pn$ was arbitrary, $a+p\mathbb{N}$ is open in $\tau$.

Here is a more "topological" proof.  First of all, the Kirch topology $\tau$ is Haursdorff (if $a\ne b$ and $p$ is a prime larger than $a$ and $b$, $a+p\mathbb{N}$ and $b+p\mathbb{N}$ are disjoint nbhds of $a$ and $b$). So any point is closed and any finite set is closed.  Now given a set $a+p\mathbb{N}$ (with $p\nmid a$) in the subbase for the prime integer topology, let $a'$ be the residue of $a$ modulo $p$.  The set $a'+p\mathbb{N}$ is open in $\tau$.  Its subset $a+p\mathbb{N}$ differs from $a'+p\mathbb{N}$ by a finite set, so it is also open in $\tau$.
