I believe "Exercise 6.3:4(ii)" would be an appropriate reference for the claim about the joint closure stabilizing at the $\omega$th stage. [Writing the semigroup operation multiplicatively, take the set $G$ in that exercise to consist of all pairs $(\{x,y\},xy)$ (forcing closure under being a semigroup) together with all pairs $(\{x\},d)$ whenever $d$ is a divisor of $x$ (forcing closure under divisors).]

For a reference, you might try Bergman's introduction to general algebra. There is a lot on closure operators...don't have time right now to look through it all...math.berkeley.edu/~gbergman/245/3.2.pdf

It is not true that $\bigcap_{Y\in C(X)}Y = \bigcap_{X\subseteq Y\subseteq S}\Gamma(Y)$ because $\Gamma$ itself might not be a closure operator. (The composition of closure operators is not necessarily a closure operator.) The set on the left is $\Gamma$-closed (and hence both $\Pi$- and $\Delta$-closed), and hence equals $\bigcup_{n}\Gamma^{\circ n}(X)$ [since this is $\Gamma$-closed by the argument I gave, and clearly minimal].

I don't know a reference off the top of my head for that last fact, The proof is super easy. Let $T=\bigcup_n \Gamma^{\circ n}(S)$. It suffices to show closure under $\Pi$ (as the argument is similar for $\Delta$). Let $x\in \Pi(T)$. Then $x\in \Pi(T_0)$ for some finite subset $T_0\subseteq T$. Thus, $T_0\subseteq \Gamma^{\circ m}(S)$ for a sufficiently large $m$. Then $x\in \Pi(T_0)\subseteq \Gamma^{\circ m+1}(S)\subseteq T$.

In general, it might take longer than an $\omega$-length union to achieve stability. But your operators are special. A closure operator $C$ is called finitary when $x\in C(S)$ exactly when $x\in C(S_0)$ for some finite subset $S_0\subseteq S$. So, when talking about algebras with operations of finite arity, the "subalgebra generated by" operator is a finitary closure operator. Clearly, the "set of divisors" operator is also finitary. Now, a general fact is that if $\Pi$ and $\Delta$ are finitary, and $\Gamma=\Pi\circ \Delta$, then joint closure happens at the union you wrote.

There are actually two closure operators in play. The "subgroup generated by" operator, and the "set of divisors" operator. I'll leave it to you to check that both of these operators satisfy the usual three axioms of a closure operator en.wikipedia.org/wiki/Closure_operator They are also both finitary. Now, the joint closure under these two operators is just the alternating application of one operator after the other (your iteration) [or the intersection you described]. It only takes an $\omega$-chain of alternations because of the finitariness.

Note an answer, just a comment: Whenever there is a closure operator in play, you have a top-down construction (the intersection you mentioned) and a bottom-up construction (closing, then closing again, etc...). So, if you are satisfied with showing that there is a natural closure operator, you could just refer to the literature on closure operators for the "iterative description" you want. (In your case it stabilizes in countable many steps because the operator is finitary.)

The Mathematica function "IrreduciblePolynomialQ" says that the answer is yes for $1\leq n\leq 300$ after about three minutes. Another six minutes or so get it up to $n\leq 400$. I'm sure this could be pushed further, but this is already quite convincing.

This question seems to be very difficult. From the existence of spoofs where the power on the spoof special prime can be nearly arbitrary, we know that there is no purely combinatorial restriction on $k$. Some congruence restrictions (as you mention) can be determined on the cofactor $m$, but not enough to rule out the possibility that $p=k$. An answer could only be found using primality in some essential way.

@darijgrinberg Journal ToCs are one tool I use, which I've found to be much more useful than daily arXiv checking (which I also do). But, honestly, to get up-to-date on a problem, it is much better to simply use MathSciNet, look through citations and the literature, talk to others who publish in the area, etc... [There are, of course, some significant counter-examples to my general claim, which is why I waste so much time checking the arXiv.]

1. My field is friendly, and I can send papers to people most likely to be interested. 2. We have regular conferences, where I can talk about my research and get feedback. 3. I have almost never received feedback from my arXiv submissions. 4. I don't like some aspects of the arXiv. 5. I post preprints on my own webpage, under my own control.

@SamHopkins It is absolutely subfield dependent. Number theory got about 75 submissions this past week, while combinatorics got 110. By the way, some researchers in my area do put all their papers on the arXiv, others never post preprints, and others (like me) put some on and others not. Speaking just for me, there are a number of reasons I don't post all my papers.

@SamHopkins In the past week, there were 26 submissions under the arXiv heading "rings and algebra". Most of these are not directly in my specialized research area (noncommutative ring theory). Every month, the single journal "Communications in Algebra" publishes around 30-35 articles. A much larger percentage of these are in my area. And this is just a single journal. The total output in good journals far exceeds the arXiv (in both amount, and in quality---depending on the journal).