Factorizations as a product of primes minus one

Let $$x$$ be a positive rational number. I am interested in factorizing $$x$$ as a product of primes minus one. In fact, I would also like make sure the primes in the decomposition are distinct, and I want to impose congruence conditions on the primes (the application is too far afield to describe). Let me break down my main question (Q3) into some warm-ups. There are also some special cases following Q3.

Q1. Let $$x$$ be a positive rational number. Are there collections

$$P = \{p_1, \dots, p_m\}$$ $$Q = \{q_1, \dots, q_n\}$$

of $$n+m$$ odd prime numbers with $$P \cap Q = \emptyset$$ such that

$$\begin{equation} \prod_{i = 1}^m (p_i - 1) = x \prod_{j = 1}^n (q_j - 1)? \tag{1} \end{equation}$$

We can call each such instance of (1) a factorization of $$x$$ into a product of primes minus one. If the answer is yes, can $$x$$ have infinitely many such factorizations?

Q2. Does Q1 still have a yes answer if we require the sets to be distinct primes, i.e., that $$P \cup Q$$ is $$m+n$$ distinct prime numbers? In other words, can we take each prime in the factorization to appear with multiplicity one? (Unlike the case of factorization into a product of primes, this is actually quite natural.)

My main question:

Q3. Does Q2 still have a yes answer if we put "legal" congruence constraints on the sets $$P$$ and $$Q$$? By "legal", I mean that the congruence conditions don't trivially make the equation impossible (e.g., a condition mod $$N$$ that makes the equation unsolvable modulo N or some divisor/multiple of $$N$$).

Example. Take $$x = 4$$ with the condition that the primes in $$P$$ are $$3$$ or $$5$$ mod $$8$$ and the primes in $$Q$$ are $$1$$ or $$7$$ mod $$8$$. Then $$(5-1) = 4$$, but also

$$(11-1)(13-1) = 4(31-1)$$

so $$P = \{11, 13\}$$, $$Q = \{31\}$$ works.

Moreover, suppose there is an integer $$k$$ so that $$13+16k$$ and $$31+40k$$ are simultaneously prime. Then $$P = \{11, 13+16k\}$$, $$Q = \{31+40k\}$$ also works. It follows from Dickson's conjecture that infinitely many such $$k$$ exist, but I believe this particular case is open. There are $$27768$$ such $$k$$ between zero and a million, $$211502$$ up to ten million, and $$1665924$$ up to a hundred million. It's pretty reasonable to guess this isn't a counterexample to Dickson's conjecture...

One could also perhaps motivate this question by recent work of Tao--Ziegler on polynomial patterns in the primes (and preceding results found in its references).

Some experimentation suggests a special case:

Q4. Can we always take $$P$$ and $$Q$$ to be of bounded cardinality to produce infinitely many distinct factorizations of $$x$$?

For example, all positive integers $$x$$ between $$1$$ and $$100$$ are of the form

$$(p-1)(q-1)=x(r-1)$$

for $$p,q,r$$ distinct.

Q5. If the answers to any of Q1-Q4 end up being no, can we characterize the positive rational numbers (or integers) $$x$$ for which the answer is yes?

Honestly, I would even be happy if $$x=4$$ admits infinitely many factorizations subject to the congruence conditions given in the example.

• I don't know if for any integers r and s there is an integer k such that both rk +1 and sk + 1 are prime. Given the Cebotarev density theorem I expect the answer to be yes with infinitely many positive integers k. If we focused on this statement alone , a yes answer would seem to take care of all your questions. Gerhard "Am I Right About This?" Paseman, 2019.06.23. – Gerhard Paseman Jun 23 at 21:19
• @GerhardPaseman That statement is certainly conjectured, but probably has about the same difficulty as the twin primes conjecture. – Will Sawin Jun 23 at 21:48
• There is an extensive literature on "shifted primes". If you search on that phrase, you may find something useful. – Gerry Myerson Jun 23 at 22:02
• @GerhardPaseman: Yes, that would certainly give it. As Will Sawin says, this is a (likely) hard conjecture. The pair of arithmetic progressions in my example is of a similar nature. – Toffee Jun 26 at 17:46
• @GerryMyerson: Thanks! I did find an interesting paper by Elliott about the subgroup of $\mathbb{Q}^*$ generated by quotients of the form $(p+1)/(q+1)$ for $p,q$ prime, which has some nice follow-up work, but nothing quite what I'm after. – Toffee Jun 26 at 17:46