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.