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.

  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Jun 23 at 21:19
  • $\begingroup$ @GerhardPaseman That statement is certainly conjectured, but probably has about the same difficulty as the twin primes conjecture. $\endgroup$ – Will Sawin Jun 23 at 21:48
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    $\begingroup$ There is an extensive literature on "shifted primes". If you search on that phrase, you may find something useful. $\endgroup$ – Gerry Myerson Jun 23 at 22:02
  • $\begingroup$ @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. $\endgroup$ – Toffee Jun 26 at 17:46
  • $\begingroup$ @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. $\endgroup$ – Toffee Jun 26 at 17:46

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