How many Primes divid two coprime numbers the same way?

Question

For natural coprime numbers A, B HOW MANY prime numbers P, Q, R, ... are there that:

A+B ≡ 0 (mod P)
And
A+B ≡ 0 (mod Q)
And
A+B ≡ 0 (mod R)

Is there any upper limit or lower limit for those prime numbers?

Answer 1

The lower limit to the number of prime divisors of the sum of two arbitrary co-prime natural numbers is one: $A+B$ could be prime. If $P, Q, R$ exist, it might be the case that $A+B = PQR$. Then there is only one solution set (of three primes) for the chosen $A, B$.

An upper limit would be the number of divisors of the next larger primorial number. For each distinct prime $P_i$ dividing $(A+B)$, $(A+B)\geq\prod P_i$.

For example, if $A+B < 11\#$ and there are a maximal number of solutions, then $P,Q,R$ must come from the set of primes smaller than $11$: $\{2, 3, 5, 7\}$.

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Answer to original question: The Goldbach Conjecture offers infinitely many counterexamples:

$$\forall \text{ even }n\in \mathbb{N}\text{, } \exists \text{ primes } p_1,p_2 \text{ s.t. } p_1+p_2 = n$$

Now let $P=2$. By Goldbach, there exist prime $A, B$ such that $A+B = PQR$.

\begin{align} P && Q && R && A && B\\ 2 && 3 && 5 && 7 && 23\\ && && && 11&& 19\\ && && && 13&& 17\\ 2 && 3 && 7 && 5 && 37\\ && && && 11 &&31\\ && && && 13 &&29\\ && && && \color{red}{\textbf{17}} &&\color{red}{\textbf{25}}\\ && && && 19 &&23\\ && &&\vdots\\ \end{align}

Interestingly, note that by setting $P,Q,R$ first and choosing any coprime $A$, then $B= PQR-A$ must be coprime to all four of $P, Q, R, A$