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We give a simple example:

$$3\cdot5 + (3+5) = 23$$ $$3\cdot7 + (3+7) = 31$$ $$3\cdot11 + (3+11) = 47$$

The expression also works if we use $pN + (p+N)$ with $N=rs$, $r,s$ primes.

Is there a simple explanation?

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    $\begingroup$ There are a lot of small primes so I wouldn't read much into small examples. Try it for large $p,q$. $\endgroup$
    – lulu
    Jan 14, 2020 at 22:09
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    $\begingroup$ The primary reason: $pq+p+q = (p+1)(q+1)-1$, so it can't be divisible by any of the factors of $p+1$ or $q+1$. This leaves a very short list of possible divisors, especially on small numbers like these. $\endgroup$ Jan 14, 2020 at 22:10
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    $\begingroup$ here for instance...I let $p$ be the $n^{th}$ prime and $q$ the $(n+3)^{rd}$ for $n$ from $101$ to $200$. Hit a few primes, but hardly anything special. You can vary the indices as you like. $\endgroup$
    – lulu
    Jan 14, 2020 at 22:11

3 Answers 3

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This is similar to Conjecture 7.1 of this paper by Carl Pomerance and Simon Rubinstein Salzedo.

Their conjecture 7.1 follows from the statement that, for infinitely many primes $p$, there exists a prime $q>p$ such that $pq-p-q$ is also prime. This statement follows from Dickson's prime $k$-tuples conjecture. For more explanation, for fixed $p$, the two linear polynomials $x$ and $(p-1)x-p$, can be simultaneously prime for infinitely many $x$.

But, Dickson's prime $k$-tuples conjecture is open.

For this problem, fix an odd prime $p$, consider the two linear polynomials $x$ and $(p+1)x+p$. From Dickson's conjecture, these are both primes for infinitely many $x$.

Combining with Hardy Littlewood conjecture (also open), we have the following conjectural formula for a fixed odd prime $p$. $$ \#\{q<X \ | \ q, (p+1)q+p \textrm{ are both primes}\} \gg \frac{X}{\log^2 X}. $$

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Well, if $p$ and $q$ are odd primes it's not divisible by $2$ or $p$ or $q$. For any other prime $r$, if $q\equiv -1$ mod $r$ it won't be divisible by $r$, while otherwise it's divisible by $r$ only if $p \equiv -q (1+q)^{-1} \mod r$, which for "random" prime $p$ has probability $1/(p-1)$.

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Part of the reason, is only one combination type or residues mod 6, that represent at least 2 primes, give back numbers that aren't in prime residues. The exceptions, are when both $p,q$ are both 1,2,3 mod 6. I included 2 and 3 in my checking.

Small examples are only great as counter examples of something holding generally. $13\times 7+(13+7)= 111=3\times 37$ for example.

of the 10 distinct cases, 7 produce prime residues that just don't produce one prime in the integers, 6 of which produce the same residue ( 5 mod 6 is higher).

Your examples are also kind of low just checking p,q both 5 mod 6 up to 521, shows only about 31% of combinations actually produce primes.

Similar results with the 1,5 combination. about 41% in the p=3 case.

and finally roughly 47% in the p=2 case.

Because of weighting that means ... under 18% of possibly prime cases are primes.(894/5092)

and 9604 cases total if you include those that can't work. which brings us under 10%

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  • $\begingroup$ Thanks, I wish we could accept 2 answers so yours could also have been accepted. $\endgroup$
    – user25406
    Jan 15, 2020 at 11:22

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