# Why does the following simple expression $p\cdot q +(p+q)$ generate so many primes?

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?

• There are a lot of small primes so I wouldn't read much into small examples. Try it for large $p,q$.
– lulu
Jan 14, 2020 at 22:09
• 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. Jan 14, 2020 at 22:10
• 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.
– lulu
Jan 14, 2020 at 22:11

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

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 $$1/(p-1)$$.

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%

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