# Consecutive primes versus prime twins

First a warm-up. Let $\ V\$ be an arbitrary set of odd natural numbers. Let $\ S(V)\$ be the generated multiplicative semi-group. What are the necessary and/or sufficient conditions on $\ V\$ for the property: $\ \exists_{x\ y\in S(V)}\ y-x=2\$?

Now real questions, all of them open to me. Let $\ \mathbb P\$ be the set of all primes. The following conjecture seems easier to proof than the twin primes conjecture(?):

Q1. Let real $\ a > 2\$ be arbitrary. Let $\ V := [a;\infty)\cap\mathbb P.\$ Then there exist $\ x\ y\in S(V)\$ such that $\ y-x=2$.

Statement Q1 would hold if there were infinitely many twin primes. Can we obtain a similar result without any help from twins? I.e.

Q2. Let $\ p-2 < p < q < q+2\$ be four primes. Let $\ V := [p;q]\cap\mathbb P.\$ Is it true that for all such $\ p\$ but a finite number, there exist $\ x\ y\in S(V)\$ such that $\ y-x=2\$?

A similar question can be asked when there only finitely many twin primes. Actually, we already have it. It is enough to replace $\ a\$ of Q1 by the largest twin prime.

Finally, these questions can be posed in terms of numerical estimates:

Q3. For every natural $\ a>2\$ compute or estimate the smallest natural $\ b:=b(a)\$ such that there exist $\ x\ y\in S(V)\$ such that $\ y-x=2\$ for $\ V := [a;b]\cap S(\mathbb P)$.

Of course $\ b\$ must be prime.

The above topic is a bit related to my earlier question: Prime residua races and two views on primes.

## 2 Answers

As far as $Q2$ , we don't know if there are infinitely many twin primes but expect that there are infinitely many prime quadruples -- $p-2,p,q,q+2$ all prime with $q=p+4.$ Then you are looking at $V=\{{p,q\}}$ and it seems unlikely that even one such pair $p,q=p,p+4$ (let alone all but finitely many) would allow a case of $|p^i-q^j|=2.$

• Indeed, I forgot that under the Schinzel umbrella conjecture a lot of more detailed conjectures (known earlier) would go against some of my attempts. – Włodzimierz Holsztyński Jun 20 '14 at 9:01
• It might be noted that 5 along with the primes in the a.p. 10k + 1 do not form a V such that S(V) contains two such numbers x and x+2. Similarly V must also not be a subset of an a.p. of the form 1 + ak for a sufficiently large. – The Masked Avenger Jun 20 '14 at 9:50

For Q1, take $x,y=A\pm1$, where $A$ is the product of all primes below $a+1$. This also gives a (likely badly suboptimal) exponential bound on $b$ for Q3.