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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.

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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 $|p^i-q^j|=2.$

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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.

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Simple and nice. Thank you Emil. – Włodzimierz Holsztyński Jun 21 '14 at 6:50

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