It is an open and interesting problem to determine whether there exist two infinite sets of integers such that , where is the set of primes. Let’s call this the prime sumset conjecture.

Observe that a set contains an infinite sumset if, and only if, there exists an infinite set such that is infinite.

We will examine the connections between the Hardy-Littlewood prime-tuple conjecture (HL), the theorems on bounded gaps between primes and the prime-sumset conjecture.

**HL and an easy reformulation**

HL may easily be seen to be equivalent to the following statement: for any finite set that does not cover all the classes modulo any prime, i.e. such that for every , there exists an infinite set such that . The condition on is obviously necessary for the conclusion to hold. However HL does not directly imply anything for infinite set of shifts .

Conversely, if the primes contain a sumset with infinite, then for any finite , HL is satisfied for the set of shifts — but not necessarily for every admissible set.

A stronger form of the HL conjecture, sometimes attributed to Dickson, is the following: let and be an integer tuple such that for each and for any prime , there exists such that . Then there exists arbitrarily large such that is prime for every .

**HL implies prime-sumset**

Granville proved that the Hardy-Littlewood conjecture implies the prime-sumset conjecture. Since his proof is more general and complicated, we explain here simply why this is so.

We construct successively integers and such that the sets and satisfy .

Start with and for instance. We can then add and . We note that and works.In general, the property we seek to maintain is that does not cover the wholeset of congruence classes for any (for, otherwise there is no risk) and neither does .

One systematic way to achieve this is to always seek in the congruence class and in the class , starting from and . Assume we have already constructed and with this property.

We seek in the form for some and . Then we must ensure that is a prime for every .

Note that is a prime greater than for every by the induction hypothesis, whereas is a product of primes smaller than , so for any . Now for the set equals the set since . Thus we can find such that is included in the primes. Let . We then define analogously.

**Bounded gaps as an ersatz for HL**

Although HL remains clearly out of reach, the breakthroughs of Zhang and then Maynard imply that HL holds for arbitrarily large sets of shifts. Indeed, Maynard’s theorem implies not only that where is the increasing sequence of prime numbers, but also that for any . That is, there exists such that for infinitely many integers , the interval contains at least primes. As a result, by the pigeonhole principle, there exists a set of cardinality such that for some infinite set . This argument does not seem to extend to infinite sets . One would need an infinite increasing sequence of sets such that the set of such that is infinite.

But even that would not be enough. Indeed, consider a set such as

where . This clearly has unbounded gaps, and for any there exists an infinite set such that is included in .

One may see that this set does not contain an infinite sumset though. First note that for any integer which is not a difference between two distinct powers of two, the set is finite. Further, if is the difference between two powers of two, it may be realised in a unique way as . Therefore, . Now take . Suppose . We may suppose that each is a difference between two powers of two, in fact , otherwise is empty. We have . Therefore is empty.

This implies that does not contain an infinite sumset.

**A simple proof that any thick set contains an infinite sumset**

It has been proven that any set of positive Banach density contains an infinite sumset. The proof, even in the simplified version of Host, is far from elementary. It may have been noted before, but I found a simple proof in the case of sets of density 1, also called thick sets. Note that such a set necessarily contains arbitrarily long intervals, sets of the form for infinitely many (say all) . Now we construct inductively and such that is inside . First, set and . Then set and . Then note that is inside whenever . As a result, we infer that is in for all .

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