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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