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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
.
Let have at least
elements, for some
(we say that
has density at least
), and very large
.
We know, by Meshulam’s theorem (Roth’s theorem in vector spaces), and even more strongly by Ellenberg-Gijswijt, who used the Croot-Lev-Pach method, that must contain a 3-term arithmetic progression, in other words an affine line.
But does it have to contain planes or even affine subspaces of larger dimensions?
First we examine the case of a random set , thus each
is taken in
with probability
independently of each other.
This implies that any given subspace has probability
to be included in
.
Now how many affine subspaces of dimension are there in a space of dimension
? This is
where
is the number of linear subspaces of dimension
, and is easily seen to equal
One gets the obvious bound
Thus the expected number of subspaces of dimension in
is at most
.
This is of the order of magnitude of a constant
when is of the order of
.
So when , there may be no single subspace of dimension
.
However, using
the lower bound
may be used to show similarly that when
the set is likely (in fact, almost sure) to contain subspaces of dimension
.
Now we prove that if a set has positive density and is large enough, then
must contain a subspace of dimension
.
The key is basically to use Varnavides averaging argument along with the Ellenberg-Gijswijt bound.
Let be the minimum
(if it exists) such that any set
of density
must contain a subspace of dimension
. The Ellenberg-Gijswijt bound amounts to
; indeed,
just has to be at least
to ensure a line. Then Varnavides argument is an averaging trick that says that a dense set must contain many lines. So many that it must contain many parallel lines. So many that the set of starting points of these lines must in turn contain a line. In which case we obtain a plane. And so on. This was used by Brown and Buhler to derive the recursive relationship
where and
is the number of linear spaces of dimension
in a space of dimension
. We can bound
by
, which yields
.
And so
Inducting, one obtains
Thus . In other words
, which is exactly the claimed result.
It is well-known that a real number is a rational if, and only, if its decimal expansion is periodic (see Wikipedia). This result is a finiteness result, as it says the decimal expansion is, in a way, not really infinite, as it consists in the repeating infinitely a finite block.
A similar finiteness result exists for Laurent series , which are the analogues of reals modulo 1. Namely, associate to
the infinite matrix
. Then Hankel’s theorem states that
is rational if and only if the matrix is of finite rank.
We show here a quantitative version of this statement. Let
The rank of this matrix is related to Diophantine properties of .
To see this, we introduce for the fractional part
and the norm
, where
is the largest
such that
. Then we have the following result.
If the rank of is at most
, then there exist a decomposition
and a monic polynomial
of degree
such that
. The converse holds too.
The classical theory of Hankel matrices (see Chapter X, Paragraph 10 of Gantmacher’s book) tells us that if the rank of the matrix is , then there exist integers
and
such that
and the
first rows are linearly independent, each of the following
rows is a linear combination of them,
and the minor formed of the first and last
rows and columns is nonzero. Let
be the rows of
. Write
where
is the last coefficient of
and
the row of the first
coefficients. By the above, we have a relation
. In fact, we shall show that for any
, we have
This equation holds for . So we argue by induction and assume it holds for any
for some
and prove it for
. To start with, the displayed equation implies that
Applying the induction hypothesis iteratively, we find coefficients for
and
such that
and
These coefficients satisfy the initial condition and the recurrence relations
for
and
.
On the other hand, we know that there exist coefficients
such that
Comparing the last equation to the one involving , and using the linear independence of the first
lines, we find that
.
Thus . But
by definition of the coefficients
. We infer that
,
which conclude the inductive argument.
To see the connexion with Diophantine properties of , notice that
is the row of the first
coefficients of
. Thus the validity of the identity for
for all
implies that
where satisfies
.
That is, using the polynomial , we find that
which concludes the proof of the direct statement. The converse is straightforward.
Let be a vector space over a finite prime field. One knows then which subsets
satisfy
, the sets that are stable under addition: these are the kernels of linear maps, that is, each such set
is the zero set of a bunch of linear forms.
Now we consider a two-coordinates analogous problem, that is, we consider a set and define the sets
and , V and H meaning vertical and horizontal. What are the sets that satisfy both
and
? Call such a set a transverse set. Natural exemples are cartesian products of vector spaces (which we shall call rectangles) as well as zero sets of bilinear forms. More generally, a set of the form
for some subspaces
and some bilinear forms
is horizontally and vertically closed. Call such a set a bilinear set, and call its codimensions the codimensions of the subspaces and the number of bilinear forms required (the dimension of the space of bilinear forms vanishing on it). What is the relationship between transverse sets and bilinear sets? Are all transverse sets essentially bilinear sets?
For a transverse set and
, let
be the fiber above x. Thus
and
is a vector space. Actually
depends only the projective class
. Moreover, the stability under horizontal operations is equivalent to the property that if
is on the projective line spanned by
and
, we have
.
The codimension 1 case
Suppose and each subspace
has codimension at most 1. Write
with
some vector uniquely defined up to homothety. It is easy to see that the set of
such that
is a vector subspace which we call
. Furthermore, if
, we have
, so that
descends to an injective map on
. Thus it is enough to study the case where
has codimension exactly one unless
. In other words, we have a map
satisfying the property that if
is on the projective line spanned by
and
, then
lies on the projective line spanned by
and
. Thus
maps projective lines into lines. Actually it can be quickly checked that such a map is either bijective or constant on each single projective line.
If we suppose it is injective, too, then by the fundamental theorem of projective geometry, we conclude that it is a projective map. Thus it is induced by a linear map . Hence writing
, we have shown that
is the zero set of the bilinear form
.
In general is far from being injective. It can for instance be constant on
(equal to
for some hyperplane
). In this case
. This is neither a rectangle, nor the zero form of a quadratic form, because if it was,
could be taken linear, and because
, it should be injective. In fact, if it is the intersection of zero sets of a family of bilinear forms, there needs to be
bilinear forms in this family, such as the forms
if (wlog)
is the hyperplane
. However
obviously contains a very large rectangle.
These are actually the only cases: such a map is either injective or constant. Indeed, if
is just a line, it’s obvious, so suppose
. Suppose that
maps projective lines into lines (in the sense above) but is neither injective nor constant. That is, there exist two distinct points
such that
, and a third point
satisfying
. This implies that
span a projective plane. Take a point
on the line spanned by
. Because
is a bijection on both lines
, you can find
on
such that
. Now consider the intersection
. Then you have
, so that on the line
, the map
is neither constant nor injective, which is absurd.
Arbitrary codimension
A more general potential counterexample, communicated to me by Ben Green, is a set of the form where
are sequences of increasing (resp. decreasing) subspaces. Thus for
, we have
where
. Again you can make it a bilinear set, but you need atrociously many bilinear forms.
However it contains a large rectangle. In fact, while
so that the density of
is at most
. Thus
. Finally by the pigeonhole principle, one of the cartesian products
have density at least
.
In general, we have a vector space . Suppose for instance each
has codimension 2. If we manage to find maps
such that
for each
, we are in good shape. Though it is not clear how to show that such a consistent choice of bases can be made…
The conjecture
If a has cardinality at least
, it contains a bilinear set of (linear and bilinear) codimensions
.
Given a compact abelian group , endowed with a Haar probability measure
, let
be the set of (Borel) measurable functions
satisfying
. The average of
is called its density.
Let be system of linear forms, i.e
and
where for any
and
. For a measurable function
, one may define the
-count of
as
Here by a slight abuse of notation, the measure on
denotes in fact the product measure
of the Haar measure on
.
Let
When a system is fixed, we may omit it from the notation
. Given a sequence
of finite groups of cardinality tending to infinity, it is natural to consider the sequence
. When the infimum is taken only over
-valued measurable functions of density at least
, this is the minimum number of APs a set
of density at least
can have. But these infimum are actually the same (basically by convexity).
The case was introduced by Croot. He proved that the minimal density
of three-term arithmetic progressions converged as
through the primes. Sisask, in his PhD thesis, was able to express this limit as an integral over the corresponding limit object, namely the circle
, thus
As customary, the corresponding problem in finite field models is cleaner. Thus we set . Whereas Croot noticed that the sequence
was in some sense approximately decreasing, it is easy to see that for any system
, the sequence
is genuinely non-increasing. This is because
embeds naturally in
; thus if
is a map from
to
, it can be extended into
by putting
.
Now , and
. This proves that
, which implies that this sequence has a limit.
Let be now fixed. We want to find an appealing expression for the limit of the sequence
. It is thus tempting to look for a natural topological compact group that would play the role of limit object of the sequence
, just as
played the role of limit object for the sequence
.
The analogy leads one to introducing
and
. The group
equipped with the product topology is a compact group by Tychonoff’s theorem, and
equipped with the discrete topology is locally compact; it is the Pontryagin dual of
. It is now reasonable to ask whether
One easily notices that for any
by extending, as above, a function
into a function
on
with the same density and the same
-count. Indeed, one can define
and then is a measurable function which has same density and same
-count.
We prove the other inequality. For this we prove two intermediate propositions. We write from henceforth .
Proposition 1
If is a sequence of
-valued functions on
that converges in
to a
-valued function
, then
.
Proof
Let . We show that
converges in
to 0, which will conclude. In fact we prove a stronger convergence, a convergence in
. We show that
. For that we simply remark that
Now, any non-trivial linear map , i.e. any map of the form
with , preserves the Haar measure, which implies that
Using this, the triangle equality and the fact that the functions are 1-bounded, we conclude.
Proposition 2
Let be the set of functions
for
of density at least
Then
is dense in
Proof
Let . We construct a sequence of functions
that converge to
We use duality and Fourier transform. We have
and
For , write
Then write
This defines a function , which obviously satisfies
.
The same holds for its extension . Notice that
is also the conditional expectation with respect to the
-algebra
. More precisely,
as both equal .
This implies in particular that , and thus
, has its values in
.
To check that , we use the conditional expectation interpretation
Now we check the convergence. It follows from basic harmonic analysis: being in
, its Fourier series
converges to
in
. This concludes the proof of Proposition 2.
Thus even for 4-APs, this shows that
exists and equals
This is in contrast to a paper of Candela-Szegedy, for the same pattern (4-APs) but with
where they obtain as a limit a much more complicated infimum, an infimum over functions
on
of the
-count, where
is another pattern. The part that holds in the setting of
and not in the setting of
is Proposition 2. The truncated Fourier series of
, which is the natural
approximation, does not give a function on cyclic groups
with the same density and AP-count as
.
We remark that the paper of Hatami-Hatami-Hirst implies that the sequence of functions admits a limit object
which is defined on a finite cartesian power of
.
In fact, if is the
-APs, pattern of complexity
, and
, the limit object
is defined on
. In particular, if
, this means that the infimum above is attained, i.e.
for some measurable. When
, it is still not known whether such a result (i.e. the existence of a limit object
) holds. However, Szegedy proves the existence of a limit object
on a non-explicit group
.
A recent breakthrough due to Croot, Lev and Pach, recycled by Ellenberg-Gijswijt and streamlined by Tao on this blog post, revolutionised the landscape of Roth’s type problems in vector spaces over finite fields. Indeed, Ellenberg-Gijswijt obtained that a set free of triple satisfying
can have at most
elements out of the
out there. Here we show the same method can deal with sets free of solutions to various other equations in any
(given that if
, we have
as a vector space, there’s no point in considering general
) and we make explicit the dependence on
of the resulting bound. We refer in the sequel mostly to Tao’s symmetric proof instead of the original articles. One great merit of this exposition is to reveal, thanks to the extension of the notion of rank from matrices (seen as functions of two variables) to functions of more variables, how the scope of the method can be extended.
Other linear equations
In fact the bound found for the sets without 3-term arithmetic progressions is also valid
for a set having no non-trivial solutions to
where are three linear automorphisms of
summing to 0. Indeed,
we can then still write
Even more generally, for any integer , if
are
automorphisms of
which sum to 0, and if
does not have any non-trivial solution to the equation
, we can still write
The right-hand side is still a function on of rank
. Meanwhile, the left-hand side may still be expanded using the formula
valid for all
, which implies that the LHS equals
where means the
-th coordinate of
; it is
thus a linear polynomial. This implies that the whole expression is a polynomial of degree . The treatment shown by Tao on his blog to bound the rank of this expression can thus be applied here too.
Roth’s theorem is simply the case where and
for
, or more generally
and
.
The more variables, the smaller the bound becomes. For instance, a set without any solution to the equation
, in particular without any 3-APs, has size at most
, less than the bound for sets without 3-APS which is
.
Even non-linear equations
Upon scrutinizing the polynomial method at work in Ellenberg-Gijswijt, one can be struck by the fact that the translation-invariance of the equation is not even used. In Meshulam’s theorem, as in all form of density increment strategy, we crucially use the fact that if
is a solution and if
is any constant, then
is a solution too. What is used in the polynomial strategy is that any constant tuple
is a solution, which is vastly weaker (though it is synonymous for linear patterns). Thus if
are any bounded-degree polynomial maps
(i.e. maps whose coordinates are polynomials of degree at most
) summing to 0, then the method should, in principle, spit out a bound for the maximal size of a set
free of solutions to
, depending on the degree. The only thing that changes in Tao’s write-up is that the polynomial
whose rank we try to bound is now of degree up to . Thus the parameter of relevance and which determines whether the method will succeed or not is
. It has to be significantly less than
. Otherwise, the method requiring a good bound (exponential if possible) on
where are independent, identically distributed variable on
, we are hopeless.
Thus, considering , we need
variables.
Equations in a polynomial ring
As advocated by Tom Bloom (for instance in this paper), it might be of interest to ask for analogues of theorems in polynomial ring. However, a relation of the form among three polynomials over
of degree less than
is simply a relation among three vectors, the vectors of their coefficients. But we can form much more general equations, such as
Dependence on
Let’s return to the case of Roth’s theorem.
Following Tao, one gets a bound of the form
where is the maximum of
under the constraints
It could be interesting to know how evolves with
.
Using the help of a computer to compute for
upto
, I got the feeling that
. This would mean that a subset
free of 3-term arithmetic progressions necessarily satisfies
.
We now proceed to show that the empirically conjectured behaviour of is indeed the true one.
The constrained extrema theorem implies that the maximum of is attained on a point where
belongs to . This condition amounts to the existence of
so that
for all indices
. So we have only two degrees of freedom for the tuple
. And because we have two equations
there should be a unique such tuple.
Putting , we transform the first constraint in
thus the second one amounts to
Here the LHS equals
Thus the equation to solve is
in other words
It clearly imposes that as , we must have
, for if it remains bounded away from 1, the left-hand side is asymptotic to
, which is not asymptotic to
, the right-hand side.
We now rearrange the equation as
that is
Let us introduce and determine the behaviour of
, or equivalently of
. One can prove by contradiction that
cannot tend to 0. Indeed, supposing it did, one could write
Thus
and comparing the term in we see an absurdity.
It can’t either tend to . If it has a limit point
, it has to be a solution to
The computer declares that the solution to this equations equals 2.149125799907062 (which is very close, fortunately enough, to values observed for for large
). Thus
tends to
.
Injecting the expression for the in terms of
, the optimum is
which we can equate, remembering earlier formulae, to
.
So let us derive an asymptotic expression for and
. We quickly obtain
.
So the maximum we are after equals
This implies that
Injecting the value we found for , the computer spits out 0.84, as guessed.
Many properties of the set of the primes are inherited by any dense subset thereof, such as the existence of solutions to various linear equations. But there’s one property, recently proved, namely the existence of infinitely many gaps bounded by a constant, which is not inherited by dense subsets of the primes. For instance, it is possible to remove basically one half of the primes and be left with a subset of primes
such that
.
So instead one could ask if when coloring the primes in finitely many colours, one is guaranteed to find a colour class containing infinitely many pairs of primes whose difference is bounded by a common constant. This does not follow immediately from pigeonhole principle, which only says that one colour class contains infinitely many primes.
However, this follows easily if we recall that the work of Maynard does not only give infinitely many bounded gaps, that is , but also for any
that
. Let
be this liminf.
In other words there are infinitely many (disjoint) intervals of size which contains at least
primes. So if you colour the primes in
colours, and you consider these intervals, you find that each one must contains at least two primes of the same colour. By pigeonhole principle again, there must be a colour for which this occurs infinitely often. Whence the existence in this colour class of infinitely gaps of size at most
.
The squarefree numbers form a set which is well-known to have density in the integers, in other words their characteristic function
satisfy
. In this set all your dream theorems about the count of linear patterns are true and easy.
Hardy-Littlewood squarefree tuples
The Hardy-Littlewood conjecture consists, in its strongest form, in an asymptotic, for every , of the form
where is a constant possibly equal to 0. It is quite far from being proven, although interesting steps towards it have been made in the last ten years, by Goldston-Pintz-Yildirim and Maynard.
In the squarefree numbers, the analogous problem is rather easy and was settled by Mirsky in 1947. He proved that
where
and is the cardinality of the size of the projection of the set
in
, that is the number of forbidden classes modulo
for
.
I show a quick way of deriving it with a much worse error term . Fix
, a large number; ultimately,
should be ok. Fix
. Then the number
of
such that none of
have a square factor
with
is easy to compute: in
or any interval of length
there are by chinese remainder theorem this many of them
Thus in , there are this many of them
There are also people who are not squarefree, but escape our sieving by primes less than . These are the guys such that at least one of the
have a divisor
with
prime. Now there are are at most
multiples of
in
. So at most
should be removed. Let’s try to balance the error terms and
; take
.
This way both seem to be .
Green-Tao type theorem
We can also easily get asymptotics of the form
where is a convex body and
a system of affine-linear forms of which no two are affinely related. Moreover
is again the proportion of non-forbidden vectors
of residues modulo
. One way to do it would be to simply use the Hardy-Littlewood type asymptotics proved above coordinate by coordinate. We can also prove it quite easily, by first observing that the set of n where at least one of the
vanishes has size
because the 0-set of an affine-linear form is an affine hyperplane. Then as
, the divisors
will have to satisfy
. So now we restrict the set of
to the ones where the forms don’t vanish. Then we partition the sum into one where
is restricted to be at most
(we fix some small
), and the remaining one. Using the well-known fact that
, we remark that the remaining one equals
Now the number of in the sum is
. Given that
, we can bound this sum by
. We proceed similarly for
, so that the sum to estimate is
Quite gross volume packing arguments indicate that the number of integral points (lattice points) to estimate above is , where
is the density of zeroes modulo the . Moreover we can extend the sum beyond
at the cost of a mere
. Hence finally for any
we can write that
Indeed by multiplicativity it is easy to transform into a product over primes.
The quest for repeated, infinitely frequent patterns in the primes, is certainly a very old one, and is often even a quest for the asymptotic frequency of these patterns, which is much harder. For instance proving that there are infinitely many primes is easy, finding a satisfactory answer for the question « How many (roughly) until N large ? » much less so. Inbetween lies the obtention of upper and lower bounds of the right shape, due by Chebyshev.
Here pattern stands for images of a system of polynomial forms. Given and a convex set
of volume increasing to infinity, we are thus interested in evaluating
for N large, where is the von Mangoldt function. The case where
and
is the original question of Hardy and Littlewood, who proposed a tantalizing asymptotic behaviour but is still completely out of reach (even the question whether there are infinitely many
such that
are both primes is not settled). But the case where the system
is affine-linear (thus the polynomials are all of degree 1) and no two forms are affinely dependent was solved by Green and Tao in the celebrated article Linear equations in primes.
Similar results for more general polynomial forms are rare. We have to mention the famous work of Friedlander and Iwaniec yielding an asymptotic for the number of primes of the form , where it appears that
for some constant .
I have uploaded yesterday an article on the ArXiv which provides asymptotics of the same shape as the ones in the Hardy-Littlewood for a few exceptional polynomial patterns. Thus for instance, I can tell how many arithmetic progressions of three primes smaller than N exist whose common difference is a sum of two squares – well not quite, because I have to weigh these arithmetic progressions by the number of representations of the common difference. Now this weight, giving a positive density to the set of sums of two squares, which is sparse, of density , just as the von Mangoldt function is a weight (mostly) on primes giving them a density, cannot be easily eliminated afterwards, in contrast to the von Mangoldt function (one can write for
that
).
More precisely, the result that naturally comes out concerning three term arithmetic progressions with common difference a sum of two squares is
where is the representation function and
are some explicit constant which I don’t reproduce here. Moreover, we can generalise to other positive definite binary quadratic forms than this one, and there’s nothing special about length three: an asymptotic is available for any length. Here we notice that in some sense, the result is only seemingly polynomial, and truly rather linear: the polynomial nature of the pattern is enclosed in a linear input into the representation function of a quadratic form.
In fact, my article contains a more general result of which the one above is only a particular case. My work consisted in mingling the von Mangoldt function with the representation functions of quadratic forms, whose behaviour on linear systems have been already analysed respectively in by Green and Tao and Matthiesen. The idea is to consider sums of the form
where can be the von Mangoldt function or a representation function, and the $\psi_i$ are linear forms. The cohabitation of both types of functions went quite well. One delicate point was to eliminate biases modulo small primes of both types functions, an operation known as the
-trick. The difficulty is that while the value of the von Mangoldt function is more or less determined by the coprimality to small primes, it is not so for the representation function, which is also sensitive to the residue modulo large powers of small primes. Once this issue is adressed carefully, it is possible to majorize them by one and the same pseudorandom majorant, which opens the way to the application of the transference principle.
Similarly, the cohabitation between the von Mangoldt function and the divisor function is quite natural, yielding asymptotics for expressions such as . This is reminiscent of the Titchmarsh divisor problem, the evaluation of
or (almost equivalently) of
, but the latter expression involves a linear system of infinite complexity, and is thus altogether out of reach of my method, just as the twin primes or the basic Hardy-Littlewood conjecture.
One may hope to extend the generalised Hardy-Littlewood conjecture stated (and proven in the finite complexity case in the paper linked) by Green and Tao to polynomial systems. For example given a polynomial and a bounded domain
(probably a convex body would be more reasonable), one may be interested in an asymptotic for
We will rather try to get an asymptotic of the form
where is basically the number of points with positive integer coordinates in
, so hopefully
, and the local factors take into account the obstructions or the absence of obstructions to primality modulo
. Recall that
classically denotes the von Mangoldt function. There are only very few non-linear polynomials for which an asymptotic for the number of prime values is available. The easiest one is obviously
. Indeed, the primes which are sum of two squares are the primes congruent to 1 modulo 4 (and also the prime 2, but it’s a single prime, so we don’t have to care), and they are represented 8 times each as a value of
. So
Now let’s check what the conjecture would say. Here . What about the
? They are supposed to be
. Now it easy to check that
if
, and
otherwise. So
in the former case, and
in the latter case. Thus,
doesn’t converge absolutely, in contrast with the traditional Green-Tao situation, where
… However, if we imagine that to each prime
corresponds a prime
with
, we could compute the product of the
by grouping the factors into pairs
. A bit more precisely,
and
which implies that
so this product is convergent, although not absolutely. I guess the constant is
, see Mertens theorem, and that
. I don’t really know how to compute this convergent product. We quickly notice that
. If the product could be equal to
, which it can’t unfortunately, being smaller than 1, we would apparently get the correct result, but this remains a quite dodgy case, which should make one cautious about stating ambitious generalisations of the Hardy-Littlewood conjecture.
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