 So, welcome everybody, I will tell you a small contribution to a famous problem, which is sometimes called the inverse Goldback problem of Osman, at least this is how Christian Elshos calls it, I don't know whether it is his invention or not. So this is about representing the set of primes as a subset, it is easy to see that the complete set of primes cannot be a subset in any non-trivial way, but maybe if we add or delete a few then we can. This is still unsolved, everybody expects a negative answer including me, what is known that if such a decomposition existed then both A and B should have about square root X element up to X, which is achieved by some version of large sieve and also Elshos has an explanation why we cannot hope a complete solution in this way. What I'm going to tell you is that the situation changes if we allow negative primes, so this is the following, we will assume hypothesis which is expected to be true and soon I will remind you what it is. So if we assume the prime hypothesis then there are infinite sets A and B such that the subset is the set of positive and negative primes including plus minus two and excluding plus minus two and plus minus three, it is easy to say they must be excluded. And I will tell you a proof of a slightly weaker version when we exclude a finite number of primes just not to, I don't want to tell about the extra trouble we have to take to include the small primes and also instead of some set I will use a different set of course we can change B into minus B which will make for a somewhat more comfortable notation. So this is the result I will, whose proof I will tell you assuming the prime number hypothesis there are infinite sets of positive integers such that the differences are, all differences are primes and only a finite number of primes is missing. Now the prime top hypothesis means that in a any linear configuration you will find primes unless there is a congruence reason to exclude it. So if we have a finite collection of linear forms A i times x plus B i then we can then find an infinity of values of x such that they are all primes except if there is a fixed prime P with the property that for all x one of those forms must be a multiple of P. Now the simplest and silly answer of the case is the twin prime conjecture nobody notes that this conjecture should be true. I will use a corollary written in a somewhat different form we want to prime find numbers x in a given congruence class module some number q such that a finite number of translates are all primes and the condition is the following for divisors of q the zero class module of P should be missing and for primes not dividing q at least some residue class should be missing and the plan is very simple and naive with have a finite collection of numbers such that the differences are primes we take a prime which is not represented and try to represent it by extending the sets. So our prime is R we want to include a set a number x to A and x minus R into B so that that difference should be R. Under the previous assumption this can be done if the union of B and the union of a translation of A by R does not form a complete system of residues module or any prime. Now the life is not that simple because if we exclude this x and x minus R then it can happen that in the next step this condition is avoid for some other prime or there is a trap that will kill the process after 10 steps and we must take care to avoid this and what we will do is that we will impose restrictions of on our sets A and B module or quite a lot of primes now we cannot fix them infinite for infinitely many primes at the starts because then there may be no integer that satisfies all those conditions what we will do is that we import some initial conditions and then as we proceed we invent new and new conditions while keeping all the numbers which are already in our set. So we list all the primes we want to represent that is all the signed primes in a sequence with increasing order of absolute value and our plan is to find sequences A1, A2 and B1, B2 such that A sub i minus B sub i is exactly the R i prime R sub i. So what in step n we will have the first n elements in each set A and B that is we are representing the first n primes and we will have also a set of restrictions u sub p v sub p of residues module of p for every primes which is less than the next than the first prime not yet represented so that all residues of AR in u sub p or residues of VR v sub p and here is the condition we need that the elements of u which are not in v and the elements of v which are not in u should have the property that their differences contain every non-zero residue module of v. So in the first step we take R1 the first prime to be represented we put it into A we put a zero into B and for the first and the first and the primes for up to k you know it will represent primes bigger than k so the assumption that p is less than the next R is exactly the same that p is at most k we put every non-zero residue into u sub B and only the zero in u in my screen somehow there is a mysterious pink line which I don't know how I got there so now this is the real plan and if we have somehow n minus one terms we want to need the ends we want to construct the ends term so we want to find the numbers A sub n and B sub n and also we want to find the restrictions u sub p and v sub p for primes between r sub n and r sub n plus one which means either one or zero prime so we want to put some x into A and x minus r sub n into B and so this represents r sub one and all the other new differences should also be primes x the new x minus the old B and x minus r sub n minus the old A they all must be primes so first we try to find residues of x modulo the small primes and here is the assumption on the previous slide comes into the picture since r sub n is not zero modulo p there are elements an element of u missing from v and an element v of missing from u whose differences are sub n and we impulse this condition on x and we will see why this is useful to apply the prime table hypothesis we have to check two kind of assumption for primes dividing this cube which will be the product of all small primes that is a condition for all small primes and there will be a different condition for big primes for small primes we need that this congruence does not spoil the possibility of the numbers being a prime that is the condition does not turn any of the difference into zero modulo p and here is how this condition is used u small u sub p is not in v sub p but the bs are in v sub p that's the way we constructed them before and exactly the same argument with different letters from for the other number now for big primes we don't have a condition we need that the numbers that we want to turn into primes do not form a complex system of residues no the number of primes the number of elements which should not form a complete system of residues is two n minus two they are the previous a and previous b translated by something so this is two n minus two number and they cannot form a complete system of residues model any prime which is bigger than that so we will need that the next absolute value of the next prime r sub n which is the dividing line between small and big primes is at least two n minus two soon we will need a bigger bound now so in this way we find our new elements a sub n and b sub n and we need or in some cases we don't need but sometimes we need new uh sets of restrictions u sub p and v sub p and here comes the only small trick in my proof so we cannot construct them arbitrary because we already have n numbers which are in u and n numbers which are in v and we can add as many as we wish but need to be careful so there are two n residues which are already disposed of and there are p minus two n for us to play with and what we do is that for each residue we toss a coin and according to result we put it into u p or into v p definitely not into both and if we are lucky these new elements uh will uh force the condition we need that is we get every element every non-zero residue model of p as a difference from an element of u which is not in v and an element of v which is not in u so take a fixed z and try to see the chances that it is represented take these pairs which differ by z z 2z and 3z 4z if the first is in u and the second is in v then we are happy this is a collection of p minus 1 over 2 pairs of there may be two n of them which escape our random choice because they are already in a or b we did not list all possible pairs which whose difference is z so from z to z we jumped into 4z and missed the pair 2z 3z this is to make those events independent so if we have a pair like that whose difference is z then if the first element comes into u whose probability is half and the second comes into v whose probability is also a half then there is a cotter probability that we get z as a difference there is three fourths probability that it is uh does not represent z and so the probability that uh none of those pairs gives us the desire the differences is three fourths to the number of pairs to p minus 1 over 2 minus 2 twice n now if this probability is less than 1 over p then those events do not cover the complete probability space there something will be out and there is a choice that works for all z we don't want a majority of them work only one so this is a condition 3 fourths to the number of pairs should be less than 1 over p and if we rearrange this we get that uh v minus 1 over 2 minus 2n is bigger than constant times logarithm of p which is p should be somewhat bigger than four times n basically four times n plus a constant times log n I'm sorry please excuse me to interrupt Anna Bromovich has a question Dan would you please unmute and ask yourself sorry um there is a question just a question did you say that the events are independent is that uh required part of the argument and if so why uh we could calculate the probability if they are not independent just we don't need it it is uh simpler to take only half of the possible elements which are independent so that the probability calculation will be uh easy those elements are independent because all because with this was all those pairs independently uh no choice not no incidence of the coin tossing uh effects two of those of those events they are independent thank you did it help Dan are you happy with the answer is it is it okay I'll have to think about it but okay so maybe I can continue yes please continue so this is an equality that has to be called from for primes which are at least as big as our sub n now if it holds for one number it will hold for bigger numbers as well so our sub n should be bigger than four n plus one plus a constant times log n which certainly happens if r of n is at least five times n and n is bigger than some easily calculated bound now our sub n in we see that our list half of our list is positive primes and half of them is negative primes so r sub n value of r sub n is the same as on the n over two the number of positive primes that is it is asymptotically n over two times log n so it is will be bigger than any constant times and after a certain limit and we choose the bound k so that it holds for numbers up to this limit and this ends the proof and this was the only proof I wanted to tell but I am going to tell you a few questions and then some observation about a similar problem for square free numbers instead of time of primes so the first problem is that the set of signed primes is a triple sum set and my conjecture is that it is and it would be interesting because we know again by a result of households that the positive primes definitely are not a triple sum set now I try to modify the previous proof to involve three or a thousand summons and I have a plan how to do it but so far I could not could not work out the details and the main difficulty is that this simple choice of residuals must be replaced by a rather more complicated choice of residuals we needed two sets U and V in our previous proof with the property that both have quite a big number of elements which are not in the other so that the difference set of elements which are unique to one of those sets represents every non-zero and for three sets the condition becomes more involved we need we don't just need that the set U minus V and W should be big we need that you should have a lot of elements outside some set of V and W actually outside the negative of those sums and similarly V and W and those three sets of remaining elements after excluding the some of the two other sets should have the property that there some includes every non-zero residue and since in the step we construct this we already have some elements of A, B and C we need sets with this property and including a number of prescribed elements this is a slightly more complicated problem in combinatorial number theory which I could not yet completely solve now here are two not closely related problems which are fun I guess one is to prove without using any unsolved hypothesis the existence of two sets A and B such that no element of A plus B is divisible by any prime congruent to one module of four and in this that if we want to residue exclude the residue class three module of four then we have an easy example namely the set of some of two squares more or less already misses three module of four except then both A and B have such a factor and we can avoid this if for A and B we only use primes congruent to one module of four so not only there are such sets but there are quite dense sets and I think this observation is also due to households definitely it is mentioned in one of his papers similarly the other problem is more or less due to him the question is that the set of some of two squares which is some set by definition is a triple some set or not and in the next some minutes I tell you a few words about the analogous study of square free numbers those square free numbers are a sort of toy model for primes and here is the first problem whether it is a set of square free numbers is a sum some set or not and here comes the complete analog of the austenland problem whether the set of square free numbers excluding some and including a finite number of more is a sum set or not now for the first I conjecture it is not a sum set and I think it probably can be proved by a finite argument so I have established some properties of possibly sets a and b and if I one could show that no such sets exist of the first thousand numbers and this would give a solution to this problem I try to play with the first few integers and they did not they did not gave a complete answer to this now so the the next problem is about a finite number of exception and if we allowed more exceptions but still density zero then the situation changes somewhat so this is a result quite easy one that we can find infinite sets a and b such that the sum set contains all square free numbers and the extra numbers that is the non-sphere free elements of the sum set have density zero so set of square free numbers can be approximated quite well by a sum set from the outside I could not decide whether it can be approximated by a sum set from inside that is that there are sets a and b whose sum is contained in s and the lm and the square free numbers not represented have density zero my feeling is that the answer should be positive and this is somewhat supported by the following results there are infinite sets a and b such that it has positive density even a density which is only a little less than the density of all sphere free numbers such that the sum set contains only square free numbers uh easy but not completely obvious so this is about positive square free numbers and if we take sine square free numbers then again the situation will be different because for this set positive and negative square free numbers I can positively solve the possibility of representation by a sum set not only three but as many as you want there are infinite sets k infinite sets whose sum set is exactly the set of positive and negative square free numbers and in the next version there are only two sets what they are identical so there is a set a whose sum set with itself is exactly the set of even positive and negative square free numbers we need even again by looking at residues module of four one sees that odd square free numbers must be excluded and the reason why the square free phase is easy is twofold the first is that we don't need any unsolved hypothesis so instead of the prime double hypothesis we have the square free double CRM if we have a finite collection of linear forms then there is a value of x for which all are square free except if there is a divisibility reason to exclude it I don't know whose result is it the similar statement is known for not only for linear but quadratic and cubic polynomials for quadratic it is still quite easy for cubic is difficult I don't remember whose theorem is it and the second reason is that to get Swerve numbers we omit all multiples of prime squares while for primes we omit multiples of p except p itself which we retain and for a similar argument of consisting constraints of residues module of prime squares the manipulations will be a lot easier and so this is all I wanted to tell thanks for the attention if and now I'm here to listen quest to questions or you can write questions and comments to my email address thank you so much