 It's a pleasure to be here. So I'll talk about, I think, the main theorem is already described in the abstract, so I'll try to start by giving a little bit of some related history. The problem is to cover almost all the integers up to some point using as few binary quadratic forms as we can. And the study of binary quadratic forms, of course, goes back a long way to Verma. And we know what integers are sums of two squares. So they are all integers which are divisible only by primes that are 1 mod 4. Or you could have primes that are 3 mod 4 so long as they appear to an even exponent. So you can count several things. You can either count with the number of representations the number of representations if I denote by Rn. We could count the total number of integers up to some point n, counted with this multiplicity Rn. That's just counting the number of lattice points inside a circle. And this is a circle problem. We know it's approximately the area of the circle plus some remainder term, which is still open what the best error term should be. Or we could count just the number of integers up to n that are represented as sums of two squares. And this also is a classical problem going back to Landau. And we know that the number of integers that are represented by this binary quadratic form is about n over root log n. And Landau's result was generalized by Bernays shortly afterwards. So the same proof goes through for essentially any binary quadratic form, let's say positive definite. So something of the form ax squared plus bxy plus cy squared. And then the relevant quantity here would be the discriminant b squared minus 4ac. And the discriminant is negative in this case if we think of positive definite binary quadratic forms. There are many other interesting results about binary quadratic forms. Let me just say a couple of them from more recent history and which are also relevant to the theme of this talk. I think a very beautiful result is due to Blomer. I think this is part of his PhD thesis from around 2005. And Blomer showed that if you look at the number of integers up to n that are sums of two square full numbers. So sums of two squares are obviously sums of two square full numbers. A square full number is one where every prime factor appears to exponent at least three, at least two. But what he showed was that there are many more integers that are sums of two square full numbers than sums of two squares. And the exponent on log n here instead of half in Landau's theorem is actually something like 0.2, one minus two to the minus one third. And what Blomer did here was to count a square full number can be written as a squared times b cubed. So what you're counting here is representations by this family of binary quadratic forms, a cubed x squared plus b cubed y squared as you vary A and B in certain ranges. There are more precise versions of this by Blomer and Granville. I think they even get the right power of maybe, maybe they refine this up to powers of log log. They also study related problems like looking at moments of the representation function of n by some fixed binary quadratic form and then studying this both in regard to, you know, how you can vary beta and also with regard to uniformity in the format. And these kinds of results have also found applications in other areas like work of Borgen and Fuchs to numbers that appear in Apollonian packings as curvatures of circles in Apollonian packings. Okay, so that's a little bit of review of related results. I want to now get to the problem at hand, which is if you take a family of binary quadratic forms, x squared, let's just say x squared plus d y squared and d goes up to some parameter delta, then the question is how large should delta be so that this family of binary quadratic forms represents almost all the numbers up to n or you could also ask how large must delta be so that you get a proportion of the integers up to n. Now, each binary quadratic form by the result of Landau or Bernays represents about n over root log n numbers. So you must clearly have delta being at least some constant times root log n if you're going to represent a positive proportion of numbers. But the set of numbers that are represented by two different quadratic forms may have dependencies and so therefore maybe this is not the right answer and we have to study how two of these sets intersect with each other if we want to determine exactly what delta is. So this seems like a fairly natural problem but it seems to have been considered only quite recently where so work paper of Brandon Hansen and Bob Vaughn from 2020 where they showed that if delta is a little bit larger than log n so maybe log n times some power of log log n log log n to the fourth, then you can represent almost all numbers up to n by this family of binary quadratic forms and they had a refinement that if you only want a positive proportion then you can save some powers of log log. This is largely as an application of the circle method computing a second moment. There's a student of Blohmer, I think part of his master's thesis, Dell, who showed that a refinement of this result of Hansen and Vaughn, he showed that you could deal with prime values of D rather than just all D up to delta. So the set of binary quadratic forms the considers has size about log n rather than log n times log log n but you still have to go up to the same log n times log log n in order to represent almost all numbers. So the result that I want to give some indication of today maybe once I state the theorem to the experts it'll be clear what the proof is is a result with Ben Green which we are still writing up which determines the threshold for when you can represent almost all numbers or when you can represent any even a proportion of numbers. So if you have delta which is approximately log n to the log two and then there are smaller fluctuations around it. So some parameter two to the alpha times square root log log n then at this stopping point the proportion of integers up to n that you would represent by this family of binary quadratic forms is the Gaussian integral. So as alpha goes off to infinity you would get almost all numbers and as alpha goes to minus infinity you would get 0% of the numbers. So in other words just to give a cruder corollary if you stop at log n to the log two minus epsilon then you've not represented many of the numbers up to n and if you stop at log n to the log two plus epsilon then you've represented almost all the numbers up to n. So where does this threshold come in? It comes from the typical structure of an integer. So most integers up to n have about log log n prime factors and it doesn't matter whether you compute count the prime factors with multiplicity or without multiplicity it doesn't matter whether you have little omega or capital omega and we can be more precise than that we have the result of Erdisch-Katz which says that you have a Gaussian law both of these parameters behave like a Gaussian with mean log log n and variance log log n. So that's where the Gaussian integral in the theorem comes in. And then even more precise version of this I've given it in a special case you can actually take any k uniformly in a wide range but let's say if we focus on the range where are typical numbers so k is close to log log n and a little bit bigger than the standard deviation but not too much bigger then you can give an asymptotic formula for the number of integers up to n that have exactly k prime factors. And that has this feature if k is one, this is the prime number theorem it's not in this range but you can see that that's the prime number theorem and it has, it's actually, you know now that one states it this way the Erdisch-Katz theorem is always stated as being a Gaussian with a certain mean and a certain variance but you could see from the way I've stated it here it's much more reasonable to think of the distribution as being a Poisson with parameter log log n this is a Poisson distribution with parameter log log n but a Poisson with large parameter approximates a Gaussian with that parameter and the same standard variance that follows just by Sterling's formula if you put in Sterling's formula for approximating gamma of k, of gamma of k or k minus one factorial then you see that the number of integers with k prime factors has this kind of Gaussian behavior so the result that Ben and I prove is actually even a little bit more precise I will state it again for just this range of typical values of integers but in fact the proof should work for k in wide ranges not just of the typical size so what the result says is that if you consider the subset of integers with exactly k prime factors then if delta is below the threshold if it's smaller than 2 to the k by let's say k to the fourth then 0% of the integers in the set are represented by one of these binary quadratic forms so we call this the upper bound result and upper bound means that we give an upper bound for the number of integers that are represented by this family of quadratic forms correspondingly if you go a little bit past the threshold if you go some power of k away from 2 to the k then the number of integers that are not represented by this family is actually small so here we save a tiny bit over the trivial bound so we save only a log log log n but it goes to zero and so this means that once you go beyond the threshold of 2 to the k then almost all numbers with exactly k prime factors are represented by this family of binary quadratic forms now from this the main result that I stated in the stated before and stated in the abstract follows because if you think of k as being log log n plus alpha times square root of log log n then the Erdisch-Kass theorem will tell you what the proportion of integers that are represented up to n is and you would get the Gaussian integral from that so that's a description of the main results and if there are any questions I can answer these. Okay, so let me try to first describe why this threshold of 2 to the k arises and then give some sketch of the proof so there's a well-known link between binary quadratic forms and imaginary quadratic fields which I would say goes back to Gauss but I have no real knowledge of history so probably that's some made up history so to recap this let's think of discriminants of the binary quadratic form so this is some negative integers since we are working with positive definite forms and the discriminant is always zero or one more four since it's b squared minus four ac and we will restrict attention to fundamental discriminants so fundamental discriminant means that you're not do so by the square of some number except for four if you have to have a four in it and it's also discriminants of the imaginary quadratic fields and we know that there's a relation between the binary quadratic forms of discriminant t and ideal classes in the imaginary quadratic field q square root of d so the number of binary quadratic forms of this discriminant is the same class number as the ideal class number of the field and when I say number of binary quadratic forms I mean up to equivalence by transformations in SL2Z the class number is about square root of d it's square root of the discriminant times l1 chi d this is true once the apart from d being minus three or minus four so most of the time it's gonna be true and this character chi d is the principle is the quadratic character to the modulus d and what it means to be a fundamental discriminant is the same as saying that this character is primitive so this is the chronicle symbol here now what we know is that the total number of representations of an integer m by all binary quadratic forms of this discriminant is the same as the well I think there might be a factor of two but it's essentially the same as the number of ideals in K of norm m the two comes from the fact that your imaginary quadratic field is going to have just two units one and minus one now the number of ideals in K of norm m that's these are the coefficients of the delicate data function with just factors as the product of data of s and ls chi d so that's just the convolution of the function that takes the value one all the time with the quadratic character chi d so this counts the total number of representations by all binary quadratic forms of discriminant d but in our family what we are really interested in is the principle form x squared plus dy squared if the discriminant is a multiple of four or x squared plus xy plus one plus the size of d over four if the discriminant is one multiple and the principle quadratic form corresponds to counting representations of m to counting ideals of norm m that are principle ideals in K so okay this works when d is not three multiple so this would be the principle form but when d is three multiple you have a slightly different quadratic form but the values of this quadratic form are related to the values of the diagonal form that we are interested in by just this identity so we are counting if you have y being even then a representation here will give rise to a representation by the form that's of interest to us so here's a heuristic idea of why two to the K is the right threshold for this problem take a number n that has K prime factors and let's just assume that n is square free for simplicity so you have prime p1 to pK and suppose I want to represent n by some binary quadratic form of discriminant d that's the same as saying that all these primes p1 to pK are split in my quadratic field q square root of d which means that all of these Legendre symbols are one and the prime pI splits us as the product of two prime ideals pI and pI bar now this splitting condition is you know if I give you a value of d then it happens with equal probability for each prime p the chance that this happens is one half so if I want all of these guys to be one then the probability that all of these K things happen is about two to the minus K so if I give you a number m the chance of finding a discriminant d for which all of these prime factors are quadratic residues have chi d of pI being one is two to the minus K which means that if I count if I run over all the discriminants up to size delta then even if I add all of these up I should be at least a little bit larger than two to the K in order for them to count a substantial proportion of numbers so that means that if I go up to just two to the K over K to the four then I'm unlikely to find a discriminant d which counts I'm unlikely to exhaust many, many numbers up to m and this can be made precise very easily and I'll show that in a bit now the lower bound of the part of the heuristic that with these many binary quadratic forms that I can represent almost all integers that's the hard part of the proof but a heuristic reason for why you should believe that is that if I am at this threshold I have gone substantially beyond two to the K so it's quite likely that I will find at least some discriminant for which all of these primes are quadratic residues now suppose I found such a d then each one of these primes factors as the product of two prime ideals which means that if I take n which is a product of p1 to pK there are many, many factorizations of n as a product of two ideals of norm n so there are two to the K such factorizations now the discriminant is of size square root of d so the number of ideal classes the class number is really of size square root of the discriminant which is about size square root of two to the K so in other words the number of ideal classes is much smaller than the number of factorizations of n so if you believe in some kind of equidistribution you might believe that every ideal class there will be some factorization or many, many factorizations that will lie in every possible ideal class so in particular there are lots of factorizations of n and it's likely that some of these ns will actually correspond to principal ideals because the class number is small so therefore you should believe that as soon as n is represented by some binary quadratic form of discriminant d it's in fact represented by the principal binary quadratic form of discriminant d and therefore this threshold should be enough so that's the heuristic motivation for the proof and now for the remainder of the talk I'll give you some idea of what actually is involved in proving it let me start with the upper bound part which is basically the same proof that I already discussed so here I'm just going to count the total number of representations of an integer by all binary quadratic forms of this discriminant d and there is no loss in doing so because we expect that if you are represented by a binary quadratic form of discriminant d then you are represented by basically every form as well now this total number we can easily count using the hyperbola method if I don't worry about error terms I don't even have to use hyperbola method I just interchange sums this sum let's just pretend this n over l and then what I have is just n times l one chi d now if I consider numbers with k prime factors and you're represented by x squared plus dy squared then this representation number must just be the total number of divisors of n and let me just pretend that n is square free so the total number of divisors of n is just two to the number of prime factors two to the k this is not strictly speaking correct since I'm writing down capital omega here but it's more or less fine so that means that the number of integers up to n that are represented by this binary quadratic form is utmost since everything here is positive is utmost the total number n times l one chi d divided by two to the k if I sum this overall d going up to delta then just taking the union bound I can see that the number of integers that are represented by some binary quadratic some form x squared plus dy squared with d less than delta is no more than the sum of l one chi d up to this range times n over two to the k and l one chi d is really a very tame object it's just of constant size most of the time and therefore that just gives me n times delta over two to the k which means that if I go up to a small up to a little bit less than two to the k then very few of the numbers up to m are going to be represented so this proves that if I stop a little bit less than sorry that if I go up to log n to the log two times two to the alpha root log log n then I get less than the Gaussian integral phi of alpha as a proportion of integers that are represented now the lower bound is a bit more involved and it's really an application of the second moment method so for simplicity, let me just restrict attention to n being odd you have to do a little bit of casework to cover n being odd and then to cover n being two mod eight and then to cover n being six mod eight in with different discriminants in different residue classes so let me just restrict give you the idea of the proof in the case that n is odd and I'm going to cover them only using discriminants that are seven mod eight and discriminants going up to delta and if d is seven mod eight then the fundamental discriminant associated to this binary quadratic form is actually just negative d itself and the principle form is this x squared plus xy plus d plus one over four y squared so what we are interested in is counting the number of principle ideals of norm n in this imaginary quadratic field which really corresponds to representations by this principle binary quadratic form but here's where this condition that d is seven mod eight allows me to finish this issue if d is seven mod eight d plus one over four is even and so if n is odd you can also deduce that y must be even here and then a representation by this binary quadratic form already gives me a representation by the form that I want x squared plus d y squared so there's just a little parity check that you can make to show that in this case we can just get representations by x squared plus d y squared which would correspond to an order in q square root of d rather than just the full ring of integers so now a natural idea to try is to compare two moments let's count here the sum of all these representation numbers as d varies and compare the first moment with the second moment now if you do that then well this quantity obviously is zero if n is not represented by one of these binary quadratic forms and it's non-zero if it is represented by some binary quadratic form so therefore if you use Cauchy Schwartz you get a lower bound for the number of integers that are represented it's just at least as large as the first moment squared divided by the second moment this will kind of work to show that you can get a positive proportion of numbers but if we want the more refined result that we want to get almost all numbers with this given number of prime factors the constants don't work out and the reason is that the representations by these binary quadratic forms are not quite independent so if you have two different moduli d and d prime and let's say one of them is one more three and the other is two more three then you're not going to represent any integer n that's a multiple of three by both of these binary quadratic forms because for one of these three will be a quadratic residue and for the other three will not be a quadratic residue so there will be fluctuations so when we compare the sum this will not if the two moments are going to match then these sums must be more or less flat and what I'm saying here is that they will not be more or less flat if there are these dependencies another reason for it is that if I take even discriminants that lie in the same progression and if I look at integers n that are multiples of three now it's kind of more likely that you will be represented by both of them because three already will be a quadratic residue module of both of them so the multiples of three will be over represented and then other non-multiples of three will be underrepresented and once again this will cause fluctuations in the two moments and the two moments will not match up as far as constants go so to refine this what we are going to do is a modified second moment method where we are going to fix the in progressions modulo all the small primes and then also take the small prime factors of n into account so here's an idea of the refined version of the proof I have a parameter w which is going to go to infinity extremely slowly it can go arbitrarily slowly to infinity but if you like you can think of it just as log log log n so the discriminants d already go up to log n so even if you take all the primes up to w and multiply them together you only get up to log log n and so that's even that is negligible compared to the size of the discriminants d and then I'm going to restrict attention to only primes and this is for another technical reason primes d that are seven more eight and I'm going to want all the quadratic characters of the primes up to w to be prescribed I want them all to be plus one so this is still a large set of primes because I have I'm only imposing congruence conditions modulo the product of all the primes up to w but even the product of all the primes up to w is very small even compared to delta so if you just use the prime number theorem in arithmetic progressions this is a big set of primes of size delta over log delta the total number of primes up to this times basically a product of a probability of one half for each of the primes up to w okay so that's fixing the discriminants d I'm also going to want to account for the small prime factors of n so I'm going to break every odd number m into two parts f and h so f is the friable part of n so if all the prime factors of f are smaller than w and h has all prime factors being at least w and so now here is the modified second moment method I'm going to define f of n to be the sum over all these good discriminants so primes in a certain residue class and then to count the number of principle ideals of norm n weighted by square root of d and then divided by the number of divisors of the small prime factors of that this quantity is positive if n is represented by some binary quadratic form x squared plus d y squared with d in this set of discriminants and once again we will compute the first and second moments of this object and if we can compute these two moments then Cauchy's inequality will once again give me a lower bound for the total number of integers that are represented just the first moment squared divided by the second moment so this is now going to be the key calculation I take f m so I have to understand something about averages of r the number of principle ideals of norm n as n varies and then if I square it I'll have to understand the product of r d n and r d prime n as d and d prime period okay so here's the calculation for the first moment I can evaluate the sum over n restricting to integers that are odd with a given number of prime factors and the asymptotic takes the following shape it's a constant depending on w and this constant is just the product of one minus one over p over all primes up to w pi is just pi times the total number of integers in this set so r d of n is on average once I divide by this divisor function of f on average it's just this quantity so the first moment of f n which had a square root of d on the appearing in the numerator here would just be this constant c of w times the total number of integers times the number of discriminants that I'm summing up to compute the second moment I expand out the square in the definition of f and so what I have to understand is a similar calculation for the product of two different discriminants d and d prime so there'll be two types of terms I'm looking here at the non-diagonal terms when I expand out the square non-diagonal terms where d is not equal to d prime and then we'll also have to think of the diagonal terms where d equals d prime with sum but if I think of the non-diagonal terms because of the way I choose these local conditions on d now I can guarantee that the answer is basically the square of whatever I had before so the previous answer had a c of w over square root of d so now it has c of w squared developed by square root of d times d prime times the number of n the answer is not quite one here but the difference is just l one of the product of the two characters chi d and chi d prime times having fixed what happens at the primes of the w and this is a crucial part of the proof this is not one on the nose but it's pretty much one on average if I vary d and d prime in this progression l one chi d is a reasonably tame object I mean I don't mean things like the class number problem getting a lower bound for l one chi d for every discriminant obviously is a very hard problem but if I want to understand statistically what l one chi d does that's a very simple problem so if d and d prime vary in this set if you have a lot of discriminants to play with then typically the size of l one chi d d prime is the same as just the product of the first few primes so it only matters here that w goes to infinity as no matter how slowly it just has to go to infinity so I can approximate it typically by the product of the primes up to w of the Euler factor but the way we chose our discriminants d and d prime chi d of p and chi d of d prime of p were all going to be one so this is just one minus one over p inverse which exactly cancels with this product of one minus one over p that I already have so in other words, this extra factor that I have is most of the time going to be one and that means that if I look at the off diagonal contribution to the second moment it just basically gives me c of w squared times the number of m times the number of discriminants that I have squared we have to add the diagonal contribution where d is d prime but there are a few terms there so you can imagine that that's under control and it's not hard to get an upper bound for that and this upper bound basically all you have to observe is that once this delta is a little bit larger than two to the k, like two to the k times k to the four the diagonal contribution will be negligible in comparison to the non-diagonal contribution so that would mean that the second moment is the same constant depending on w squared times the number of integers with this given number of prime factors times the total number of discriminant squared which is very closely related to the first moment which is the same constant times the number of m times the number of discriminants so now if I use Cauchy-Schwarz the first moment squared divided by the second moment squared everything here will cancel out these will cancel out and what I get is asymptotically the number of integers up to m that are odd with a given number of prime factors which means that almost all the integers that are odd with a given number of prime factors are in fact represented by one of these binary quadratic forms of discriminant d going up to this threshold so that's a very quick sketch of the what the proof relies on but of course what I haven't told you is how one does the first and second moment calculation so I'll give a very brief indication of that so the proof to try to understand the number of ideals principle ideals of norm m in these imaginary quadratic fields we use this is a very natural idea we use the class group characters of this quadratic field use square root of d so these are just characters from the class group to the non-zero complex numbers or unimodular complex numbers and so you can think of these this of psi as a function on all ideals and it's a multiplicative function if you like on all the ideals taking values on the unit circle corresponding to each one of these so the class group this is the class group characters themselves of course form a group it's the dual group of the class group it has the same size the class number as the class group does now corresponding to each one of these characters we can form the representation number r and psi which is the number of if I sum over all the ideals of norm m and then I weight them now by the class group character psi of a we always have the principle character here which takes the value one on all the on all the ideals and then the number of representations by this principle character is nothing more than the total number of ideals of norm m the orthogonality relations for characters now tells us that the number of principle ideals of norm m is simply the sum of these representation numbers as I vary over all class group characters psi now what you would like to show is that the term here where psi is the principle character dominates and you get the principle class gets its fair share one over the class number of the total number of ideals of norm m and that all the other non-principle class group characters should contribute a negligible term so by the way this is also in my previous discussion I said that we would restrict attention to d being prime and this is kind of the reason why I'm doing it because if I restrict to d being prime I can avoid having to discuss genus characters where also something slightly different can happen for these class group characters all the class group characters that are non-principle here are non-genus characters because the discriminant is just the negative of a prime in this case now if I so for the non-principle characters we have a classical zero free region for the Hecker-L function associated to these characters and you can analyze the some corresponding sums that we need here and show that they are much smaller but if I just focus on the principle character let me just tell you how these sums are dealt with I have some kind of multiplicative function here but I just have to understand it on the integers with k prime factors and this is basically a very classical object now it can be understood by Selbert's method or the Selbert-Delange method so to understand the sum you first understand another function which is just a multiplicative function with z to the omega of n we just write down what the generating function is going to be its generating function will naturally be associated to something like zeta of s to the z times this Dirichlet L function ls chi d to the z and then you can compute what this answer is going to be so z here might be some complex number so it doesn't have a polar singularity but the answer works out essentially to be the same as if you had a pole over log n some power of log n to the z and instead of a z minus 1 factorial the only difference is that you get a gamma of z once you understand this then you can understand the sum where you restrict to integers with exactly k prime factors by using Cauchy's theorem you can take this formula and then integrate it out in z of size 1 with dz over z to the k plus 1 and that gives you the formula that I wrote down for the number of principal ideals of norm n divided by the divisor function tau of f summed over all m in this with k prime factors the second moment calculations are exactly the same of the same flavor the what we have here in that case would be two different objects r and y not the principal character for one discriminant and the principal character for another discriminant that can be understood in terms of the the data can zeta function of the bio quadratic field associated to those discriminant so it will be like zeta of s times l 1 chi d l 1 chi d prime and then l 1 chi d d prime that's where that l 1 chi d d prime term comes from and for the class group characters that are not the principal character this amounts to a rank and sell book type calculation but once again there is a zero free region for those and the rest of the proof goes through easily so that's some kind of rough indication of the proof thank you very much