 Thank you to the organizers for inviting me. And yeah, on this on this point about questions, please feel free to interrupt me. Any any questions great, including as we were talking about the question just why why is this true. So okay, I'm going to be talking about sums of two cubes. I'm going to be talking about sums of two rational cubes. But maybe I'll kind of justify why two cubes rational cubes, etc. in the very beginning. So I'm going to be talking about the point of this talk is going to be to produce lots of integers which are or are not the sum of two rational cubes. So I guess at first I'm just going to focus on producing lots of integers which are not the sum of two rational cubes. Okay, so okay, I'm going to justify why I'm talking about sums of two cubes right now, or for a little bit. So barely any integer is the sum of one integer cube, or you know, even one rational cube that doesn't change anything. On the other hand, everybody every integer rational number is a sum of three rational cubes. And you know, here's an identity you get to pick your favorite why pick it to be absolutely ginormous so that the denominator doesn't vanish. Oh, I should use the pointer. Yeah, I can do that. So the denominator doesn't vanish. And you have a representation of x as the sum of three cubes. And I think a good reference for, you know, the beautiful geometry involved in finding this identity why such an identity should hold, etc. is a book by Monion on cubic forms. But anyway, so okay, every every integer rational number is a sum of three rational cubes. Therefore, anything more than three rational cubes as well. And let's now talk a little bit about sums of three integer cubes. So first off, if I have an element of Z mod nine, its cube is automatically in zero or plus or minus one. So what does that mean? This is a test that my handwriting works. That means that Oh, sorry. x cube plus y cubed C cubed is in zero plus or minus one plus or minus two plus or minus three mod nine. And in particular, this does not include plus or minus four. Okay, so that gives an obstruction mod nine to being a sum of three integer cubes. Because yeah, when you reduce mod nine, you can't be plus or minus four. So that's kind of the obvious obstruction, the only local obstruction there is. And Heath Brown conjectured that if you have an integer which avoids that obvious local obstruction, then that integer is the sum of three cubes, which I represented like this because why not, I found the cube macro or something latex. Okay, so the immediate thing to ask and Heath Brown within the conjecture, conjectured something much more precise. The immediate thing to ask is how many solutions are there? How many ways are there of writing and is the sum of three integer cubes? And okay, the natural thing to do is to say, suppose I bound the size of each of the cubes, like if it's x cube plus y cube plus C cubed, let's say x, y and z are bounded by this parameter capital X. So how many solutions do I expect? You expect infinitely many and not only that, Heath Brown conjectured a precise asymptotic, but the point I want to just quick thing I want to quickly point out is the asymptotic, the number of solutions of size at most X, let's say, grows like log X. And it's hard to get the circle method or kind of the standard ways to produce solutions to spit out something like log X. You usually get like a polynomial or at least you could get a, in principle, log X times polynomial or something. But anyway, that's why I wrote, it's not really a mathematical justification. I wrote in quotes, hence the difficulty of the problem. But anyway, the conjecture is you expect infinitely many solutions when you avoid this obvious local destruction. This is not what the talk is going to be about. I just, I love writing huge numbers like this. So anyway, this was obviously in the news recently. We all saw it and it's really cool. And I just, I wanted to write the solution. So one thing that was not known for quite some time was, okay, the number 33. So when you have such a conjecture, you immediately say, let's search for the first couple numbers, search for a solution. And 33 resisted progress, famously. And 42 also resisted progress. But recently, Booker and Sutherland found a representation of 42 as a sum of three cubes. And the cubes are just gigantic. But they're gigantic, but kind of, the fact that there's a log here suggests that you should be, you may get some huge first solution, so to speak. I don't actually know if this is the first solution, necessarily. I don't know if it's been shown. But anyway, let's act like it is. All right, so that's sums of three integer cubes, a very difficult problem, but very much in the news recently. I'm just going to talk about two cubes. And the first claim, which I guess is already in the abstract, is just that there aren't really that many integers, which are a sum of two integer cubes. And yeah, the reason is, if let's just say x cubed, I'm going to put a minus sign. This obviously doesn't matter, but for psychological reasons, x cubed minus y cubed is at most x. Well, this is like x minus y times some positive definite guy. Then, well, okay, the point is cubes repel each other. If I had like a y squared minus x cubed for all those who love integer points on more del curves. Well, okay, squares and cubes kind of do repel each other, but that repulsion comes from bakers bound from linear forms and algorithms. And yeah, you're not going to get a great lower bound on this sort of difference. But in this case, yeah, cubes really do repel each other very well. So x minus y is at most this parameter capital x over, I'll just write it like max of little x little y squared. So first off, what does that imply? So if, so if this thing was, let's just assume it's non zero, because it's really easy to figure out if zero is the sum of two integer cubes. And that implies, since this difference then is at least one, that implies that the max of these things is at most x to the half. So automatically, we have like we have an a priori bound on the size of a representation of a given integer by some two cubes, both of the, both of the, I don't know, cube roots or whatever are not really that big. And so yeah, so bound number of x y pairs with max at most, let's say kind of the obvious thing x the one third, because if I wanted to get some sum of two cubes to be at most like x, well I could plug in little x and little y of size like x to the one third and yeah, it's size of most x. So that bound that trivially, trivially meaning well, how many possibilities are there, how many integers are of the size x to the one third. So there's x to the one third choices for x, x to the one third choices for y. So there's x to the two thirds choices for those guys together. And, and for the rest, without loss of generality, let's say, oh my goodness, x is the bigger one. And then, okay, so first you fix the bigger one, in this case x. And then you say how many possible y's are there that satisfy this inequality. Well the number of possible y's, well they all sit in some interval of this length. The interval, this bound has a size at least one, that's from this upper area bound. It's a common mistake I think, where you say, oh, how many integers are in an interval of size is epsilon, oh, it's the most epsilon. Oh no, you could have one integer in there. Anyway, so that allows generality x is the bigger one and then the number of possible y's, y's is x divided by this squared. So the bound you get is the kind of non-trivial solutions is x to the one third goes up to x to the one half, x divided by x squared. Okay, which is x to the two thirds because of this lower bound. All right, good. So I just wanted to, it's like a little elementary dinky thing, just wanted to put it, get it out of the way. So there aren't that many integers that there's some of the two cubes, two integer cubes, excuse me. And you know, there's not that much multiplicity. There aren't that many solutions. So given n, n non-zero, I guess the statement's true for n-zero, but anyway, there isn't that much multiplicity among, there aren't that many representations of n as a sum of two cubes. You can imagine doing this by just factoring and saying, well, this is one of the divisors of n, there aren't that many divisors, for example. And then this is kind of determined, x is then determined to two choices, or you could use a bound of a verte and just totally destroy it like that. Okay, so all right, now I'm going to get to the point which is we're going to be talking about sums of two rational cubes having kind of put aside all the other cases. But I want to just quickly note, because you know, if you know me, you know, I love the word effective. So anyway, if you give me n, do I know how to answer the question of whether or not n is a sum of two, for example, rational cubes in finite time, you give me n and you say, I'm going to give you a million years to come back with an answer. Am I confident that I will be able to answer that question within a million years? Well, if it's two integer cubes, then yes, okay, let me try to scroll back up. If it's two integer cubes and this kind of a priori bound tells me there's not that much I have to search, I just search for, you know, I start plugging in all possible values of x and y and I either find n as a sum of two integer cubes or I don't. Sum of two rational cubes. Well, I'll put it this way. I mean, we do know kind of, but we don't mathematically, you know, provably know the answer to this question. So we don't know a finite time algorithm that will even an n produce a representation of n, for example, as a sum of two rational cubes if there is one, or even just determine if there is one, and then say there is no such sum of two rational cubes representation event if there is none. Well, okay, the reason I said we do kind of know is we pretty much know is this is okay. If you assume the finiteness of Shaw and or the Bershwin-Schwinn-Dyer conjecture, I guess it's kind of part of the BSD conjecture. And actually, if you look in Tate's, I think it's an ICM address. Anyway, Tate's first writing where he conjectures the finiteness of Shaw. He immediately notes, yeah, if you have this finiteness, then you can just run this like iterated higher and higher descent algorithm to determine whether or not an elliptic curve, so this thing I should have said earlier, to determine whether or not an elliptic curve, like this one, this is an elliptic curve because it has a point in infinity, has a rational point or not, has an infinite order rational point or not. But this whole question basically is about infinite order rational points. All right, so I just like noting that. Now we're going to get to the actual stuff. All right, so the main theorem of the talk is, this is joint work with Manjul Bargava and Arish Nibbin. And the main theorem is that at least the sixth of integers are not the sum of two rational cubes. And on the other hand, I kind of sneak a factor of half in here. But anyway, on the other hand, at least a sixth of odd integers are the sum of two rational cubes. Okay, so that's what we're going to be talking about. So in other words, this produces the correct order of magnitude. This doesn't get the, I'll mention in seconds, this doesn't get the optimal constant. And the method will not give the optimal constant unless you really improve it very significantly. But anyway, it produces the right order of magnitude for these numbers. And I think that, yeah, I'm pretty sure the best lower bounds that were known, so to speak, were both zero. Okay, so yeah, so you expect like a half half, you expect half of the integers to be a sum of two rational cubes and half of the integers, sorry, not a sum of two rational cubes and half of the integers to actually be a sum of two rational cubes. That's one of these standard heuristics. Okay, now I'm going to say something a little bit more about these curves. So this is a family of elliptic curves. Again, it's smooth cubic in P2 when you homogenize. So it's a genus one curve. Okay, when you put a z cubed here and set z equals zero, then there's an obvious rational point, one minus one. So this is a family of elliptic curves. These elliptic curves, as we so to speak know, but can't prove whether or not the elliptic curves have an infinite order rational point. Let me just ask you to believe me that there's not interesting torsion that will give solutions. So whether or not the elliptic curve has an interesting as an infinite rational point is governed by the order of vanishing of its L function at the central point of its functional equation. Okay, so that order of vanishing, certainly the L function is forced to vanish if the L function is odd around that central point. If the functional equation is an odd functional equation. And so that's like some, whether or not some root number is a minus one or a plus one, that sort of statement whether or not the functional equation is odd. And the root number equidistributes. You know, you can actually show it really easily in this family. So because the root number equidistributes, okay, half the time the L functions forced to vanish half the time it's not necessarily forced to vanish. And then so that's kind of where these halves are coming from. And then kind of 100% of the time if you give me what the root number is, you expect that the L function shouldn't vanish. And so if it's not forced to vanish by root number considerations, 100% of the time, the L function shouldn't vanish. And so the rank should be zero 100% of the time because of BSD that that's the, you know, crucial relation. And similarly, 100% of the time the rank should be one, because it's forced to vanish once and, you know, 100% of the time it shouldn't vanish anymore. So that's why this half half thing happens. Okay. All right, so now let me get to how do you prove or how do we prove a result like this? Okay, so this is going to be some geometry numbers for a while. So okay, I wrote like this. Let V be, I wrote this kind of strangely V is the space of pairs of binary cubic forms. So what's a what's a typical element of V? It's an eight dimensional vector space. Actually, I'm thinking of it as a z to the eight or something. Anyway, typical element is what's a binary cubic form or a pair of binary cubic forms. So this is what the elements look like. So there are actually write that. So there are four, sorry, why do I try to write four dimensions? Three, there are four coefficients a zero through a three and there are four coefficients b zero through b three. So you get kind of eight dimensions. And so that's the kind of space in which I'm going to talk about geometry of numbers. And this the relevant group is going to be so this is going to be kind of a arithmetic statistics type discussion for a bit. Relevant group is kind of an SL2 squared. Okay, so how does SL2 squared act? Well, one thing you can do is if you give me two binary cubic forms, I can replace them by like linear combinations. df2, cf1, oops, that's df2. Yeah, that's one legitimate thing you can do. You start from a pair of binary cubic forms and end with a pair of binary cubic forms. So that's one kind of action of SL2, like on the outside, in this weird looking notation on this outer two. And then you can also change variables, so to speak, inside the binary cubic form by changing x and y by replacing x and y by a linear combination, pair of linear combinations. Okay, so that's kind of the action on the inside. And, well, okay, I kind of hid something here. I wrote like sim lower three. That means it's threes in so to speak. So whatever it doesn't matter, these are just constants. You include the binomial coefficient. So like the if one, three, three, one is okay, if threes doesn't doesn't make sense just ignore the constants, but it's like, you know, Gauss's notation for binary quadratic forms. Okay, anyway, so that's the space that I'm going to be talking about. The threes do matter because we're talking about integer points. So anyway, so g is acting as an SL2 squared in the way I mentioned on this space. All right. So I set that up. Okay, so let me just mention this crucial parameterization is the following. So I'm going to be talking about, let me just scroll up for a second. So how am I going to get a result like this? This is something like a lot of these curves, these elliptic curves have ranked zero. One way you can show that a lot of these curves have ranked zero is, well, you know that, let me write it like this, two to the rank of the curve is bounded by, bounded above by the size of the two-cell moment group of this elliptic curve. So how do you get lots of curves with rank zero? Well, you can try to get lots of two-cell moment groups that are of size one that are trivial. And how are you going to get lots of two-cell moment groups that are size one? Well, it's like this probabilistic method thing. You can show maybe the average size is not really that big. So I'm going to start by, so to speak, talking about how do you try to do that. So you're trying to show the average size of some two-cell moment groups is not that big. And apparently, I'm going to talk about two-cell moment groups of these elliptic curves and in a second, in a bit connected to, but okay. So I'm going to talk only about two-cell moment of these elliptic curves for a bit. So if you give me such an elliptic curve, this two-cell moment group is something. And to find the two-cell moment, you can go instead and look for pairs of binary cubic forms, this space that I just talked about. And if you give me this particular elliptic curve, so it has two coefficients that are allowed to vary. A2, so to speak, and A6, just notation. The two-cell moments are in bijection with pairs of binary cubic forms with given invariance. So it turns out that the space of binary cubic forms, where do I write this, pairs of binary cubic forms, maybe I'll write it like this. Just understand what I'm writing, I guess. So if I take polynomial invariance, polynomials in these coefficients ai and bi that are invariant under the action of g, then that's generated by, it turns out, just two invariance and no relations between them. And the two invariance, the notation suggests this one has degree two and this one has degree six. I'm actually going to write this invariant down in a second, but I'm not going to write this degree six invariant down. So okay, two-cell moment elements of these elliptic curves are in bijection with pairs of binary cubic forms, integer coefficients, with given invariance. And then there's this condition local solubility. Okay, so actually take those. And if you have such a thing with given invariance, well, it's obviously you can change variables by like SL2z cross SL2z, because these are supposed to be invariance for the action of this group. So you'd get an infinite answer here. So okay, it can't possibly even be even bijection with this two-cell moment group. So you have to mod out by equivalence under the rational points of this group, g of q. Okay, that's not so deadly. Just imagine I'm saying mod out by SL2z cross SL2z. It's pretty much the same thing. All right, and then there's this local solubility condition. Well, in the definition of two-cell moment group as some Galois homology classes that are locally soluble, it has a local solubility condition there, and there's that same condition here. All right, but anyway, just basically imagine that two-cell moment elements are pairs of binary cubic forms within integer coefficients. From that, these are theorems of Manjul Bargava and Weihou, also a theorem of Manjul Bargava and Weihou in a different paper. From that you get, or they got eventually after more work, that the average size of two-cell moment groups in this family have looked at curves, which I still haven't connected to the family we care about in this talk, is at most three. Okay, so it's just, anyway, they got some bound for the average size of two-cell moment groups in this family using this parameterization. I'm going to talk very quickly about how the parameterization allows you to prove something like this, because, you know, if you imagine trying to do this by saying, oh, I'm going to look at Galois homology classes and impose local solubility, et cetera, et cetera, I don't know where you'd get started, but this is kind of a much more concrete object that you might be able to count. Okay, so let's talk about that. So how do you, how can you possibly get a bound on the average size of these two-cell moment groups? How can you possibly have this clean bound three? So roughly speaking, once you parameterize things by binary cubic forms, very concrete objects, you use G, it's like kind of SL2Z cross SL2Z, to put all the binary cubic forms you're trying to count, you're trying to deal with in some decent normal form. And now, so for example, decent normal form would mean like you're trying to count pairs of binary cubic forms. What you don't want to have to try to count is binary cubic forms, pairs of binary cubic forms, where all the coefficients are like one or two, and then one of the coefficients, only one of the coefficients allowed to be like a trillion. You don't want to try to count those things kind of on average, because that's that's going to live in some horrible set that whose integer points you cannot count easily by saying, oh, they're almost the volume. The number of integer points is almost the volume. So anyway, you see this kind of in Gauss's work, where he tries to use, this is called reduction theory. He uses SL2Z to change variables so that his binary quadratic forms are decent. They satisfy some clean inequalities, and then he's able to count them. Yes, yeah, so I guess I should have mentioned this. This is a universal, sorry, this family of elliptic curves is a universal family of elliptic curves with a Mark III torsion point, namely x equals y equals zero. I should have mentioned that there's an obvious point on these elliptic curves, and that point turns out to be three torsion. So yeah, anyway, so the counting goes by, okay, you figure out how to change variables just by kind of theories of fundamental domains, how to change variables to put the relevant concrete objects into a decent form and then count integer points in decent sets. And I'm just going to kind of put down the main integral that you face or main point counting problem that you face. You don't really need to know, I'm not going to explain all the notation because you don't need to know it, I'm just going to explain like what you need to, the problem that you really need to think about. Okay, so, all right, so how to do this? All right, so okay, first the usual, I'm going to draw like a usual picture of the fundamental domain for SL2Z acting on the upper half plane. I'm just going to put two things down. It's kind of like a U and I like T, just like T squared. So I have like a pair of, sorry, sorry, sorry, I have a pair of like unipotent elements. So there's like a unipotent part of SL2. And there's a torus. This is kind of my notation, you don't really need to know this. Notation, but just to kind of give you an idea, this is stuff going on in the group. Inverse T2. Okay, so kind of what's going on is lambda is going to be like scalars. These U's are going to be like unipotent elements and these T's are going to be like the torus. And we're kind of writing SL2, so ignoring lambda, we're kind of writing SL2 is like the NAK decomposition. Because K is compact, I can just act like it doesn't exist. So N is like the U's and A is like the torus, it's like the T's. All right, so the, but now you can just kind of ignore that. It's just some integral I'm going to tell you in a very concrete way, what you really have to count. So you're really trying to count the number of points, number of, let me write it, number of A0, A3, nope, B0, B3, Nz to the 8, such that, okay, inequality is like this. A0 is at most like lambda times T1 to the minus 3. A1 is like, at most like lambda times T1 inverse. I'll just write the general inequality down in sec. B0 is at most like lambda times T2 to the minus 3, etc. So the inequalities are actually, this is basically the counting problem. You're going to have to trust me that it's actually, it's a 3, sorry, I'm just 3 plus 2i. So you're trying to count integers in this region. This is, I mean, obviously this is, it's like significantly more difficult, but kind of given the techniques that we know now. This is the intuitively correct thing to think about if you trust me. So you're trying to count integers, 8 tuples of integers that satisfy these inequalities. Let me actually color them in. So they satisfy these inequalities. Okay. And the Ts kind of, each ti is at most like lambda to the 1 third, just so the right-hand sides of these inequalities are always at least one. Because I guaranteed that and lambda is kind of a big parameter, just think of it like that's the relevant big parameter. And you can forget about anything everything else that happened before. It's easy to count integer points in this kind of box. Okay. So that's intuitively, you're reducing, obviously it's in reality more difficult, but intuitively speaking, you're reducing to counting integer points after a lot of rigmarole in some box, which is potentially skew. And T, the T parameters kind of control how skew the boxes. So like, you know, if T were huge, then A0 would be ranging over some kind of small interval, but A1 is ranging over a bigger interval. And then the later ones are ranging over a much larger interval. So, but still you can count integer points in a skew box by saying, okay, there's, you know, roughly the volume of the box or alternatively count points in an integer points interval by saying roughly it's the length of the interval. Okay. So that's intuitively speaking, that's how you should think of it. You kind of have to integrate these point counts whatever over Lambda, U and T. And you kind of just ignore you, just basically the issue is integrating over Lambda and these parameters. All right. So now I'm going to like put that aside for a second and explain to you quickly, because we're going to come back to this in a second. I'm going to explain to you quickly what this counting problem had to do with our problem. Okay. So what, why do these elliptic curves, you know, what do they have to do with what we are trying to do? Okay. So the family of elliptic curves we're actually interested in is this family x equals y cubed equals n, n varies. That's an elliptic curve, may as well put it in virus trust form. The virus trust form is y squared equals x cubed minus 432n squared kind of famously. And so it's a Mordell curve. And anytime you have a Mordell curve, y squared equals x cubed plus k, that's three isogenous to y squared equals x cubed minus 27k. So this teaches you this three isogeny. Okay, where does it come from? It comes from, if you take x equals zero, and there's like a complex, sorry, a Galois conjugate pair of rational points, y equals plus or minus the square root of k. And those are both three torsion points. And so along with the identity point, that gives you a subgroup, a Zmod 3 subgroup, stabilized by the Galois group, preserved, I don't know how to say, kept inside itself, so to speak, by the Galois group. So that's, it's a rational subgroup and so you can mod out by it and you get an isogenous elliptic curve. Here it is. So this teaches you that, you know, this 432, for some reason, maybe doesn't look satisfying to you. You can eat any factors of minus 27 with a three isogeny. So, okay, this elliptic curve family that we're interested in is three isogenists to this family. y squared was x cubed plus, I brought the four inside, four n quantity squared. Okay, and now you can kind of uncomplete the square to write it in the form that we were just talking about. So I'm going to write it as y squared is zero times xy plus eight n times y is equal to y squared plus that is equal to x cubed. All right, so this is kind of the form that we were talking about before. So remember, the coefficient is like zero and eight n. So I'm going to just scroll up to show you. So if you don't believe me, y squared plus a2xy plus a6y is x cubed. So these elliptic curve family are the same as, this elliptic curve family we're interested in is the same as before, except all of them have this a2 equals zero. Okay, but like kind of forget about the eight, these sorts of converse conditions don't matter. So roughly this parameter is allowed to vary arbitrarily, but we're varying in this family where we restricted one of these AIs to be zero. So what we are going to be interested in is to get an average, a bound on the average size of these two summer groups in our family. And to do that, well, we could try to run this, parameterize, put a normal form, count argument for this family. But what would we face if we did that? We would face, well, okay, so we would be trying to do Bargava and Ho's work for our curves. But to do that, we would have to impose, so a2 was like the value of this quadratic invariant on the pair of binary cubic forms and that parameterization. So to do that, we would have to impose the equation that the quadratic invariant vanishes for all the pairs of binary cubic forms that we're interested in. I'm going to write it down, but that's a, you know, catchphrase maybe, quadratic and arithmetic statistics. Okay, so imagine, now I'm going to write down the quadratic invariant. I'll just write AI, let's forget about the factors of three, xi, y to the three minus i. So bi, xi, y to the three minus i. Yeah, so what is this i2 on this pair of binary cubic forms? I hope I've been saying binary cubic instead of binary cubic forms. It's the following thing. There's always an invariant bilinear form on the space of binary enix, I think, so maybe n should be odd, let's just say, just to be safe, because I don't want to think about it, but anyway, here it is for n is three, m, a, i, b, three minus i. So this is kind of like a0, b3, minus a1 times three, sorry, a1, b2, plus dot, dot, dot. Yeah, so this is a quadratic form in the a's and b's. Okay, so let me just go back to that integral we were facing. We're going to try to run Bargo Ho's argument, but with this extra condition imposed that all the invariants, quadratic invariance of all the pairs of binary cubic forms that we're trying to count have to vanish. So we have to impose this kind of equation in our inside the counting argument. So I just kind of rewrote the exact same integral as before, except I added this condition. Okay, so now remember that this set, again, this looks like some mess, but this set was basically like a0 through, you know, eight tuples of integers. The a's weren't like allowed to be skew by t1 was allowing them to be skew, and the bi's, well, t2 was allowing them to be skew. And now we have to also impose that i2 of a's and b's is zero. Okay, so we're trying to count inside this kind of box, which potentially is skew. The number of solutions, integer points on this quadratic, this very explicit quadrant. And that's not a problem if the box is not skew. It can even be a little bit skew. It's easy. I mean, yeah, from the analytic perspective, you have eight variables and you're trying to count integer points on a quadratic. That's like a, by now that's like a joke for the circle method. Okay, so if the box is not skew at all, like if the t's are like a lambda to the epsilon, lambda, just forget about now the integral, lambda is our big parameter, so to speak. If the t's are not that big so that the box is not really that skew, then yeah, the circle method says, oh yeah, you have a degree two polynomial and eight variables. Of course, I can count the number of solutions. Asymptotically, you get your beautiful product of local densities, you get the correct asymptotic, you get a power saving error term. Everything's great. But that's, yeah, so anyway, that's so to speak, not that high in the cost. Remember there was this kind of, I drew the picture, I never actually went back to it, but sorry, this ugly picture that I drew. Let me just say the u-parameters kind of the real part parameter and t is kind of like telling you how high you are on the cusp. So if you're not that high in the cusp, you're allowed to be like sub polynomial high in the cusp, and actually it'll work if you went up to like lambda to the 10 to the minus 10 or something, lambda to the 0.01 or whatever. If you went that high, then the circle method would still be fine and still, you know, laugh and produce the correct answer. Now if you are, if your box is skew, you're not in such bad shape either. This is one beautiful thing about trying to count integer points on quadrics. You can get down to almost the correct asymptotic by doing nothing. So here's our quadric. It doesn't really matter that it has this beautiful form, but anyway a0p3 minus 3a1b2 plus 3a2b1 minus a3b0, that's supposed to be equal 0, yeah? So we can get down to the correct, so let me just explain to you what you expect. So each of the ai's, let's act like each of the ai's, even though I keep talking about the skewness business, each of the ai's and bi's are roughly size lambda, okay? So you have kind of eight things of size lambda. The possibility, the number of possibilities like lambda to the eighth, the left hand side is of size like lambda squared. So what are the chances that this left hand side of size lambda squared is the number, the single number zero? Well probably like lambda to the minus two. So you expect, so the circle method will give like constant product of local densities times lambda to the eighth times this probability lambda to the minus two, so you get lambda to the sixth. Okay, so that's the expected number of solutions. You sort of speak, save two variables when you have a quadric and if you had a degree n polynomial, you would expect to save n variables. If you had like tons of, tons of variables. So anyway, in this case you expect like a lambda to the sixth many solutions. And that's what saving two minus like epsilon variables means. You can get easily, so number of solutions, ignoring skewness or whatever, is it always at most lambda to the sixth plus epsilon? Why? Well, let's fix six particular variables, fix everybody but these two variables a1 through a3, b2, sorry b0 through b2. In other words, let's just fix this part of the quadric. All right, so then the equation is the product of these two variables is some kind of constant that we've fixed. I'm going to hide the word non-trivial, but for those who notice this subtlety, that's how I get around it. So the number of divisors of a fixed integer is like a, like if the integer is n, it's like n to the epsilon, the divisor bound. Anyway, yeah, that's how many divisors there are. And a0 and b3 are both divisors of this integer. So there's n to the epsilon, many choices. This to the epsilon, many choices for a0 and this to the epsilon, many choices for b3. So that's how you get lambda to the epsilon, many choices for a0 and b3 given the other six variables. So the number of possibilities is certainly at most lambda to the sixth plus epsilon. Okay, so it's easy to save. It's easy to kind of almost get down to the correct asymptotic without doing any work. So that you get down to the number of solutions is the most like lambda to the sixth plus epsilon, whether or not the box is skew. And that's actually enough. You don't have to do anything if the box really was skew, because we're not actually trying to get the asymptotic for every single one of these point counts. We're just trying to get the asymptotic for the integral of these point counts. And yeah, let me just say if you imagine integrating, how to say this? Anyway, if you let me just say it quickly, if you are following and then if you're not following, it's okay. But if you imagine like integrating, yeah, if you imagine integrating here and putting a lambda to the, you know, 0.01 or something, then the integral is converging so quickly because these like ti's to the minus two are way more than you need for convergence of the integral. So the integral is converging so quickly, like power savings quickly, that even though you lost an epsilon on the asymptotic, you're going to gain it back from the integration. And so that's going to tell you that the part where the boxes are skew is inside the error term. Okay, so anyway, a little bit of a use of the fact that we're averaging the point counts. All right. So let me just keep moving because maybe that's a bit technical getting inside that. So the result is that you can run the barge of a whole argument in this family two. And we get that the average size of two summer groups of these elliptic curves, x cubed plus y cubed equals n, that average is also three. And I mean, I wrote the average over all integers, but just like in kind of all these geometry numbers type theorems, you can impose finally many Congress conditions no problem. So for example, you could say n is divisible by, you know, 1729 or something. Okay, or n is congruent to 1 mod 691. But okay, so now we're going to get back to this. Can you restrict to n prime? That's a fantastic question. I'm going to say no, not in this argument, certainly. And if you restrict to n prime, there's let me try it. Oh, man. Okay, I don't remember about like the exact Congress classes. Maybe I should avoid like 1 mod 9. No. Anyway, I don't know. But there's work of Sylvester, which does a dissent on these curves. And anyway, I'm going to avoid that question, but not in the argument. You can't restrict to n prime in the argument. Quickly, why you'd be trying to say count points on a quadric, but also the degree six invariant, you'd be like multiplying by Mobius of a degree six polynomial. And I'm scared of doing Mobius of any polynomial. That's not like I'm scared of doing Mobius of a linear polynomial, but not that scared, but definitely scared of doing Mobius of a degree six polynomial. So that's, that's kind of what you'd have to face. And that's the reason I think no, you can't face that at least at the moment in the argument. But that's a great question. And I'm going to just refer to you refer you to Sylvester's work and his conjecture, but I don't know much more than the fact that Sylvester is involved. So, okay, so, but even restricting to all integers or not restricting, but running over all integers, we get the average size of two summer is three. So let me just, we were trying to produce lots of ends for which this is a one. And two summer, the size of two summer. Well, it's a power of two always. It's like a one, two, four, eight, you know, these possibilities. And so we get some average of these powers of two, which is a three. But that's not good enough to produce lots of ones, because everybody could have been a two and a four, for example, you can have like 50% twos and 50% fours. And, and so we produced no just looking at this and staring at it, we produced no nobody who not necessarily anybody who has trivial two summer group. And therefore, we're not produced, we don't necessarily just by staring at this produce any of these elliptic curves having lots of elliptic curves having rank zero. So three is not small enough for the, for the conclusion. But it's, you know, this, this bound is the truth. So what are we going to do? We're now going to look at root numbers. Okay, so I'm going to explain the second, why we're going to look at root numbers. But let me just mention that, I think the right people to cite our Rorlich and really Alvarado the problem is, I don't remember where the accents go, but anyway, Rorlich and Verily Alvarado, which says that the work that work says that the root number of x u plus y q equals n our beautiful elliptic curves is I'm going to use garbage in a technical sense right now garbage two and three, which I'm just going to ignore times roughly times minus one to the number of prime factors, dividing n, which are two mod threes. So minus one to the to the parity of the number of prime factors, not priority of the number of two mod threes dividing in. And this is, let me just note without multiplicity, it really is the count of the number of primes. So for example, if n is square free, so that that multiplicity comment I just made, you can ignore that then finish square free, and we ignore the garbage of two and three, then this function minus one to the number of two mod threes. Well, that's just n mod three, that's, you know, here's another a better way to compute that function you compute n mod three, because prime factorize and you really are two is minus one mod three. So okay. So let me just say it like this, the root number is like n mod three. The real thing I should be saying is, because you take into account the garbage of two and three, if n is square free, and you can impose a congruence mod, I think nine. You see, I'm always, I always try to be safe by writing everything ridiculous 10 to the 10s everywhere, but I think you can impose a congruence mod nine. If not, you can impose a congruence mod this finite large quantity. You can impose a congruence mod nine that lets you, that forces the root number to all be plus ones or or be minus ones. Okay, so, but let's ignore the garbage of two and three, and then we can just impose a congruence mod three on n, so long as it's square free, and then the root number is forced to be plus one. So let me write, I wrote it here so we can restrict restrict to root number plus one curves by imposing congruence conditions. Okay, infinitely many congruence conditions, because yeah, you impose like n mod three, but then you have to impose the congruence condition n is square free. So that's infinitely many congruence conditions, because you're saying at each prime, p squared is not dividing you. Okay, but, but at least they're congruence conditions. So if I did the full, so to speak, full family of elliptic curves, oops, then what's the root number? It's like, if I remember right, it's like you take the discriminant of this, so some kind of polynomial in a and b, like a polynomial like a cubed and b squared, you take the discriminant of that and you look at like the parity of the number of prime factors of the discriminant of this elliptic curve. That's kind of a much crazier function to try to face than, than just looking at, or anyway, it's, you know, much crazier to face that than just try to impose the congruence conditions that n is square free. So as family, some, you know, a miracle has kind of happened. So now I'm going to talk about quickly why imposing these infinitely many congruence conditions is not really an issue, because they're congruence conditions. All right, so why? So imagine, so b is like, b is like, imagine my set of square free numbers ends. Forget about this congruence condition mod three, because I'm going to talk about, you know, why imposing Congress conditions is not such an issue. So just imagine b is like square freeze. So we got, we have to impose infinitely many congruence conditions. And I told you before that I'm going to scroll up, scroll back down a second. Told you before that in this average, yeah, I wrote n is running over z, but I could impose finally many congruence conditions if I so desired. And I would have still gotten an average of three. This three is not really, you know, it doesn't care about the Congress conditions. The three is coming from the one identity element, which is always there and like a two from the tamagawa number of a group. So I remember SL2 squared mod of mu two, the size of that mu two is the two. So anyway, that's where the three comes from. It does not care about the Congress conditions. So going back, so B is my kind of set of square free integers. I'm trying to average the size of two somewhere over that set of square free integers imagine. And I can imagine trying to impose only the first trillion congruence conditions. So I can cut off the kind of imposition of Congress conditions at some parameter t. Let's see how this affects the average. So the average, okay, it's a numerator divided by a denominator. The numerator I'm averaging, you know, a non-negative quantity. The numerator can only go up if I increase the set. Yeah, if I take the all square free integers, will they certainly sit inside the integers which are square free at the first million primes? Well, that's a larger set. And so the numerator goes up when I, yeah, so I can I can up or down the numerator by the numerator for the cutoff set. But I know count the denominator correctly. And the denominator is like with an epsilon of the denominator for the cutoff set. So what I learned is that the average of this two-cell mer or whatever non-negative quantity I want is asymptotically, you know, the limit of the averages of in this case two-cell mer over the cutoff sets. But all those averages are three. They don't move when I impose Congress conditions. So what I learned is that the average size of two-cell mer is also at most three over curves with constant root number. So root number plus one or number minus one, because to impose the fact that the root number is plus one is it's congruent conditions. All right. So why is that relevant? Well, it's relevant because of the theorem of I'm going to try to pronounce this correctly, Nicovarj and Dokchitzer, Dokchitzer, maybe Dokchitzer and Dokchitzer, I don't know. People, please correct my pronunciation after because I'm actually really interested in how to pronounce this. Anyway, Nicovarj and Dokchitzer and Dokchitzer, that the reason it's relevant is that the root number is controlling the kind of two-cell mer rank, the parity of the two-cell mer rank. It's a theorem, a very serious theorem. I wrote it very strangely maybe. So if the root number is like plus one, oops. Okay. So like, you know, this is like minus one to the minus one to the that is the root number. Hopefully that makes sense now. It's like something to find mod two. And this is like, you know, two-cell mer is a power of two, which power of two. So the parity of the root number controls the parity of the two-cell mer rank. Okay. Given this theorem, three is now small enough. Yeah, because I just explained if I average over a root number equal to plus one curves, which I can, for example, I can average over the family where n is square free and n is given like mod nine or something. If I average the size of two-cell mer, well, that's three or at most three. Anyway, it follows it's three, but and now in that family, the size of two-cell mer is not just any power of two. It's kind of an even power, even power of two doesn't make sense. Two to an even, even number. So one, four and bigger. Yeah. And if I have an average of ones, fours and bigger that manages to be three, well, there have to be plenty of ones. So that's why three is small enough. We kind of improved our situation by just doing this one bit of root number analysis. All right. So three is small enough. So I'm just going to say, hopefully it makes sense how we include. The average is at most three. Two-cell mer size is always one, four. In this family, it's either one, four or larger. And so since the average is three, there must be a decent proportion of ones. And because the two-cell mer size being one implies that the rank is zero, that means that these, there are plenty of n's in this family, but that's a positive proportion subfamily. Plenty of n's where the rank is zero and therefore there's no solution for of x cubed plus y cubed equals n in the rational numbers. All right. So that ends the kind of rank zero production. And now, how do we produce things of rank one? Let me just note, I could have also done the same argument with, root number is minus one. Since this average is at most three, with the minus one root number guys and their summer groups have size in now two, eight, so to speak, odd powers of two, if that makes sense. Again, three is small enough for them too because, you know, it's an average of twos, eights and bigger. And for the average to be at most three, there have to be plenty of twos. So what this teaches us is that there are plenty of n's for which the two-cell mer size is two. In other words, there are plenty of n's for which the two-cell mer group is a z mod two. We want to conclude that that means that there's a rational point on the elliptic curve. So I'm going to write this a little higher. So I'll do this slide. So two, shot two, zero. It's like the defining exact sequence of shot. So we conclude that there are plenty of n's for which the two-cell mer group is like a z mod two. If we knew that the take shot for average group was finite, then if we knew that this thing, the take shot for average group was finite, then this sort of thing would be a square as a group from like the castle state pairing being alternating. So this is like a square as a group. So how could it possibly contribute to only a z mod two? It would have to contribute at least a z mod two squared if this thing were non-zero. So if we knew that shot was finite, we conclude that this is zero and actually this z mod two was coming from the so-to-speak rank part. So we would conclude that there's actually a point. Yeah, I don't know how to prove shot is finite, but there are people who know in these situations how to prove shot is finite. And so we use a theorem of Ashe Brungele and Christopher Skinner that is as follows. So PB a prime number, it's really crucial. And these are like technical issues in these techniques. Not that I have any expertise whatsoever on them, but anyway, I'm going to mention the two key technical issues that they overcome. So let PB a prime number, PB being allowed to be two is the key for us, and they do allow it. And let E be cm with, well, okay, in our case, the elliptic curves have good reduction. So n odd gives it x cubed plus y cubed equals n, certainly has good reduction, reduced to x cubed plus y cubed is one. But the reduction is super singular, and that's a key technical issue in these techniques too. Yeah, you can count very easily the number of solutions to x cubed plus y cubed is one mod over f2, and you'll find it's super singular elliptic curve. So the reduction is super singular, the reduction at p, so p is r2. It's super singular, that's an issue. So okay, such that these two hypotheses hold. So the p-salmar groups, in our case, the two-salmar group is like a z-mod b. You can just ignore torsion because there's, in our case, there's not interesting torsion. So the two-salmar group is like a z-mod 2. And there's this extra condition that the image of this two-salmar group, now I have to mention these elliptic curves x cubed plus y cubed equals n are all isomorphic over q2 because, I write it strangely, but anyway, everybody is a cube in every unit in z2 is a cube. Just a fact, you know, everything's a cube in f2. So because of that, because that means all n's are cubes, you know, too adequately. And so because of that, they're all isomorphic to the curve x cubed plus y cubed is one. So kind of over q2, this thing is like some kind of constant group, size like 4 or 2, something like that. So yeah, just I want to note that because I'm going to say something about it in a second. So anyway, the image of this localization map at 2, so if you have an element of the two-salmar group, when you localize, you get something solvable, so it lands in here. That image is not contained in the image of the local two torsion, but whatever, that's some condition. Then they can prove, then they do prove that the rank is one, Shaw is finite, the analytic rank is one. And anyway, enough for us to conclude, so we certainly conclude that there's a solution. So yeah, now I'm going to conclude on this slide. I just want to say that this condition is kind of weird looking, but it's not such a problem for us because it's some local condition at 2. This group is not moving, this group is moving, and we can just prove because we weren't counting only salmar elements, we actually started counting pairs of binary cubic forms and then imposed congress conditions. So we can prove that instead of sieving to locally soluble forms, we sieve to, you know, we impose different congress conditions, we can prove that the image of this group echo distributes inside this kind of fixed guy. And so in particular, there are plenty of ends for which the image is not contained in this subgroup. That's all I wanted to say. I'm just going to end by putting this slide up. And thank you.