 for the invitation. So the last thing about this web seminar is that you have a much larger audience and you would have usually in the seminar. Before I start with the talk in earnest, let me indulge a little bit in the past. So what you see here is the top part of the title page of my very first published paper. I can see it was fairly short, six pages. You can see my mouse cursor here. And it appeared in Arifta Mathematik in 1992. So I wrote it while I was still an undergraduate in Munich. Cornelius Greiter, who was an assistant at the time, pointed me to this problem. So you can see it's about iterates of polynomials. You have polynomial x squared plus a with an integer a, you look at the iterates of this. And the question is about the Galois groups of these polynomials. And there was a conjecture I think by Odoni who conjectured that the Galois group, when you take the constant term to be one, so iterates of x squared plus one, it'll always be the maximal possible Galois group you can have for iterated quadratic polynomials. Concretely, this means that the nth iterates should have Galois group, the pseudo two subgroup of the symmetric group acting on two to the n letters. So two to the n roots of the nth iterates. And so Cornelius thought that I might be able to do something about this and turn out that he was right. So I did this. And so you can see the main theorem of the paper. So I was able to show that for, well, infinitely many integers, say integers that are positive and one or two mod four or negative divisible by four and such that minus a is not as clear. These Galois groups really are the largest possible ones. And of course, this includes the case that a is one. So the original problem. Now, at the time, this looked like a kind of isolated curiosity, maybe. And so the paper didn't have much of an impact at first. So I think got its first citation 10 years later, 2002. Um, but then, um, yeah, so two things happened. Well, yeah, the one is that people started to look at, um, I will read a lot of our presentation. So that's a term point, I think by natural Boston, you look at, um, how the Galois group of the rational numbers, for example, in this case, um, acts on rooted infinite tree. So in the case of these quadratic polynomials and the iterates, you get a natural action on an rooted infinite binary tree. And then you can look at the actual the Galois group. Um, and so one can phrase the theorem, I prove there that for, for these values of a, what you get is a subjective representation. So the Galois group gives you the full automorphism group of this infinite binary tree, which is maybe an interesting result if you look at it in this way. And, um, also there's this field of arithmetic dynamics that's become fashionable. Things was pushed down to some extent by Joe Silverman, um, which studies arithmetic properties of iterates of maps of an algebraic nature. So for example, polynomials and of course, questions about Galois groups or irreducibility, which is what we will look at in this talk, they naturally fit in there. And so this little paper of mine has been picking up citations at ever increasing pace in the last 10 to 15 years. And so last time I looked, and this was yesterday was just one citation short of making the list of my top five most cited papers, which I think find rather surprising. The nice thing is that what I did back then actually has some connection to things I'm doing now. And so what I'm doing now is mostly related to questions about rational points on curves and hyperliptic curves in particular. And, um, yeah, we'll see how hyperliptic curves come into this picture. So let me switch back to the present, um, and sort of define the notions that I'm going to use. And I want to say that if you have questions during the talk, please do not hesitate to raise your hand virtually or write in the chat. And then I'll try to answer them. So we want to look at iterates of polynomials for the polynomials, um, in particular. And then I mean, easy to see that up to some, um, FI linear transformation to conjugation by FI linear transformation, you can always write polynomial in this form, they just have x squared plus a constant. So in my paper, I used a for the constant. I'm switching here to C because it's of more usual notation. And I now see to be any rational number, not just an integer. So fc of x, this is polynomial, and then I can define the iterates. So I start with the zero iterate, which should correspond to the identity map. So it's just given by the polynomial x. And the first is the polynomial itself. And then the second I obtain by plugging in the polynomial into itself. So I take x per plus C or squared plus C. And then I continue. And in general, I have the n plus first iterate of fc. Well, there are two ways of writing this down. I can sort of plug in the nth iterate into fc or plug in fc into the nth iterate. And of course, I get the same result, but sometimes it's more helpful to look at it in this way. And sometimes it's more helpful to look at it in this way here. So I get a sequence of polynomials. And of course, the degree doubles in each step. So the degree of the nth iterate will be 2 to the power n. And then, of course, I can ask questions about these polynomials. So the question I was dealing with in this paper was about the Galois group of these polynomials in the sequence. In this talk, we will ask a kind of weaker question. So the question will not be whether we get the largest possible Galois group, but just whether the Galois group extensively on the roots or put another way, whether these polynomials are irreducible or for which values of the parameter C and which indices n is this nth iterate irreducible. So if I fix C, then I get this sequence of iterates of fc. And then looking at this way of writing the n plus first iterate, it's pretty clear that as soon as the nth iterate is reducible, then you have a factorization. And this factorization is inherited by the next one, since we just plug in something into the polynomial. And so there are basically two possibilities. One is that all the polynomials are reducible. And the other one is that they are reducible up to some point. And then the next one is they are irreducible, sorry, up to some point. And the next one is reducible. And then it has to stay reducible. So there's this kind of dichotomy. And well, in the case that not all the polynomials are irreducible, of course, one can ask at which point they stop being irreducible and become reducible. And then if you look at examples, you may come up with this conjecture. So as far as I know, this is compatible with all the evidence that's available. And this is sort of all the best possible conjecture that you can make in this context. And it says that if you only require the second iterate to be irreducible, then it will never happen that at some point later, in the sequence, the polynomial becomes reducible. So if the second one is irreducible, then all the later ones have to be irreducible as well. So why is this best possible? Well, it definitely does occur that, well, the first iterate, so the polynomial itself is reducible when you take c to be a negative square. And of course, you can factor fc. And then all the polynomials from the first one will be reducible. But it's also possible that the first one is irreducible and the second one is reducible. And it's not too hard exercise to work out for which ways of c this occurs. And this is this remark down here. So the first case is when the first iterate, the polynomial itself is already reducible. And the second case is when the second iterate is reducible, but the first is irreducible. So there's an infinite family of parameters where this happens. Now, but conjectually, these are the only two possibilities. And then as soon as the second one is irreducible, it should stay that way. Okay, so that's the topic. And maybe to take out the suspense immediately, I will not prove this conjecture in this talk, but I will improve what is known about it. Okay, so to be able to say something about this, we need to be able to detect if at some point in our sequence, we switch from irreducible to reducible. And to do this, we have this lemma here. So if we are at the point in the sequence where we have an irreducible iterate, so the end iterate is irreducible, but the next one becomes reducible, then this implies that the constant term of this next iterate has to be a square. So that we give a proof because it's not so hard and every talk should contain a proof, I think. So the first thing to note here is that our n plus first iterate is an even polynomial because we plug in x-vector c into the nth iterate. So it's a polynomial that only depends on x squared. And so if I think of the factorization into reducibles, and I substitute minus x for x in the factors, then up to changing the sign of the degrees odd, I get another irreducible factor of the same polynomial. So I get an evolution on a set of modic irreducible factors of my polynomial, the n plus first iterate. Now I can look at fixed points. So if I have a fixed point of this evolution, then there are two cases, the polynomial that's fixed is an even polynomial or it's an odd polynomial. And it's an even polynomial. Then it's a polynomial that depends only on x squared. So I can also write it as a polynomial that depends only on x-vector plus c. And then I have a factor here that's a polynomial x-vector plus c. So this gives me a factor of the nth iterate, which by assumption does not exist. If my fixed point is a proper factor, which it must be because I assume that the n plus first iterate is reducible. So I'm looking at the irreducible factors, and so the whole thing cannot be an irreducible factor. When my fixed point is odd, the only odd irreducible polynomial is x itself, and then for degree reasons it follows that x squared has to be a factor, and then I can argue in the same way. So whenever I have a fixed point, I get a factorization of the nth iterate, which I don't have by assumption, so there are no fixed points. If I have no fixed points of my evolution, then all the orbits have size two, and so I can pit one representative of each orbit and take the product, and I call this h. And then the other percentages I get by substituting minus x for x. Take the product, I get h of minus x, so one integer thing here is one has to note that the degree of h is even because the degree of the nth first iterate is at least four, power of two that is at least four. So I get h of minus x and not minus h of minus x here. And so the actual is that I can write this polynomial as the point of h of x and h of minus x, and then I can just plug in zero to get that the constant term of f n plus first iterate is the square of the constant term of h, and this proves the lemma. So I have a way of detecting when I switch from irreducible to reducible, but I should add here that this is only one direction, so it's not an equivalence, but so it gives me a necessary condition for switching from irreducible to reducible, and therefore a sufficient condition for staying irreducible when I turn this around. So here's an example that shows that it's not an equivalence. If I take for the parameter c, the value one-third, and look at the second iterate, I get this polynomial here, which has square constant term but still is irreducible. Okay. So if we use this to show that, well, if the second iterate is irreducible, then some others also have to be. So what is known about this? So we have to look at the constant term of these iterates. That's why I introduced this notation. So capital A n of c is the constant term of the nth iterate of fc. So starts with zero. I mean, the zero iterate is just x has no constant term, then f itself has constant term c, and then I get c squared plus c, and so on. And I mean, I can define it by this recursion here. And then the question is, well, is this, or for which c is this a square? And so the first, well, I will assume that the second iterate is irreducible. So the first interesting case is when n is two in the lemma, which means I have to look at the constant term of the third iterate, which is a three of c. And yeah, it's, I'd say something about this, what, in a few seconds. So one can show that a three of c can be a square only when c is zero. Of course, when c is zero, then all the a's are zero and are squares. And a four of c's is square only when c is zero or minus one. Yeah, when c is minus one, then all the even ones are zero and therefore squares. Yeah, how can one see this? So for a three, I mean, from what I've written here, you can see that a n of c is a polynomial in c with integral coefficients of degree two to the power n minus one. So for a three, I get the polynomial of degree four. And so this equation here defines the genus one curve. And it has special points, so it's an elliptic curve. And then one easily can determine that it has rank zero and five is racial points. And so I mean, it has three racial points, but two of them are at infinity. And so the only racial point is not infinity is the point that corresponds to c equals zero here. For a four, we get the polynomial of degree eight. And so we get, I mean, this equation is y squared equals the polynomial x defined hyperbolic curves. So when the degree here is eight, then this is a curve of genus three. And yeah, one can look at the Jacobian variety and the being variety of dimension three in which I can embed the curve. And so the group of racial points on this being variety is a finally generated to be included by model A. And it turns out that this group is finite. So there are only finally many racial points on the Jacobian and embedding the curve. They only find the many racial points on the curve. And I can find them by looking at the pre-images of the finite many points on the Jacobian. And it's not too hard to figure out that the only points here are given by these two x coordinates. So this is in a paper by Rave Jones, Nate Hins and a few other people. And at some point I also joined as a co-author. So I mean, you can find it in the archive and it also has appeared by now in the New York Journal of Mathematics. So this implies that when we have the parameter c such that the second is irreducible. And I should mention that when c is zero or minus one, then this is definitely not the case because then fc itself or f zero is x squared and f minus one is x squared minus one, they are both reducible. So in this case, all the polynomials are reducible. But if it's irreducible, then c is not zero or minus one. And then by the lemma, or then I mean, this proposition tells us that a three of c and a four of c both are non-squares. And then by the lemma, this implies that third and the fourth iterate are also irreducible. So it can go from two to four. And the main thing I want to do in the talk is to go one step further. So from four to five. And this basically means that I have to figure out for which there is of c a five of c can be a square. And the result will be that this only happens when c is zero. Again, now similarly to a three of c. Okay, so that's the goal. Now a five of c is a polynomial of degree 16, but it has no constant term. So maybe I just go back quickly to this slide here. Obviously, I mean, fc c squared plus c and then the square of something that ends in c plus c will always have zero constant terms. So it has a factor of c. And there is more convenient to work with an equation for the amplitude curve for the polynomial on the right hand side says odd high right hand side has odd degree. And so I basically just turn it around replaced a polynomial capital A five by lowercase a five bridges x to the power 16 times capital A five evaluated at one over x. This is now a polynomial of degree 15. Because the constant term of capital A five is zero. So the degree drops by one. And then the question is equivalent to to this. So I mean the points where C zero for y square equals a capital A five of x now move to infinity and the points that were infinity before now now move to zero. And so I have to point that infinity on the curve that comes from the capital A five. They move to zero. So they will be there. And then the other point moves to infinity. And what I want to know is that there are no other points. Yeah. So if my lowercase a five I can write in this form, I mean, it just comes from the recurrence. And so I'm looking at this curve, which I call a C was that equals to a five of x, which is this polynomial test degree 15. So you get, you know, seven here. And I want to know what the rational points are on this curve. The usual way of of trying to do this is to look at the Jacobian variety again. I mean, I've told you that for a four, they have this three curve and it's a common variety and it turns out that the group of rational points on the Jacobian variety is finite. Um, in general, when the rank of this group is finally generated within group of rational points on the Jacobian is strictly smaller than the genus of the curve, which is the dimension of the Jacobian, then you have a chance of applying Shabuti's method. Shabuti proved more dense conjecture that they're only finding many rational points on the curve of higher genus. In this special case, when the rank of the model of a group of rational points on the Jacobian is less than the genus. And one can make this approach at least say effective in practice. But this requires some some knowledge of the model of a group. So the first thing one has to do is one has to figure out what the rank is, or at least show that it's smaller than the genus. And the way of doing this is that one finds an upper bound for the rank by computing a certain suitable Selma group. And for hyperliptic curves, for the curvature of hyperliptic curves, one can compute the two Selma group relatively easily. And so in this case, if we do this, we get an upper bound for the rank, which is two. So we know that the rank of the group of rational points on the Jacobian is at most two, and two is less than seven. So in principle, Shabuti's method should work. When there are general results, it would allow us to deduce a bound on the number of rational points from this information. But this will not be enough. I mean, what you want to show is that this curve has only three rational points. They are the one point infinity. And the two points, when x is zero, then I mean, the right hand side simplifies to one. And I have two points, one with y equal to one, and one with y equal to minus one. And I want to know that there are no other points. But the bound that I get just out of knowing that the rank is at most two will never be three, but always be larger. So to really get at the points, I would have to know what the rank is, not just an upper bound. And then I would also have to know the right number of independent points in this group to do the computations I have to do. But the problem is that I cannot find points of infinite order in this group. So I mean, yeah, this is a seven-dimensional object. And searching for points there is not so trivial. And also, I just have this upper bound. So it's perfectly possible that the rank is actually zero. But I have no good way of proving this. So I'm sort of stuck here because I cannot go further and apply the method in the usual way. So that's what I mean by standard shabu-t techniques here. I also have to do something different. And this is this so-called Zener group shabu-t method that showed up in the title, which I will explain on the next slide. But first, I want to note that, I mean, the three points we know on the curve, the three racial points, the point infinity and the two points with x squared at zero, they all have the property that if I embed the curve into the Jacobian in the usual way here, where we take the point infinity as the base point, then, well, of course, the point infinity, the base point will map to the origin. So it has order one and the other two points also have finite order and the order is odd. I mean, it's 15 as one can figure out here. So we have three points of odd order. And in fact, these are the only points of odd order on the curve. Okay, now, what's the idea for the Zener group shabu-t? So I mean, basically, we don't have points in this monorail group. We do have the Zener group. And so basically, we would like to use the information we have. So the Zener group has a kind of proxy for the monorail group. And so we look at the following diagram. So for the time being, I mean, you don't really have to know what the two Zener group is, only that there's a map like this. So I mean, they have the curve, the racial points on the curve, we have an embedding of the curve into the Jacobian given by mapping the point infinity to zero. So this also sends racial points on the curve to into the racial points on the Jacobian. This is a group. So I can take the subgroup that consists of doubles of elements and look at the quotient. And then the Zener group, the two Zener group, there's an injective homomorphism from this group into the Zener group. So the two Zener group is a finite, a billion group that's killed by two, so vector space over the field of two elements. And so this is just the f2 linear map here. Then the next step is to look at the same thing over the two added numbers. So of course, we still have the embedding of the curve. We have the group of q2 racial points on the Jacobian, and so this quotient. And the Zener group, the two Zener group has a natural map into this quotient. So not just for q2, but for every completion, but we only needed, every completion of q, we only needed for q2 here. Now, the q2 points of the Jacobian form a two-adic league group. So I mean, I can do analysis over the two-adic numbers. And this two-adic league group has a normal map that I can arrange so that it has imaged the two-adic integers to the power g, g is the g is seven, in our case. And then I can, I mean, I can take the reduction module two of something in here, I get something f2 to the g, and sort of completing this query here, I get the kind of reduced logarithm map on this quotient. I'll just take it pretty much here, take a look at the reduction. So if we work with this diagram, and so the sketch of the proof is like this. So we need to know that this map sigma here is actually injective, which we can check because we know the Zener group, so we can compute it. And then our goal is to show that the set cq odd that I defined on the previous slide, which consists of the three known points, almost all of the set of virtual points. So I want to show that if I have a virtual point, it's not one of these. I get the contradiction, the point cannot exist. So how can I do this? Well, I take my point p here, I map it into the Jacobian, so this is i of p. In the simplest case, this point maps to something non-zero here, which means that it's not divisible by two, and then it will map to something non-zero here because it's injective, and also here because sigma is injective, and I get something non-zero here, hopefully, and I can check that what I get here is a lot in the image of stuff that comes from the curve down here. That's the idea, but in general, of course, it's possible that my point is divisible by two, and then I would get zero here, and this would not be sufficient to compute something. So what I do is I divide the point by two as many times as possible, so n is maximal, and this makes sense because the only points in the Jacobian that I can divide by two arbitrary often are the points of odd order, and these are the points I have excluded here. Yeah, I mean, I didn't write it on the slide, the previous slide, but these three points are the only points on the curve that map to something of odd order in the Jacobian, but any other point will not be a point of odd order on the Jacobian, and so can be divided by two only finite number of times. Now I can play the game I sketched before with Q, so Q is the point here, and basically I map Q from here to here in the two possible ways, going this way or this way, and then going this way, I see that it has to come from some element in the same group, so it maps to something that's in the image of this map, composed with sigma. On the other hand, yeah, so that's what I know, and the logarithm of Q is the same as two to the minus n times the logarithm of I of p, because the logarithm is the linear map. And now to get a contradiction, I consider points, Q two points on the curve, map them over here, write them in this form, and then figure out what they map to here. And if I can show that no matter which point I take here, which is not one of these points of order, of course, ends up here in some element that's not in the image of this map, then this shows that, well, since every rational point also is a Q two point, this shows that these rational points cannot exist. Now, so that's the idea of the proof, and so we use the two same group here, which captures some information on the model Baye group, to replace the model Baye group, which you don't know explicitly enough in the Schabutzi argument. Okay, so I have to say a little bit more about the same group, and how we compute them. So we have this polynomial of degree 15, which is on the right hand side of our curve equation. It turns out to be irreducible. So I write theta for a root, and then I take the number field that's generated by this root, which is a number field of degree 15. Yeah, then for every field extension of Q, which I call L here, I can look at the rational points on the Jacobian and this quotient, which we have seen for Q and Q2 on the previous slide, and then there's a certain map, which I can write down, which I will tell you how I can define it into, well, K is the number field here. So I take L tensor K, and then the multiplicative group modulus squares. So this does not have to be a field, but it will be the direct product of fields. And the point is that the two cinema group can be identified with a subgroup of this when I take Q here. So then I just get K. It's sort of L tensor K, so subgroup of the square classes of K, such that when I take the obvious map to the same thing, when I take QV here for a completion of Q, so I take K tensor QV over Q, the square classes in that I have this map over QV. And then the cinema group, there's the subgroup of this consisting of elements that map to the image of this map for QV, so that I sort of locally in the image. Now, I mean, this is a condition for infinitely many completions. And so if I want to compute something, I have to cut it down to something finite. And the way to do this is to say that the elements in the image or the elements that belong to the cinema group have to be unarmified in a certain sense at most primes. And in this case, so because the discriminant of this polynomial is odd in square three, and the class group of this number field is trivial, as one can show by computation, one gets that the two cinema group actually sits in the square classes that come from units, also set of just elements of K. And the only condition that I have to satisfy then is the, essentially the condition that comes from the completion at two and at infinity. So if I remember the previous slide, one thing they had to check is that the map sigma from the cinema group to the group jq2 mod 2jq2 is injective. And this is something they can sort of check on the right hand side of these maps. So we know the cinema group is contained in here. And if I take q2 here, I get a similar thing on the right hand side, which is just this ring. So two electric integers. And then I have to join theta take the units when you lose squares. And so I can just check that this map is injective, that's a finite computation. And this implies that sigma is injective as well. Okay, but I still have to figure out what my image is of a q2 to really determine the cinema group. And so maybe that it's a little bit technical, but that's a way of writing down this map pretty explicitly given a certain value of a presenting element of the Jacobian. So maybe I don't go too much into detail here. I mean, if you want to know more, you can ask. But so one, one thing I have to do is I have to figure out what the image is of jq2, or 2j of q2 in this group here and z2 replacing L here. And so basically what I need is the basis of this group here as a vector space over f2. And because the curve has a fairly special shape, it's possible to write down such a basis. So if I take this polynomial, it's roots as x-coordins of points on the curve, and then pick suitable y-coordins and take the images of these points in Jacobian, and then I take their sum, I get the point in jq2. And if I do this for all these polynomials, I get the basis of this. So it's possible to show this by some very explicit computation. And this allows me to figure out what the image of this map delta q2 is and therefore what the two-seller group is. And in this way, I see that the two-seller group has rank two, and this gets really bound on the rank. But I also have an explicit representation of the two-seller group that I can use for computations. Okay, so they have this, and then they have this logarithm map. So we need to be able to compute this. And by the logarithms, the logarithm is obtained by integrating differentiations on the curve or invariant one-forms on the Jacobian. And so one way of doing this is that we define the logarithm by integrating differentials from this point infinity to the point p. So one needs a periodic integration theory for that, which does exist. And which I cannot go into further here. So this defines my logarithm on points that are images of points on the curve. And then the image of the curve generates the Jacobian. And so I get it on a Jacobian by extending it linearly. Okay, now, I mean, our goal is to show that the three points we know are the only three rational points. And the way we will prove this is we will look at each of these points and then show that there's no other point that's too adequately close to it. So we begin with this point p0, which is the point 01. As I said, it's a torsion point of odd order even. And so the logarithm is 0 because the logarithm is a homomorphism into a torsion free group. Now, near this point, we can expand this map log as a tuple of formal power series in the uniformizer at this point, which we can take to be the x coordinate. And then I mean, looking at this, one can show that one can determine the ways of convergence of this power series. And so this process converge when the two edit valuation of the x coordinate is a little bit larger than zero. So definitely for all rational numbers, that whose numerator is divisible by two. And using this, because the points that are involved in this basis of jq2 over 2jq2 that you saw on the previous slide, they all satisfy this condition here. They can just compute the logarithms by evaluating the power series. And this way, we can determine the basis of, so I mean, the other way I have to find it, the logarithms maps into q2 to the power g. But by evaluating the logarithms at this basis, representatives of this basis, they get a z2 basis of the image. And this allows us to identify the image of the logarithms with z2 to the g. So now because the points that are close to p not in the two edit sense. So I compose the logarithms with isomorphism in my head in the previous slide to get the version of the logarithm which I now call log prime that surjects onto the z2 to the g. So that's the logarithm that was in the diagram on the slide before. And yeah, then I have to figure out the image of this map log bar composed with sigma. And I can just do that. And taking the suitable basis of my space here, I have these two explicit generators. So this is an f2 vector space, g dimensional, the fg tuples of elements of f2, and both these generators have the property that the first and the last coordinate are 0. And so this is the only thing that we need to remember here. And then, yeah, so computing this power series that gives this modified logarithm, so it's the ocean logarithm, and then I multiply by some matrix. You can compute this and we get this up to o of two squared terms. So we have a t squared here and then everything else has more powers of two and higher powers of t. So this is, I mean, I'm looking at points that are we'll have the same reduction as p0 mod 2, so they have an even x coordinate. So I write this as two times t. And so, well, I have to work out what the images are for all these points. So when t is odd, then I have something odd here, something odd here, and everything else is even. So I get this point if I reduce mod 2, 1, 1, and then zeros, but it has a one in the first component, so it's not in this image. And then t is even and nonzero. Then, whether I have this t here and everything else in most of these t squared, or two times t, or actually four times t. And so this term here will be the one that has the smallest triadic variation. And if I divide by the highest possible power of t, this gives me something odd and everything else is even. And so I get this element, and I reduce mod 2, and still has a one in the first component, so it's not in the image of the same apple. And this shows that there's no rational point other than the ones with x coordinate zero that has positive triadic variation of the x coordinate. So that is with these points, and then we do basically the same kind of thing with the pointed infinity, and then we still have to look at points that have odd x coordinate. So let's do that, but it will be, I mean, yeah, so for the points infinity, it will be very similar. So by this, I mean points that have negative triadic variation, so the x coordinate has a two in the denominator. I can write down a uniformizer at this point infinity, which is this here. And then again, write down my log of the midterms of power series in this uniformizer. And well, this gives the following. So I get stuff that involves of higher order things. And then the last coordinate, I get minus t plus 2t squared plus something divided by 4. And so this makes it pretty clear that regardless what t is, this last term will be dominant. We have the smallest triadic variation. And so if I divide by power of 2 and reduce, I will always get this element. And this is the one in the last component and therefore will not be in the image of the same apple. And so by the same reasoning, this shows that the point infinity is the only universal point that has x coordinate with negative triadic valuation. So this leaves points where the valuation is zero. And then by looking at the equation module eight, you see that this can only be the case when the export is going to minus 3, module eight. And all these points give the same image under delta q2, which is not zero. And this is not in the image of the same apple. So this shows that these points also cannot exist. And this proves what we want to prove. So the only point coordinates are the points with x coordinate zero. The only point with x coordinate has even denominator is the point infinity and there are no other points. So this finishes the proof that A5 of C is a square or even C is zero. And this extends what we know about the conjecture from the beginning by saying that if the second iterate is irreducible, then the fifth iterate is also irreducible. And now we want to go a little bit further. So what about the sixth, seventh, and so on. And yeah, so one, one idea here is to reduce the question whether A n is a square, the question whether A m is a square for a divisor of m. So then I have a divisor of m, and A n of C is a square, then A m or minus A m has to be a square. And if m and n have the same parity, then you can get rid of the minus sign here. So this allows us to reduce the case n equal to six, basically to well n equals three, and then the combination of n equals three and n equals two. So if I take n equals six, n equals three, then A six is a square. I know that A three is a square or minus A three is a square, but A three is a square when even C is zero. And for minus A three, I also use that A two of C has to be a square and combining these two things, I get a genus two curve of rank one. And it's by now pretty easy to figure out what the rational points are on such a curve. And I mean, this gives the point minus one here. The next one is seven. So seven is a prime number. So they cannot use divisors of seven. So we have to work with this curve, y squared equals A seven of x directly. But in this case, it turns out that the two same group is trivial, at least assuming the generalized Riemann hypothesis, because the polynomial on the right hand side of the curve now has degree, well, if I take the little a seven as degree 63 is irreducible. So I need the class group of a number field of degree 63. And this is only feasible assuming grh. Now, but otherwise, I mean, if I assume this, the seven group is trivial, and therefore the group of rational points is finite. And we can argue in the same case as for A four, if you remember, so we have just finding many points on the Jacobian, which we can pull back to the curve and see that this is the only possibility here. Yeah, eight and eight, nine, we can deal with using three and four or four and three. And then 10, we have to look at the case that we have that minus a five is square. And so we have to look at this curve, which is a quadratic twist of the curve we have been looking at as a morphic or to it over to a joint I and I mean, we can use that to do something similar, which I will not go into here. What the upshot is that if you assume that the second eternal is irreducible, what we've done here, so this and this implies that the sixth eternal is also reducible, and then assuming your age to get to seven, we can even go up to 10. So the 10th eternal is irreducible. And then we're going up further. So one would have to look at a 11. But this is a polynomial of degree 1024 or 1023. If I, if you have turned it around, to get something of odd degree, and it's irreducible. So we would have to work with a number field of degree 1023. And this is completely impossible, at least with the current technology. So this is what we can do regarding this conjecture. So I guess I should finish, but let me just mention this that the same kind of method does work for curves or more generally for curves of this form. So if you remember the expression we had for the right hand side, which was x to the 15 plus x to the seven plus blah blah blah square. So we had a square here of lower degree than this. And the concentration was one. So we only needed its odd because we do things too adequately. Then one can try to use the same kind of approach to show that the curve has only the obvious three points. And for this, it's essential that one can write down this basis of this group here. And just as an illustration, I've implemented this and then run this on lots of curves. For example, I've taken curves of genus five with coefficients of h between minus three and three and h of zero or h of one even, positive leading coefficients efficient of h, sorry. And then while you can see that most cases the method is successful, it shows that these are really only points. And then the remaining cases split well roughly evenly into curves that actually have more points and then curves where it's not successful. Okay, but that's that's all I wanted to say. So thank you very much for your attention.