 Thank you very much for the invitation. This talk is joint work or based on joint work with Nunufleitas, Alan Klaus and Haluk Sengun. And the title is the Fermat equation and the unit equation. So to start us off, everyone knows the statement of Fermat's last theorem proved by Andrew Wiles says that this equation here has no non-trivial solution. So every solution has to satisfy ABC is equal to zero, where if you take your ABC to be a rational numbers and the exponent L to be at least three. You may also know this other theorem by Jarvis and Meakin. They proved Fermat's last theorem over Q2. And you notice that there's a slight difference apart from the field. There's a slight difference in that their exponent starts with L equals five. So they don't allow L equals three. And the reason for that is that this equation has solutions for L equal to three when you take your field to be Q2. So you can ask what should the statement be for a general number field? What should the conjecture be? And here's an attempt at formulating a conjecture. So this is, I've called this the asymptotic Fermat conjecture and it says that if K is a number field and it does not contain the cube roots of unity. So Zeta three is a primitive cube root of unity. Then there's a constant BK such that if your exponent L is bigger than BK, so you take a prime bigger than BK, the only solutions to the Fermat equation with exponent L and unknowns belonging to K are the trivial ones that satisfy ABC equal to zero. Okay, so I'm going to call that this, the asymptotic Fermat conjecture or AFC for short. And you might be wondering why we have this restriction on the field, cube root of one is not in K. And so as you know, if you add the cube roots of one, you get zero. So as a consequence of that, for every L bigger than three, you're going to have, if your cube root of one is in K, you're going to have a non-trivial solution to the Fermat equation for all large, for all exponents bigger than three. And so maybe the statement of the conjecture should be, let K be a number field, then there's a constant L bigger than, there's a constant BK such that for exponent L bigger than BK, the only solutions to the Fermat equation are either the trivial ones or permutations of this. Okay, so the talk is going to be about the relationship between this conjecture and the unit equation or the S unit equations. I'm going to get back, get to that eventually and attempts at proving not this conjecture and complete generality because it's unrealistic, but proving this for some interesting infinite families of number fields. Okay, so to start us off, we're going to have a crash course on what you might call the modular approach to the Dierfantian equation or the proof of Fermat's last theorem, but in a slightly broader context. So we look at not exactly the Fermat equation, but the Fermat equation where we've put in an extra coefficient, so P is going to be some fixed prime, some fixed odd prime and yeah, I'm going to look at this equation and what I'm trying to do is to apply the strategy of the proof of Fermat's last theorem to make deductions about the solutions. Okay, so it's following Sehr and Mazer. Now, the first step you do and you do this because it makes things slightly easier later is you take the three terms and one of them we're going to call capital A and one of them we're going to call capital B and one of them we're going to call capital C, but we permute them in a way so that the middle one, capital B is the even one and because everything is co-prime, the other two will have to be odd and one of them will have to be minus one mod four, so that's my A and the one that's one mod four will be my C. So we do this permutation and then we write down this elliptic curve, it's called the fly curve that yeah, okay, so there is a mysterious step that happens after this or several mysterious steps. So theorems of Mazer, Wiles and Libit tell you that somehow this elliptic curve that we've built from an unknown solution to this equation is related to a modular form, so an U form of way two and level two P. So there's no point for the purpose of this talk, it is to sort of delve too deeply into what is going on here. So we'll take this as a sort of a starting point and see what it tells us. Now, here we have a congruence of two Galois representations, but I want to get rid of this relationship straight away and write down a more elementary consequence of this. So by the way, this is the hardest part of the talk after a couple of slides will be reduced to doing very easy manipulations, okay. So this relationship, whatever it means, it implies a congruence between the traces of Frobenius of this fry elliptic curve and the coefficients, these are the coefficients of the cusp expansion of F. So I get a relationship for every prime Q that I choose, I'm going to get something like this. And what's lambda? Lambda is some prime ideal that divides my own energy L and it's a prime ideal of the ring of integers of the field of coefficients of this uniform. Okay, so to summarize, if I pick a P and I look at this equation, somehow a solution is associated to a new form of weight two and level two P and this is the relationship and there's only finitely many new forms of weight two and level two P. So if I didn't have the P, so I'm just looking at the Fermat equation, there would be no new forms of weight two and level two and I would say contradiction and the proof of Fermat's last theorem, that's the end of it. But in other cases, in some other cases, you're going to have finitely many new forms, you can compute them by algorithms of Climona and Stein and you can say, well, now that I know the new forms and I know their coefficients, what do they tell me about the solutions? So here's what happens next. This prime ideal lambda, which I said is a prime ideal that is above L, it's going to divide some explicit constant that I build from F. For every prime Q, I get this constant here. So because, well, you're in one of two cases. So if you're in the second case, then it divides either this one or that one. And if you're in the first case, you'd say, okay, AQF, I know it because I have the new form. But I don't know what this is, this integer is, but this is something in, I know that this is something in Z and it lives in the Hasse interval. So it's an integer whose absolute value is less than twice the square root of Q. So lambda must divide, divides this difference, so it divides one of these differences. So I've got this number for every prime Q I choose. I can calculate it and I know that lambda divides it by for some mysterious reason. And because lambda is also divides the prime L, what's L? L is the exponent in my original equation. I know that L divides the norm of this. So this is living in a number field that's generated by the coefficients of L. L divides the norm of this thing. So this is something I can calculate and you might think about, you look at this and you say, oh, great. I've bounded the exponent L in this equation. I have a bound now because it divides some norm. And that would be true if this number here is non-zero. So you can say, well, when should it be non-zero? So one situation it should be non-zero is if this AQF, if I have a prime Q for which AQF is an irrational number, okay? It's not in Z, it's an irrational number because here I'm subtracting something in Z from something not in Z. I will get something that's non-zero and likewise here. So this would be non-zero. So the conclusion is if F is irrational meaning that for some of its coefficients are irrational numbers, I'm going to get a bound, an explicit bound on L which is the exponent in the equation that I'm working with. And so I'm reduced now. So I say, good. Well, maybe I no longer want to solve this equation. I want to do something realistic. I just want to bound the exponent. So if I'm bounding the exponent, I'm now happy to delete all the irrational new forms from my list. And I'm left with the rational ones. And rational new forms, so by something called the Eichler-Shumor theorem. So this is the converse of the modularity theorem. Modularity theorem tells you you start off with elliptic curve, there's an associated eigen form with rational coefficients. Okay, Eichler-Shumor, it's the other way around. You start off with this modular form with rational coefficients. It's associated to an elliptic curve. So there is an elliptic curve E. It's not the same as our fly curve, something else, but it's something that you get from this rational new form and it will, its conductor is equal to the level of the new form. So it's going to be 2P. And because I've started with an explicit, with a fixed P as in my coefficients of this twisted pharma equation, I can go away and look in Cormoranus tables and list, let's say, all elliptic curves of conductor 2P and ask, what does that tell me? But you can take the argument a little bit further, this kind of argument here. And you can show that there's an even bigger constant, so it doesn't matter what we call it, an even bigger constant such that if L, if the exponent L is bigger than this even bigger constant, then you can assume that the elliptic curve E not only has conductor 2P, but full two torsion. So if I don't have full two torsion, then I'm going to find some prime Q for which this constant is non-zero. And I'm going to get a bound on the exponent L. That's how the argument works. Okay, so now you're sort of reduced to saying, if I want to get a bound on the exponent L, I have to rule out elliptic curves of conductor 2P and full two torsion. And well, let's try to write them down. So here's how you do it. You say elliptic curve with full two torsion, it has a model like this where U and V are integers and the discriminant, because we have conductor 2P is going to be a power of two times a power of the prime P and yeah, but the discriminant has this form here. It's 16 times the discriminant of this cubic polynomial. And so you say, well, U must be plus or minus a power of two times the power of P, V must be a power of two times the power of P, U plus V must be a power of two times the power of P, up to side. We take these two and we substitute them in here and we get an equation. And now you can go away and solve this. Okay, it's very easy. And it's an easy exercise to show that when you solve this equation, the prime P has to be either a mercenary, or you can substitute them in here. And the prime P has to be either a mercen prime or a Fermat prime. Okay, because, you know, mercen and primes and Fermat primes are one away from a power of two. So you can see that they're going to give you a solution to this equation. Let me just say that this equation has other solutions. For example, one plus one equals two. But if you work backwards, if you take the solution one plus one equals two, you put it in here, you're going to get something like conductor 32, not conductor 2P. So although we've reduced to solving this equation, we are not interested in all solutions. Not all solutions when we go back will give us conductor 2P. Okay. And so here's a theorem of Sarah and Mazer, which when this was proved, it was assuming Sarah's modularity conjecture, but it's now unconditional. If P, you know, thanks to theorems of Wiles and the Bit, or thanks to the proof of Sarah's modularity conjecture by Cary and Vandenberg. So if P is neither Mersenne nor Fermat, then in this equation here, the prime exponent is bounded by some explicit constant, which is going to be something like, I mean, this was worked out by Alan Klaus and something like P to the 12P, if I remember correctly. Okay. So as a summary, if I have a Fermat type equation and I want to use this Fry curves and Galois representations of elliptic curves and modular forms to say something about it, then you end up having to do one of three things, either computing all new forms of a certain level or computing all elliptic curves of a given conductor and full two torsion, or computing all solutions to an S unit equation. But well, maybe the word all is wrong here for the last one, because as we saw, okay, well, what's an S unit equation? We'll come back to that. So here is an example of an S unit equation. So I'm going to define it in a minute, but notice that here we weren't interested in all solutions. We were only interested in some solutions that satisfy some further conditions, because when we work backwards, not all the solutions to the S unit equation are going to give us elliptic curve with the correct conductor. Okay. So what's an S unit equation? Let's talk about that. So here's the general setup. You have a number field K, S is a set of prime, I should say finite set. S is a finite set of prime ideals. And I have this thing called the group of S units. So this is the notation, OS star. It's elements in the multiplicative group of the field that are only supported on the prime ideals appearing in S. So if I take any, if I, it's alpha, such that the valuation at any prime, not in S is zero. Okay, so if I take alpha and I think of it as a fractional ideal, I factor it, I'll only see things in S. And the S unit equation is this equation. So epsilon plus delta equals one, where epsilon and delta are S units. And so here's a very simple example. If you take your K to be Q and S to be two, then the S units, well, they're the numbers whose, the rational numbers whose valuation is zero at all odd primes, the primes not in S. So you get powers of two, up to, up to side. And if you look at the solutions to this equation in powers of two up to side, well, you get three solutions, half, half, two minus one, minus one, two. And if you take, okay, if you throw in another prime and odd prime P, then your S units are these, and the solutions to the S unit equation will be these three, plus if P is a firmer or mercent prime, the things that you get from that. So firmer prime minus a power of two is one. All right. So more of a sort of a little bit more background on S unit equations. So by the way, we finished the hard part of the talk. So everything is going to get much easier now. But, okay, so we do know, thanks to a classical theorem of Siegel, that S unit equation has only finitely many solutions. And there's this really beautiful theorem of Iversa. This is I think the most famous result in the subject, which is the number of solutions to the S unit equation is bounded uniformly in terms of three things. So it's the R1 and R2, that's the signature. So the number of real and complex embeddings. And S is the number of prime ideals in the finite set S. Okay, so you have this bound. And there's other bounds, there's bounds on the heights of solutions. For example, you to be Joe and Giori, but they take more notation to write down. So I've missed them out. That's a famous, famous result. And there's algorithms. So if you start off with an explicit field K and an explicit set S, there are algorithms that find all the solutions to the S unit equation. Starting with D-Vegas. So these, you know, D-Vegas algorithm. And I think everything that follows it uses essentially Baker's bounds plus LLL to compute the solutions. And this is still, you know, a lot of progress with these algorithms have been made in the last few years. So, but things are still happening. So here's a theorem which really surprised me when I saw it. And it surprised me because it's, the statement is very simple, but in a subject that's 100 years old, you don't expect a new discoveries of this generality and elegance. So this is a theorem due to the Antifolu. And it says if you have a number field K, N is going to be the degree. And there's two assumptions. One is that three doesn't divide the degree. And the second one is that the prime, the three splits completely in K. Then the unit equation has no solution. So here the unit equations, S is the empty set. Okay. So my OS star that I had before is the units of the field K. It has no solution. And notice that somehow to the Antifolu theorem is saying I have a restriction on the degree. I have a local condition on the field K, not on the unit equation, just on the field K. And these two together, they tell me that the unit equation has no solutions. So one of the things that we're going to be doing a little bit later is looking at similar theorems. Okay. That has been discovered since of this type. Okay. Restriction on the degree plus local condition imply that the unit equation has no solutions. I'll also say that I believe that, and maybe some people might correct me, but if you fix a number field and let's say, you know, of degree five, so the first can, well, let's go a bit higher. Let's say degree seven. So the first condition is satisfied. And you look at the set of number fields of degree seven where three splits completely, I believe that set should have positive density somehow. But I think we do not know how to prove that yet. So potentially to the antifluous theorem, saying for every degree other than the one, okay, for every degree, not divisible by three, we have this set of positive density of number fields, conjecturally, for which we can rule out solutions to the unit equation. And so one of the natural questions, now that we've said that there's a connection between the Fermat equation and the S-Unit equation is can we prove results of the type, restriction on the degree of the number field, local conditions on the number field implies, instead of unit equation has no solutions, the asymptotic Fermat equation, Fermat conjecture. So a statement about the Fermat equation, not the unit equation. So that's what we're trying to answer. And so here's the main tool. So we spoke a little bit, we spoke about this example of cell and laser. So here's some kind of generalization to number fields. It says that if you take K to be a totally real number field, and you, so we're going to have these two sets of prime ideals. S is going to be the prime ideals dividing two. And T is going to be the prime ideals dividing two of residual or inertial degree one. So what this means, the second condition is, I'm taking the prime ideals where the residue field is F2, they go in T, everything else, all of the prime ideals above to go in S and the ones with residue field F2 go in T and I'm assuming that T is non-empty and now I suppose that every solution to the S unit equation, this S unit equation here with this S satisfies some extra condition, which is if I take a solution here, I can find a prime Q in this set T where the valuation of epsilon and delta is not big. It's bounded by something explicit. So that's the condition. Then the asymptotic firmac injection holds for K. So that means that the exponent L in my Fermat equation over K, that's bounded, is bounded. And so this theorem is one of several theorems of the form. Local conditions plus assumption on the solutions, okay, so assumptions on the S unit equation, assumptions on the solution of the S, local assumptions on the solutions of the S unit equation implies the asymptotic firmac injection. There's other theorems like that, but we're just seeing a specimen. So how is this proved? Well, there's nothing actually very clever. It's just following exactly the same strategy as SEAR and MESA, but of course, because you're dealing with elliptic curves over number fields and module forms over number fields, you need different, you need, yeah, you need contemporary versions of the theorems of Wiles and MESA and Dribbit. And so here's some of the, so one of the things you need is Merrell's uniform boundedness theorem. Another, so that's sort of, you can think of this as a, something that replaces MESA's Isogony theorem in the proof of Fermat's last theorem. And you need, so these are, you could think of them as some vast generalizations of the theorems of Wiles and Dribbit. So modularity, lifting and optimization theorems due to many people. I've mentioned some Kissen, Barnett, Lam, G, Galati, Broi Diamond, and so on. So, and there's also, so the theorem that I've mentioned on the previous slide, it assumes that the number field is totally real. And the reason for that is that these modularity and level optimization theorem, they're sort of, yeah. There are, there are the sort of, so we do know, for example, thanks to recent theorems of Alan, Harry and Thorne, that a positive proportion of elliptic curves over each imaginary quadratic field are modular, okay. But that's, but we're still missing something called level optimization. That's the analog of Ribbit's theorem over general number fields. So there's a verge, but if you assume some standard conjecture in the Langlands program, then you can just write down almost exactly the same theorem for general number fields. Okay, so here's a, here's a sort of a very explicit example. You take this quartic field, this is totally real and to ramifies in this quartic field. So S and T are going to be just this set. And the great thing about this is that Nigel Smart figured out the solutions to this S unit equation, which is exactly the ones you need in the thing that you need to apply the theorem mentioned a few slides ago. And you find, so there's 585 solutions, but you find that they all satisfy this condition, that their valuations are sufficiently small and the theorem tells you that the asymptotic firma conjecture is true for K. Okay, so what I'm going to do from now on is go through some elementary calculations with S unit equations and to try and sort of come up with a theorem like the Antifolu's theorem for number fields where some prime ramifies. So Antifolu's theorem we had three totally splits. Here we've got a prime P that isn't necessarily three and P, the assumption is that P is totally ramified. So when I factor P in the ring of integers, I get this Gothic P to the N where N is the degree. And so I'm going to do some very explicit algebraic number theory. So the first observation is that if I have something in the ring of integers, epsilon, then its norm is going to be congruent to epsilon to the N modulo P. Okay, so let's see how this proof goes. Well, totally ramified, one of the things it tells me is that the residue field of Gothic P is the same as the residue field of the prime P. So it's a field Fp. And so that means when I look at epsilon modulo P, it must equal something in here. So it must equal something that's in Z. So epsilon is equal to A, it's congruent to A for some A in Z. And now, I mean, I'm going to, I need a definition for norm. So we will know, we used, well, I often think of norms as products of conjugates, but that is only convenient when you're dealing with a Galois extension. So here's one way to define norms is you look at this linear transformation, okay? Which takes an element of okay and multiplies it, okay, and multiplies it by epsilon. And so this, if you think of okay as a Z module, so it's something like Z to the N once you fixed a basis, then you can represent this T epsilon by a matrix and N by N matrix. And the norm is defined as the determinant of this matrix. But because of this here, epsilon is the same as A mod A mod P where A is something in Z, our matrix is going to be, when I think of it mod P is going to be A times the identity matrix because both of these on an integral basis, reduced mod P, they just do the same thing. And now when I take norm of this, that I'm going to get the norm of this one, AIN, A times the identity, which is A to the N and A is the same as epsilon. So I've proved this, what I was trying to prove. Okay, now let's take this result and apply it not to arbitrary elements but to units, okay? So here's what I get, is if I have a number field, same assumption as before, okay? I've added P is odd, that's not important. P is totally elaborate and the GCD of the exponent N with P minus one over two is one. So I'm assuming that these two things are co-prime. The degree N is co-prime to P minus one over two. I'm taking now not an arbitrary algebraic integer but a unit in okay. And the statement is that under these assumptions, the every unit is plus or minus one, modulo gothic P, the unique prime ideal above my small P. Okay, so how do we prove this? Well, the previous labor tells me the epsilon to the N is congruent to the norm of epsilon but epsilon is a unit so it has norm plus or minus one. And now I'm almost there. What I do is I think of epsilon to the N, I think of the image of epsilon to the N in this group here, okay? So this is, I'm taking the multiplicative group of the residue field of P and I'm courting out by plus or minus one, the subgroup plus or minus one. So epsilon to the N maps to the identity element in this group because it's congruent to plus or minus one mod P. But if you think about this group, it's the same as this one here because the residue field of a totally lamified prime is the residue field of the prime underneath it. And this thing has ordered P minus one over two and I've conveniently assumed that the N, the exponent N here which is the degree of the number field, I've assumed that it's coprime to the order of this group. So because epsilon to the N maps to the identity element, so the order of the image of epsilon divides N and divides P minus one over two and these two are coprime. So epsilon is the identity element, okay? In this group. So the identity element. So it has to be congruent to plus or minus one mod P. Okay, so I think this is, you're probably thinking this is all ridiculous for a web seminar, but here we get a conclusion, which is that I impose these two conditions I've seen before. Okay, so I'm going to have P's, I'm going to assume that P is not three, P is at least five. It's totally a lamified and the degrees coprime to P minus one over two and the statement is the unit equation has no solution. So assumption on the degree, local assumption on K, and we get the same conclusion as in trying to lose a theorem. And again, I mean, it should be true. It should hold for positive proportions of number fields where this condition on the exponent is satisfied. So how do you prove it? Previous lemma tells me epsilon has to be plus or minus one because it's the unit. Delta has to be plus or minus one mod P. You take these and you substitute them into this equation and you get plus or minus one plus or minus one is congruent to one. And that doesn't work unless P is three. So we get a contradiction for P not equal to three. Okay, I'm almost embarrassed to call this a theorem after I've shown you the proof anyway. So, well, can we do something like this for the Fermat equation? And this is what the next theorem is about. And it says that if you have a totally real number field of degree N and you take P as a totally alamified prime, any totally alamified prime that's at least five, and you have an assumption on the restriction on the degree, which is the GCD of N and P minus one is one. And you suppose two is totally alamified or inert in K, then the asymptotic Fermat injection holds for K. So here we have what we asked for before. We have to get the asymptotic Fermat injection, it's enough to have only restrictions on the degree and the local behavior of some primes in K. So there's, I mean, obviously infinite families of number fields of every odd degree that satisfy these conditions. And this is, again, one of several theorems. So there's theorems for even degrees that do similar things. So let's have a look at the proof of this. Well, I'm assuming that the prime P is totally alamified. So it looks like this. If I factor it, I have a gothic P to the N and two, it's either inert or totally alamified. So there's a unique prime above two, which I called Q. So I have a prime gothic P that's above P and a prime gothic Q above two. And the exponent here is one or N, because it's either inert or totally alamified. And I'm thinking not about the unit equation, but about the S-unit equation, because that's what we need to deduce the asymptotic Fermat injection. And recall that we're not trying to prove anymore that there are no solutions, because there are solutions. I can take two minus one equal one. That's a solution to this equation. But what I'm trying to do is to control the valuations of the solutions, because that's the theorem that we saw earlier, is if the valuations aren't big, then you're fine. And then asymptotic Fermat injection hold. Okay, so here's a typical case of one of the several cases you're going to have to look at where epsilon is a unit and delta, so if epsilon is a unit, delta is one minus epsilon is going to be an integer. Delta has high valuation, okay? So it's divisible by four. That's already big valuation in the case where the exponent here is N. Okay, so let's try to, let's see how we will out this case. This is the typical case of the proof. So you say, ah, so epsilon is one minus delta. Delta is divisible by four. So epsilon is one mod four. And okay, so that tells me that the norm of epsilon has to be one mod four. But epsilon's a unit. So it's norm is plus or minus one. So I get plus or minus one is congruent to one mod four. So I see that the norm will have to be one. So using this condition, I've deduced that the norm of epsilon is equal to one. And now we switch primes. We're not thinking about the prime of two. We're thinking about the prime above P. And we remember that the norm of epsilon is congruent to epsilon to the N mod P. But we've already said that the norm of epsilon is one. So we get epsilon to the N is congruent to one mod P. And now I use this condition, condition three. Condition three is saying that N is coprime to the order to the, to the, to the, okay, the order of the multiplicative group of the residue field of P. Okay, which is P minus one is prime to P minus one. So I deduce the epsilon is one mod P. Okay, so that's my second deduction. And now I look at Delta, what's Delta Delta is one minus epsilon. So Delta is one minus epsilon. That's one minus one modular this prime ideal. And I've written contradiction. Why is this a contradiction, because this prime has to divide Delta, and Delta is an S unit. It's only allowed to be divisible by Q. It's not allowed to be divisible by the other prime P. Okay, so this is a typical case also all the proof is just elementary algebraic number theory like this. And you deduce this, this theorem, which gives local sufficient local conditions for the asymptotic firm architecture to hold for a number field K. Okay, so let me, in the last few minutes, let me give another interest. So an interesting family of Dumber fields for which we can say something about the unit equation and the asymptotic firm architecture. Okay. And this is the ZP extensions of Q. So these are very popular and you are sour theory. And basically, if you take any, well, I said odd prime, that's the important. Okay, so let's take an odd prime P. And you look at the field of the Pn plus one lutes of unity. It's Galois group is isomorphic to this. And you can take the fixed field of this subgroup here. Now, you're going to get something whose Galois group is this said more P to the NZ. So this is this we denoted by Q and P has nothing to do with periodic numbers. It's, it's a number field that is totally a real it's Galois it has degree P to the end. This is it's Galois group. So if you fix your P, okay, P totally ramifies in here, because it totally ramifies in this one. All the other primes are unlamified, and you get a tower of you fix your P value and you get a tower, and you take their union. And this thing Q infinity P is called the cyclotomic ZP extension of Q. So in EOS our theory. People study things like the ranks of elliptic so you fix an elliptic curve let's say over Q, and you go up this tower, and you want to know how the ranks behave. But you can ask sort of similar questions for other diaphragm teen problems. And as a corollary of the theorem that we wrote before, because in the in the theorem for for unity equations. We had that the prime P is the prime P totally ramifies well that's true for this thing. And P had to be at least five so you have to rule out three, but it seems that turns out that two is allowed. Okay, and the degree had the degree had to be co-prime to P minus one or the degree here is P to the end, it's co-prime to P minus one. So you did use the unit equation has no solutions for this family of number fields. So point out that this doesn't work for P equals three, and that there are solutions so it's an open problem. You know, can can you determine the solutions for the unit equation for Q and three as n varies but but you know if you fix if you fix and you can you can just run a program and you get you get the solutions. And I'll just say that this corollary is in the spirit of this conjecture. So it's a conjecture of solid and passion that says that if you if you have any curve of genius at least to over rationales. Then the, as you're going up the tower, the set of rational points stabilizes. So X Q infinity is finite for all primes P. So if you, it might look like these two are not connected, but there's a common generalization pointed out to me by Min Yong Kim, where instead of curves you look at hyperbolic curves, and then this one becomes a special case of this conjecture. Okay, so let's look, let's let's go to further and finish off with this slide. So for these ZP extensions or the layers of the ZP extensions. It turns out that the what the theorem I wrote before immediately immediately gives you this theorem or this corollary I should have called a corollary, which is, if you take a prime P. That's not three. Okay, maybe you can include two. Okay, P is at least five. And for odd primes you need an extra condition, which is P is non V fridge. Non V fridge means that two to the P minus one is not one more P squared. Then the asymptotic firma conjecture holds for each one of these number fields. Okay, so maybe I should point out a few, a few things. So why, why do we need this non V fridge assumption. Well, it's equivalent to two being inert as you go up in this tower. So, so in the theorem that we had before for the firma equation over number fields with local condition wanted to to be either ramified or totally ramified or inert. So this gives you this condition. And so this, I mean, so I think a very vexing open problem is to show that there are infinitely many non V fridge primes. Okay, it's, it's really surprising that we don't know how to do that. But it seems that almost all primes are, are non V fridge. The only ones that we know to be V fridge are these two 1093 and 3511. And this condition also pops up in, in, in, in the classical theory of firm as last theorem. V fridge showed the first case. So P, okay, the exponent P does not divide any of the variables, like in the first case of the firm as last theorem holes for exponent P. And we sort of notice the word effective here. So, for the constants I'm talking about the, the, the bound on the exponent in the firma equation for that to be effective, what you need is modularity of elliptic curves over the particular number field that you're talking about. That, so for, so if activity makes use of the following beautiful theorem of jack thorn, which is the elliptic curves over these ZP cyclotomic extensions of q are modular. Okay, that's an excellent place to stop. Thank you very much for your attention.