 Thank you very much. Do you hear me? Yeah, okay. Thank you very much for this invitation to this very prestigious seminar series and thank you Philip for helping me out with advice. I'm speaking to you today from my Basel flat and I hope the Wi-Fi keeps working because I get the Wi-Fi through the windows and through this window here and the buildings you see outside the windows are other university institutes and so I get edge your own through there through the windows and if I move too much the walls are very thick in this building and if I move too much I wouldn't get it. So let's hope this is not going to be the talk that goes wrong. This flat that I'm in is a building dating back officially to 1394. Erasmus lived here from 1522 to 1529 and he died in Basel. The guided tours come by and the guide stands outside the building and points up and mentions Erasmus, the wise man. Once I open the windows and shouted very loudly, is my dinner ready Mrs. Erasmus? More relevant, the building was owned by the Benuri family from 1685 for 100 years and in particular Johann Benuri I used to give, it's known that he used to give oiler weekly lessons and in those days there weren't any institutes and there weren't any offices and so it's almost, it's fairly likely that the lessons took place in the building I'm speaking from. Basel is very proud of the Benulis, there are at least eight or nine altogether that were mathematicians and Daniel Benuri lives up the road from me, he's not a mathematician, he's a geologist. They weren't so proud of oiler because after he, well he didn't get a chair here and then he left and he never came back to Basel, I think never came back to Switzerland. So let's start now, this is, so this here here, pencils of norm form equations and a conjecture of Thomas, joint work with Francesco Amoroso and Donberto Zagne. All about diaphragm equations and we already heard a lot about this, let me just get that over, yes. We still don't know, we heard a nice talk early on, we still don't know if there are integers x, y, z with x cubed plus y cubed plus z cubed equals 114. We know that x cubed plus y cubed plus z cubed equals three has at least ten solutions, x, y, z integer triples and we also heard some heuristics for the existence of infinitely many but it seems nobody has any conjectures about any possible structure. I think it's believed that somehow they grow exponentially but more than that I don't, I'm not aware of. Swinnett and Dyer used to, he once wrote that a particular problem has only not been solved because no one has seen how to apply modular forms to it. So this, he was, I think he wrote this before Fermat and he was certainly right about Fermat so possibly that might be an approach to this very difficult problem. This one is not a norm form equation, now I hope this works. Yeah but this one is x cubed plus 2y cubed plus 4z cubed minus 6x, yz equals 165 and the left-hand side factorizes over the field obtained from the rationals by joining the cube root of 2. Another example is this quartic ternary x to the fourth minus 2y to the fourth minus 8y to the fourth minus 8, yz times x squared minus yz equals 1 and there you have the fourth root of 2. Now the first equation actually has infinitely many solutions in integer triples and they have a very nice multiplicative structure, it comes from units but the second is different, it has most finitely many solutions. You see a couple of solutions by putting y equals z equals 0 and x equals plus or minus 1. The finiteness of the number of solutions lies pretty deep, it's the Schmidt subspace theorem, we did have a talk about that as well and this is incidentally not effective so it may be rather difficult to find them all in the current state of knowledge. So later on we'll see more details about what I'm talking about here. So the individual study of individual equations is sometimes easier and sometimes harder and what we're interested in here is pencils of diphanton equations, a fancy word for one-dimensional families and I use a parameter t and this parameter t is also taken to be an integer. Now if you take a general kind of family it's not possible to say much and the first non-trivial diphanton equation in two variables is probably the per equation, x squared minus ty squared equals 1 so here comes the here comes the parameter and for each fixed t positive which is not a square there's an infinite set of solutions and it has a very nice multiplicative structure but if you start varying t then there may be no no no simple description of of what the solutions look like but if I fiddle around and replace the parameter t by t squared minus 1 then it is possible to solve this equation for actually for all t. The solutions are given by this formula x plus y square root of t minus t squared minus 1 is plus or minus t plus square root of t squared minus 1 to the power plus or minus n and that's for all n 0 1 2 and that's an infinite set if t is bigger than 2 in absolute value at least 2 in the absolute value so this can be this can be written we can get rid of the square root here and write it in a rational way in the following sense x uh hey oh what's happened here ah panic oh panic no i'm panic it's a shopping list it's a shopping list that's all it is uh uh they say that the jk rolling uh who wrote the harry potter books could make a best seller out of her shopping lists uh but this is not very exciting it's a sort of slightly embarrassing scones jam cream and earl grey tea uh and the fish uh oh dear this is embarrassing uh how can we get rid of this uh fish this reminds me of michael mcquillan who used to uh punctuate his lectures with slides of fish dinners oh sorry about this let's let's go on let's let's go on to the next page ah so i made a double mistake here i've got that page twice and now the thing is gone so where were now ah now how do i how do i go back ah ah now it works so i have to just recap this these integer solutions here were the solutions of the the modified pel equation and uh x and y then become after getting rid of the square root x and y become polynomials in uh t with integer coefficients um i think the first i won't write any down one here is one and zero and another one is t and one now uh so things are interesting even on the pel level if you go further to higher degree then you're no longer pel you have a 2a equation and uh if i take simply replace two by three i get this equation and then i one can say a similar thing uh for every t not equal to one the only solutions in integers are one zero and t one so the same as as one has for this modified pel this is a deep result of uh benet known as mike to his friends who prove much more including the higher degree things and uh actually an even more general result where you have coefficient a here and a coefficient b here now the proof uses hyper geometric identities and linear forms and logarithms and some very sophisticated estimates for divisibility so uh our theme today was started by emory thomas uh who was an american mathematician rather than original mathematician he he found some nice results on 2a equations with few coefficients uh before that in 1990 he considered this equation so this is a again a 2a equation a cubic form here and you have a parameter here t sitting in a nice linear way as opposed to any cubic way here so this is a quite an elderly equation and he showed that this equation has only the solutions uh there's the well no one and zero and then this and this solution provided t is sufficiently large 13 million or so probably this has been improved a lot the solving 2a equations has been a very been developed to a high degree of sophistication recently this was the first result of its kind i think he used linear forms in logarithms uh linear forms in logarithms tend to throw out very big numbers but not if you use the rather fine versions due to minot and balschmidt and others so a bit later 1993 thomas proposed the general concept of stably solvable we write our equation out as a polynomial x y and t so uh as in this one here uh then we can ask all to to solve the corresponding equation when you put big letters in big x big y and big t where big t is now a variable corresponding to little t and you're trying to solve this and x y as polynomials so you've gone kind of polynomial or uh let's call these functional solutions and then uh why would you like to do this if you can do that then by specializing t uh 2t big t to little t then we get into the solutions here so uh sometimes you can get the inter solutions from the functional solutions but not always already the pell equation has no functional solutions so this doesn't work always and then thomas proposed a general kind of uh idea that uh all integer solutions in vague terms this is just a vague formulation all integer solutions little x little y come from specializing functional solutions big x big y at least if t is sufficiently large uh you must expect that t can be must be taken large to avoid special cases for small t uh this does hold for this modified pell equation as we said and also the high degree analogs cubic and so forth he gave a nice example where some functional solutions are easy to see uh and this is a kind of typical example you take d polynomials over z and uh you write down this two a equation here it factorizes nicely but then you add y to the d and then the factorization is gone so uh you have this equation uh then you have for there the functional equations x y is one zero which is an old friend now and a one big t one a d big t one so if you put y y equals one then x equals a one for example then this this is zero and then you get y equals one here one here so he considered this kind of family and he conjectured two things so first of all this general principle star local to global you can think of it as this holds if the degrees of these polynomials are different and then he also conjectured part b that these above these functional solutions are the only functional solutions and uh he had a lot of partial results on a but soon afterwards b was disproved by sigler with even with d equals three and these values these polynomials these three polynomials so uh b is more problematic so since then there was a lot of work uh and i don't give a detailed history i wrote down uh the names i'm sorry if i've missed out some names uh sometimes the names have initials uh for certain reasons uh there's more names than this on the next sheet du jella a halter koch hoiberger li ej lemmermeier lettl uh here is a now i have a request um uh this ej lee was the last research student of allen baker and she's mentioned in the literature uh but i can't she has no internet presence and uh cam stewart and uh rob tiderman drew my attention to this to this woman uh and if anybody knows the full name i would be grateful to to know this full name so uh levec mignot petr roth r so this is not the roth but a roth tiki uh i can't i don't know how to pronounce this name sanakis buti wakabayashi and balshman so uh what's new here um a general method um instead of stating any general theorems which which get rather technical i'm going to stick in this talk to um some examples which may or may not be interesting so for example we we were able to prove this star thing this local to global uh for uh this equation when d equals three uh and not just for the degrees of the a is different but the polynomials themselves are different it probably holds for all polynomials but we didn't go quite that far so the zealot counter example is no longer interesting because this star says nothing about the functional solutions the method extends to this uh this two a equation but it doesn't give uh t all t not equal to one just t sufficiently large it also extends to certain non non non inhomogeneous equations uh which don't seem to have been in the subject before uh so we look at this x q minus t q minus one y q plus three t q minus one x y plus t q minus one squared equals one so this was before and we've added this and this and this makes it then non homogeneous so you can't transform this into a two a equation with linear transformations so this is uh slightly different and now the finalness for each t at least not equal to zero one is guaranteed by seagull's theorem on uh curves of genus one and that's effective because of baker and codes in 1970 for this equation we term did we determine all the functional solutions and also uh the arithmetic consequence and the upshot is that for t at least t sufficiently large the only interest solutions are given by this t squared two t and t squared minus t further we're not restricted to two variables as earlier work was uh here is an example of a of a quartic ternary quartic so it's a little bit like the other one i had before a little bit like this one here but we got the parameter in it so here they turn out to be infinitely many functional solutions which i don't think was envisaged envisaged by thomas thomas in his conjecture writes down a final list here there's infinitely many functional solutions but still uh the local to global holds and the upshot is now that for t sufficiently large all interest solutions are given by x y z equals plus or minus x n t squared zero plus or minus y n t squared the varying n and these are the x n and y n that come from the pale equation which i wrote down but not explicitly and then uh the uh sporadic solutions you have 12 sporadic solutions uh in the sense that uh well they're just 12 of them and this is one of this is the this the most complicated one 40 to the 6 minus 4 5 t squared 40 minus 40 so these hold identically in t and so they come from the functional equation the functional solution uh in all the results this t zero that comes up is effective so but we for no single equation did we calculate it explicitly so what is this method about uh let's uh go through uh how it works using this uh uh this uh two a two a nonform with the extra thing here now it has an aspect of the two a equation in that the left hand side does factorize x minus s y plus s squared x minus omega s y plus omega squared s squared and x minus omega squared s y plus omega s squared where s is the cube root of t cube minus one we can expect this s to come up from this part here so that's the factorization of this left hand side um omega is the cube root of one e to the two pi i over three i recently had to uh look again at some papers of seagull and um and uh aduring on the class number problem and they came across seagull's notation which greatly improves this notation one to the one over three so one is of course e to the two pi i so this makes perfect sense i don't know why this this notation didn't didn't catch on uh but it seems to have been forgotten okay so how can we go forward here we have the factorization here and that's equal to one and therefore if you have a solution then all of these things are units and uh let's it's enough to take the first one that's a unit in the field uh generated by s now s is the cube root of t cube minus one and it is rather easy to see that if t is sufficiently large then q of s is a real cubic field and it has a identifiable unit namely t minus s you just verify that and this is the uh crucial point of the whole method that you have to be able to see the units and you have to be able to see the units in a functional sense as well here we're still in the arithmetic t t is a t is an integer the unit has no one and therefore this x minus s y plus s squared is some power of this with no sign t minus s to the power m where m is an integer so there's not a lot new so far it's all looked very familiar i think to experts uh what does what do people do now they eliminate x and y from these oh well sorry i forgot to say that now we now take conjugates s is this cubic cube root of something and so putting the omegas in give you two more equations omega omega squared and then omega in here omega squared omega omega squared in there so we now have three equations in two unknowns s and y m is of course a well known and now we eliminate x and y through seagull's identities this is not the last appearance of the name seagull this is known as name dropping and if you do this elimination it's rather simple and you get the t minus s to the m that's this one plus t minus omega s to the m here plus multiply by omega plus omega squared t minus omega squared s to the power m is 3s squared and the 3s squared comes from this stuff here so uh this is an s unit equation but it's got four terms uh that's an s unit this these are s units the m's it just doesn't involve m but uh if you want to make it an s unit equation you have to give it allow it four terms the subspace theorem applies for each fixed t but as i said this is ineffective so it's useless in our context where you want to keep track of the t how am i going where's the i'm going far too fast what to do play for time uh so there's the equation so we now write this in a more general form um incidentally on if on the on for the usual two a equation you wouldn't have these terms here so you wouldn't have this term and it would be an s unit equation with three terms and there are many many ways of handling that effectively here this this here is the joker so what do we do well we immediately make it more complicated we write it out as uh in uh sort of geometrical language um which is not too critical to think geometrically but seems natural here we write it out as a k term equation with k equals four p is now a point t s on a curve c defined by this equation corresponding to the equation connecting little s and little t and uh the alphas are the coefficients in here so the alphas are one omega omega squared this one and the f's are fixed functions on the function field of this curve over q bar because we've got omega floating around everywhere and q bar of s is just of course q bar a join big s and big t so f one s t is t minus s here that's this one uh alpha one is one and that's this coefficient now we uh apply one of the main results of our paper with uh namorosa and zany bounded height in pencils of finally generated subgroups that's duke 2017 and this gives you an estimate for the height of p in terms of the various objects here and it gives the following estimate that the height of p is the most k times the height of the coefficient vector divided by m plus h zero uh i'll go to the next slide immediately pretty soon this m here is in the denominator which is important and one can imagine that this should be true these are logarithmic heights i'm not going to define heights too much in detail these are logarithm although i could to make up for i could to to spin out the time these are logarithmic heights the height of and then so this m here if you take it over here you get m uh height of p which is height of p to the m and this corresponds to the exponential there so you may expect a denominator of m here if things go well and this you do have there are some conditions here no vanishing subsums that's um that's a kind of universal condition a knee jerk condition this has to be there and m is sufficiently large m at least m zero and m zero is effective and this h zero is also effective and they depend on yon c and the s and so for our original equation here they would be absolute constants so we have to say something a little bit about the heights because we've got the two of them here what's the height of p here so you see p is t of t and s the height of t is by itself log t t is a positive integer so the height the height in here would be log t asymptotically or exactly and the s here is uh that was the analog of this equation s little s cubed equals little t cube minus one so the height of s is again about log t and if you bunch them together you get depending on how you do it you get log t or two log t two log t would be the crude bound so the hp is asymptotically two log t the h alpha one up to alpha k is is as you see here the the coefficients that's constant that's constant they're even roots of unity so the heights of these coefficients are zero but here you have this thing and so this is going to give you another two log h log t so the height of the coefficients is asymptotically two log t so here you have two log t here you have up here two log t and if m is then bigger than k which is four then is bigger than four then you have more log t's on the left hand side than you have on the right hand side so you can take the right hand side over and you get a bound for t the bound for t then depends on h zero but that was effective so as soon as m is bigger than four and m is bigger than m zero we get t bounded and uh that is that is the basic strategy of the proof um there are these uh snags here uh no vanishing subsums uh they can be handled by induction on k they don't turn up in in this particular four term equation uh but here is another snag uh m bigger than four and m bigger than m zero so the small m's have to be handled and indeed something has to be done because we haven't seen anything about the functional equations yet the functional solutions uh we just appeared to get no solutions now what to do about small m or let's say a fixed m so if you if m is fixed then this equation or even go back to this equation is simpler if m is fixed here this is a fixed equation for t s is an algebraic function of t so you're going to get a fixed equation for t here and so you can just solve it for t and that's going to give you another upper bound for t but there's a problem here slight problem here that the equation could end up as zero equals zero so the equation for t just disappears and then you're back then you're home because if it's zero equals zero then the equation holds for all little t in other words identically in big t and you've ended up with a functional solution and you don't have to even find what it is if you just want this local to global situation oh here I here I have made a mistake here there's two pluses uh this reminds me of uh Seier wrote to once that uh if he sees a mistake on the blackboard he he he experiences physical discomfort so I'm sorry if any of you experience physical discomfort with this term here so how does one prove this this seven let's go back into that uh still far too much time left how to prove this height boundedness result uh there are some uh height boundedness results already in literature um whose point is to prove that that in merely that the height is bounded from above we want a bit more than that we want a specific bound here and we want this really want this m here this is this is crucial so it's a kind of uh very sharp height up a bound estimate so I'm going to take an example I'm going to simplify this and take an example which is much simpler than this uh two a equation still a four term one and still three powers three things raised to power m but instead of t and s just a tau a one minus tau and a one plus tau and that's equal to one and now m is any integer positive or negative tau is not allowed to be zero one minus one otherwise these won't make sense but it's allowed to be any algebraic number and we want to get an upper bound for the height of tau which is an absolute constant so uh why should it be bounded the height of tau if you rub out that term and rub out that term then you just get tau to the m equals one so tau is a root of unity that's in the set one to the power q in this equals notation uh when tau is a root of unity then the height is zero and you've got uh the best possible upper bound uh if you rub out this one and this one then it's a bit different you get the one minus tau is a root of unity so tau is a root of unity minus one and the height of that is not zero but is at most log two so uh that's that's almost convincing that the height should be bounded but it's non-trivial it's non-trivial even if you rub out this term here if you rub out only this term then you've got tau to the m plus one minus tau to the m is one and they were shown by burkas in 1997 uh that the height of tau is at most six uh again using uh hyper geometric functions uh we don't know if these can be used here probably they can't be probably there are no uh pad A approximations appropriate pad A approximations so but it's possible uh we don't know how to do it and so we use another approach which goes back to uh what touay was doing after he uh failed to use hyper geometric functions in a different context uh we use seagull's lemma now to construct polynomials a b c d over z with a certain identity a x x to the m plus b x one minus x to the m plus c x one plus x to the m equals dx this x should really be capital t i uh but i switched into the uh automatically into the x notation uh so you construct polynomials like this with seagull lemma and their degrees are not too big i won't say exactly what they are but it's such that uh these exist and they their coefficients aren't too big if you substitute x equals tau in here then uh you get another relation between these three quantities but now with some coefficients in there that's not enough so you differentiate this thing once and when you differentiate the x to the m more or less stays x to the m and you get a third relation between these three quantities also with coefficients so you've got three relations between these three quantities and you can hope to solve for tau to the m so there we are three relations between these three quantities and if you do solve this then you get an upper bound for the height of tau to the m which is as i've indicated before absolute value of m times the height of t height of tau and the bound will involve the coefficients of a b and c and these are also exponential in m by the construction and that means you get a a bound for h of tau it doesn't involve the denominator m anymore because i've changed the context slightly uh there are various snags on the way uh you have to actually differentiate um a number of times which is a small proportion a small multiple of m and then you have to check that you can do the elimination all these snags were were removed by 2a and 2a is strongly associated with ineffectivity but just to show that this is this is this is effective and can be made by excellent a master student of mine andre and dense who very shrewdly left mathematics to go into infectious diseases uh and i think he's flourishing there now he calculated that the in this uh this three term situation here the height of tau is less than 600 uh another nice result that he had um was if you if you eliminate this thing but replace this exponent m by a different one n then the method still works tau to the n plus 1 minus tau to the n equals one and uh in that case the height of tau is at most 1300 i didn't write this down i provided m and n and not both one because then this would be a functional solution so finally how does one find the function of solutions if one wants to uh for our example uh for our cubic non-form we end up with the functional equation functional version of five so that would be the functional version of this which is just big t and big s here with big t a variable and big s the cube root of big t cube minus one so you've got this equation and now you're in abc country abc territory or abcd here and this is all well known to be effective and you can use various results in the literature but uh in practice it's these involve rather complicated definitions of heights and considerations of genus and huvits and one can just throw maple at it um you can argue that these functions the the corresponding functional versions of the things i had before the three the three things occurring in five here are these three these four functions are linearly dependent over c so uh erotskin is zero start differentiating with with respect to big t and the ronskin is zero there will be a huge power in this ronskin the whole business complication of abc abcd stuff is that the if what happens if the ronskin is zero but if you take an example then the ronskin won't be zero very unlikely and you remove these large powers and you get a very simple equation which leads immediately to m equals minus one zero one or two uh there's a curious so ah time is i'm making a respectable pace here there's a curious side issue here um our results solve the equations in a uniform way for t sufficiently large which is effective and uh what about t small the methods don't work and uh we see no way of using of uh of solving these quartic ternaries or small as t because one has to uh one has to use the subspace theorem so this is a curious outstanding problem uh so let me see the time yes uh here is a bit here is what i call the baker quintic from 1967 which is a lovely result i remember first coming across this it's a ternary quintic of the same kind of shape and t is sufficiently large so this is rather thomas thomas thomas thomasy this if t is large and you have this equation here this is a form now equal to n then the solutions x y and z are all at most t to the power 2500 times n squared and uh you have this polynomial dependence on t and this may suggest that the thomas uh the thomas uh uh philosophy or strategy or conjecture applies to this at least if you um take n equals one and again small t makes somewhat uh the same the same trouble uh the dependence on n is uh extremely good this is done by pad air approximation uh this i think i'll skip oh here's here's my fish dinner now i'm now i'm making up time here is my fish dinner here are some oysters they've they've gone or the remains of oysters this is uh this is the fish called john dory these are potatoes and this is spinach and this is uh this is the accompaniment uh what i did on my holidays uh here is bulbaki one of these gentlemen is uh is general uh claude denny bulbaki a french general and he was in command of a french army which was under attack by under attack by a german army near the swiss border and he uh knocked on the door and the swiss let him in and uh there is a huge panorama of the whole thing the battlefield is is to the left is to the right here well there was no battle that's where they were quartered this is uh now this is switzerland uh lever air and there's a huge field there where we're the kept and there's a magnificent panorama painting which you can see in leucerne i took this photograph last july uh when i was passing through lever air the site is is rather impressive the connection with the mathematical bulbaki is extremely tenuous and uh not really known it was suggested that some french mathematical student heard that this seminar was running in lung in uh in paris and uh thought it a bit crazy and said oh this guy bulbaki was crazy why not call it that and this this uh this somehow caught on there seems to be nothing more than some student suggestion uh he was a slightly um full-on figure uh bulbaki uh after this debacle here he tried to shoot himself but the bullet flattened against his head apparently uh so he came out well in the end uh i almost still have time this is what i did also i i visited the gravestone of reman uh i can't magnify this this is in latin anyway this is uh in italy overlooking lago lagomogore and uh he died here at the age of 39 of tuberculosis uh his his actual grave is not there his actual grave was somewhere near but uh it seems to have been destroyed they moved they moved the cemetery and what became the original site is no longer known what became of the actual grave is also not known it was not so easy to find i found myself uh cycling down this weird lane not knowing if i would find what i wanted or not uh this is what i had prepared in case something went wrong this is hopeless the server keeps going down but of course if something went wrong i wouldn't be able to show this uh okay so that's practically all i have to say i i mentioned two of the organizers by name so uh maybe i should uh mention the third organizer by name this i saw carved uh at the top of a castle this is uh a couple of miles from basil this is i was out during the lockdown when i shouldn't have been uh quite a sunny day there and i saw this carved on one of the beams there uh and uh here is the castle here is the flag the swiss flag or the basil flag uh okay i think that's all i have to say thank you for your attention