Andrew Wiles: Fermat's Last theorem: abelian and non-abelian approaches

Well, thank you very much for that introduction and for the kind words. It is a great honor and a great pleasure to be here in Oslo and receive this award. I have realized that there may be many people in the audience who are not fully trained. number theorists let alone fully trained mathematicians, so you may not understand more than the initial descriptions of the problem and perhaps the first things I say about them, but there is no better way to understand the mathematical experience than to simply understand the question and opening statements that is what we suffer every time we are faced with a new mathematical problem, so I hope you will join Jane and enjoy the genuine mathematical experience that will be involved, so this is the family occasion when X to the N more and to the N equal said to the N they should not have solutions as everyone knows FEM a copy of a Greek mathematics text is annotated and in the margin he wrote it is impossible on the other hand to divide a cube into two cubes a fourth power and to the fourth power or also to any power greater than the fourth in two similar powers.

For this, I really have a wonderful test, but this margin is too small to contain it. He wrote this, we don't actually know when because it was only after his death that his son republished the original Greek text with these marginal notes in 1670 and only through his son's intervention can we really know what is ephemeral that it's getting worse so we can't date it precisely, we don't really know if it started crossing this. but for a long time people assumed he had a solution and tried to find that solution including myself so the case N equals two of course there are many solutions we learn in high school 3 squared plus 4 squared equals 5 squared 5 squared plus 12 squared equals 13 squared and so on, these were known to the ancient Babylonians.

The work on this problem of trying to extend this or rather refute this in the general case to show that there are no solutions for n greater than or equal to 3 generated an internal. was inspired by many developments in number theory, the field of nama theory in a sense began with Famer himself and there is a clear split in attempts to solve this problem that paralleled a split in the subject of algebra that was studied by tree so that you know well that I will demonstrate the ability to solve the general equation, the general quintic and, in general, the equations of higher degree, and that marked the end of hundreds of years of study of algebraic equations to solve equations using radicals, but this equation solving problem for radicals is an algebraic answer to solving equations and what number theorists are looking for is a much more arithmetic question, but in reality we are stuck at the same point that was solved by tree, So what if Emma really makes it?

So why have we written evidence of the case? N is equal to 3 and N is equal to 4, so in the case of three, although it is not entirely obvious, you can reformulate the problem by showing that there is no right triangle where the sides have rational lengths, the three sides and where the area is one, so it's not obvious, you use the formula for the area of ​​a triangle and you can deduce this using the Pythagorean

theorem

, which is equivalent to the N equals three case of the famous loss

theorem

, so we stated this strongly in letters several times in his life, particularly in letters to Digby in khakha V the case N equals four showed and in fact described the proof of the case actually something even stronger that the sum of two fourth powers is not a square Thermo describes the method and called it the use of infinite descent so unfortunately we know quite little about exactly how he wrote his proofs, who thought of his proofs because he still lived in an era with quite limited publications, so that in the Middle Ages, when the equations before the tree were studied in the 14th and 15th centuries, mathematicians in Italy challenged each other. having mathematician joules where they would pose problems against each other and their reputations, their status and their careers really depended on the outcome of these duels and I'm sure you know the famous story of the solution of the cubic well, there was a certain leftover inheritance from this , in the way the family discussed mathematics, would pose challenges to other mathematicians, especially English mathematicians, who failed to understand or solve most of the cases, so that is part of the reason we don't have The detailed tests because he said his results are challenges instead of explaining well the tests of this infinite descent method.

I tried to trace where it came from and, in fact, in 1636 the family knew nothing about number theory. You must have found some problems from Diophantus' book and the first question from him actually. Stealing to evolve from another French mathematician was actually an extremely elementary question where I didn't understand the proof, but it actually boils down to a very famous ancient classical proof of the irrationality of the square root of two, so just to give a idea of ​​how infinite descent works. I'll show you how I could have thought of that proof in terms of infinite descent, so here's an actual proof for your theorem.

The square root of 2 is not a rational number. So what you need to do is start by assuming. The square root of 2 is a rational number, so you can write it as a quotient a over B, where a and B are integers like 7-elevens. Whichever you choose, a solution where a and B are positive, you can rewrite it as a square. is equal to B squared and now a is even because it is twice B squared, so a is twice another substitute integer which in a is equal to zero in the original equation and you get 4a 0 squared is 2 B squared cancels out the 2 you get 2 so 0 squared is B squared which now tells you that B is an even number so B is twice B 0 ok now you do the same thing again so you substitute in B B for zero and you get a zero squared is zero squared and Now you have that the square root of two is actually a zero of B zero, so what you've done is take a solution to the equation and show that there is actually a smaller solution and This new one has a zero. strictly smaller than a well, then it says well, a smaller solution creates a contradiction because you can't keep getting smaller solutions forever, there's no infinite descent, so that's not how it's normally presented, but I imagine This is how Pharma came up with the idea of ​​infinite descent and it's very surprising that actually this idea of ​​infinite descent was basically the basis of all attempts by one family until the end of the 19th century, so now I'm going to mathematically jump to the next attempt. successful partially successful that was by coma. in the 1840s, there was a lot of drama in this period where at the French Academy people were announcing proofs that were fallacious and there was a lot of enthusiasm, but it actually turned out that they were wrong and they were wrong in a way. way that had actually already been explained by comma and previously published a few years before, so the idea is that you try again to find, suppose you have a solution to the equation and then try to deduce from it a smaller solution, so what you do is you take the equation the general equation then it said that P is partial of 1 so this is a use of complex numbers, we can factor this as a product of linear factors.

Now the idea is that if you take a number that is a power p f and write it as a product of other numbers if those numbers are co-prime to each other then each one has to be a p f-- power, for example if I take In terms of cubes, if I take 8 times 27 that number, there is no other way to divide it as a product of cubes, what does that do? That means that, essentially, if these numbers are co-prime with each other, then each one should be a power piece. Well, what we mean by P powers, we have to work in Z Zeta arithmetic, so we have to allow not just ordinary integers, but combinations of Zeta powers and then miraculously what you can do is have this identity completely formal X plus Zeta Y multiplied by 1 plus theta minus X plus Zeta squared Y is Zeta X plus and well, what's the point of that?

Well, the point is that if each of these factors were powers of peeth, then this is one piece of power multiplied by 1 plus theta minus one piece of power is Theta multiplied by one piece of power. It's very similar to the original equation and Kama's idea is a good one, assuming we start with a solution here and now. we will probably start with a solution in Z Zeta instead of a rational solution, we can deduce that each of these factors of P power and therefore we get another equation that is very close to the original, maybe we can use infinite descent if we start With A solution we deduce in this way that there is actually another solution that is smaller than the original and again we try the infinite descent.

At first glance, this seems very complicated because this 1 plus Zeta is in the way, the satyrs are in the way, but it works out. that if you really study the arithmetic of Z Zeta you can make this work, these are units 1 plus Zeta and Zeta are actually units on what we call a ring of numbers and he said that Zeta provided something that is true about ideal class groups, so the problem is this principle that if you take a number like 72 it is 2 times 2 times 3 times 2 times 2 times 2 times 3 times 3 there is a unique way to factor any number in our arithmetic unfortunately in general in this arithmetic there is no there is a unique way to factorize So this brought a great development to the invention and study of groups of ideal classes in general, so the idea is to take ideals in this ring, Zeta said, that is,

abelian

groups that fall between some multiple of Z Zeta and said that they tell themselves so that They are conserved under the action of the ring so many times alpha that a billion groups are again within the

abelian

group and then you have special ones that are called principal ideals that are simply multiples of a particular element, so this way you have certain things attached to it. to this new arithmetic, this arithmetic of Zedd Saito, well, what he showed was that if you study what is called the ideal class group, which is the quotient of the group of ideals that were principal ideals, it is not exactly a group, but The quotient is a group of ideals. mod principal ideals, first proved that the famous

last

theorem is true if P does not divide the order of this group, so he gave a very explicit criterion to prove Fermat's

last

theorem well because it could be quite difficult to calculate this size of this group of classes. then he gave a second criterion that is even more surprising in some ways: that P divides this edge of this group of classes if and only if P divides the product of these Bernoulli numbers b2 to be P minus 3, these Bernoulli numbers are the numbers. is obtained by expanding T over e to t minus 1 as a power series and are the coefficients of that power series 1 plus the sum of B MT to the factorial M over N, so he gave this amazing analysis of not only the equation but Actually, the obstacle to solving the equation was that all the attempts with the latest pheromone serum until the end of the 20th century really depended on this technique and trying to refine these properties of these class groups was the something about class. missing groups, can you show more about them because unfortunately this was not enough?

It's very, very difficult to determine when P divides them and it certainly happens often, in fact one of the things I studied in my last adventure, so to speak, with guys. Last theorem. Using the old methods, when I was working with Mazur at Harvard, we analyzed the fine structure of this group of classes to actually determine a relationship between each individual binary number and some part of this group of classes, but these methods were not enough to master the problem. original. there was still a gap between what we could hope to do and a resolution of the problem by this method seemed completely inaccessible, so, as I said, how his topic proceeded for a long time, I have now called it the Abelian approach to Farmers' Last Theorem, So why do I call it a billion?

What does it have to do with our ball? Well, it has to do with abelian extensions, so the tree called abelian is attached to extensions of the rationals or the parent fields where AG is our group and the action. is a billion, so for example, here we were using arithmetic with this Zeta or Zeta P. I'm writing here P up to one, so that's the root of this equation X to P minus one over X minus one, just like that , it's just another way. of writing it from the root of that serum, so it's a billion because the gawe group of signals a to P over Q are the rational ones and join the Zeta P is described as the abelian groups said mod P said through that is the multiplicative group of said mod p see how it works wellthat, a typical fall orphism here ten will take Z P to some other P through divinity, it has to be a V side power, write it as the alpha power and that goes to L mod p, so this describes Galois. group explicitly and the Gower group is abelian, this describes it as an abelian group and this Sigma L became known much later if L as a prime is Frobenius in L I usually write it from Bell, well these are abelian extensions and this for a long time. time became perhaps the central study of number theory, inspired in part by a theorem of Kronecker and Weber, they did not work together, in fact they were separated for quite some time, this took a long time to be approved properly, so done, all abelian extensions. of rational numbers are generated from roots of one, so it's actually two statements, one that feels generated by roots of one, our abelian and that's the easy part and the hard part is that each abelian extension arises that way, well, this brings us to the Kronecker-weber theorem.

Until almost the end of the 19th century, the first part of the 20th century was spent generalizing this Abelian theory to other fields and this is known as class field theory and this is really the kind of centerpiece of number theory for much of the first half of the 20th century, so to describe billion extensions of any number field, let me take an example, sorry! An example of F is the field generated by the cube root of two, so what you want is a billion extensions of that field, so it's actually very difficult to write a simple description like using roots of unity, so We don't really master it. any generality although it is still a problem an open problem what we can say is what exists so we can say that there is an extension that is similar to Q's a tipi plays the same role and has the property that the GAO is a group of this field which I have called F P, so FP just means that it is just a label for this field that I claim exists, the GAO, a group of FP over F is isomorphic to the integers of F, modulo P multiplied by the numbers integers of F, the multiplicative group of that. module the image of the units then Oh F of the integers that said cube root 2 and this group Oh F crosses more for the units or the positive units of Oh F so this took the first 25 30 years of the 20th century, but to formulate really Well, it took at least until 50, even 60, it's a lot more complicated when Oh F's ideal class group isn't one, so in the cases where family worried me, I mean, it wasn't a fair mile, but the common thing was that I worried, it would be much more. complicated, so of course when you've spent so much time developing the theory of abelian extensions, one wants to move on to non-abelian extensions and this is the great divide and in the second half of the 20th century a huge program was developed to try and I understand these extensions, so they are basically equations over rationals over another number field where the Gower group is not billion, where maybe it's even a simple non-Abelian group and the fundamental description of this, what it should be, It's still a guess. what is called the Langlands program, there was a decisive shift towards this in the 1960s and there are many problems in this area that many people work on and we have made a little progress in this context, but not much, it is actually more easy to describe. the parallel progress that this generated in the Fermo problem itself, so I'm going to go back now, that's how number theory was developed, but now the non-Abelian approach to FEM I itself, so this breakthrough occurred in 1985 and was suggested. first by Gerhard Frey and then the actual connection was completed by ribbit a year later, so the idea is this and it's a completely different idea, assuming you have a solution for a company, our problem, okay, new idea, suppose eight people rush towards the PSC. al P P I'm just going to assume that P is prime and hunger is enough to do it with prime exponents, so we generally restrict ourselves to those and then you artificially fabricate this new equation, which is the elliptic curve and squared is X X minus a al P , B to P and then a to P.

P plus P to P, which is C to P by hypothesis, so the discriminant turns out to be a perfect power piece. Well, elliptic curves have been studied a lot and it seemed very strange to find a discriminant that is a perfect piece of power and in fact, what Frye realized was that such a curve should not exist discriminant, it should not be powers of peeth, for which proposed a way to try to test this. It was a little faint at first, but like I said, the ribbits confirmed that it was, in fact, assumed. what is called the modularity conjecture, the taniyama-shimura conjecture has several names, so in fact you would have a concrete way to prove Fermat's last theorem if you could prove that there is no elliptic curve.

In this way, you prove Fermat's last theorem and to show that there is no elliptic curve. curve this way you can use the modularity conjecture, so this gave a completely different approach, so just to give a little hint on this, I need to say a little bit about elliptic curves again. The origins of this topic are in arbol arbol was the first. To really understand and talk about these doubly periodic functions, he died at 26, maybe he could have done all of this. Pete lived longer, he was very fast, so let's say you take an equation of this form and squared is X cubed plus ax plus B, what elliptic functions do you get? what you have to do is describe the complex solutions of this as a group, it is a C module, a network, the network is just an abelian group of the form set plus Z multiplied by another generator, this tower is not rational in itself and then this quotient actually forms the complex points of one of this curve, well, one thing that is very strange about this is that it is actually a group that is not at all obvious when you look at this equation, so the solutions of this equation actually form an abelian group and if you draw it looks strange the lattice in the plane and it's basically the interior points with some of the limits described like this, well in particular you can ask about the points in this group of order finite and the points of order P to N had the structure Zed mod P to N Z plus Z mod P to the N said, interestingly, this was something that puzzled tree for a while, he didn't understand why when he looked at this he got N roots squares instead of n roots.

I mean, it's clear now, but he wrote to her in his correspondence. He was puzzled by this, he understood it after a while, but if this equation is really defined over the rationals, you can actually get the Galois action at these points, so the gamma group is the colonel from P to N, that is the mod C of p a anvils see my PT in the group gawe gives an action so he injects and she L a of said mod p TN said then, what does this do? Okay, this is background on elliptic curves, we have good enough to understand how to apply this to the FEM I. equation yet, so I just want to specialize in the case P equals three, so this made sense, the first big step .

Well, the first step that I described yesterday in approaching fair MA was to understand that choosing an arithmetic approach, which, as I said, there are different ways. of addressing this modularity conjecture through geometry through arithmetic through analysis, so I chose nur arithmetic, that's what I knew, but then the big choice was to go to P equals three because if you look the three points of division, I put them in these are the nine three points of division here algebraically, you can actually describe them if you take the equation and squared is X cubed plus ax plus B these are actually the turning points just in terms from classical calculus if you calculate the turning points of that equation remember that you are not just talking about real turning points, there are actually eight of them and then you get a ninth their infinity, actually they themselves form a group, there are nine of them and generate the field Q III and the Gawe group of Q III. on Q then acts on GL two of said set mod three, so this is the same as in the case of couch P.

I'm just specializing in P equals three. What's special about this is that okay, it's not a billion anymore, this is not an Abelian extension of rationals, but luckily for me it's also a group with a solution, this group also said that mod three said that actually is a double cover of the symmetric group in four letters, it is a group with a solution, so as a tree he thought in terms of there being a division between equations up to the fifth up to the quartic in which the equation had a solution and high degree equations in the Galois language that might not have a solution.

I'm actually considering three distinctions: Abelian groups are solvable groups and non-solvable groups Groups that have no solution are extremely difficult and, as I said, arithmetic hasn't gotten there yet, but I realized that actually this intermediate stage is a stage with a solution that maybe you could pass well, so it's the elliptic curves and then the modularity conjecture, so I have To say what it is, try to link what are called modular forms or modular functions with elliptic curves, so modular functions were studied in the 19th century, they are actually a generalization, if you will, of the exponential function, the exponential function has the periodic property if you take e to 2 pi I said take the same value when you move from Z to Z plus 1 I said plus 2 to Z plus 3 and so on and this is the basis of Fourier series and Fourier analysis and so on but again, that is the Abelian situation and the situation non-abelian is when you go to functions that are invariant not under Zed under the abelian group but under a non-abelian group, in this case a subgroup of SL is established so that these functions do have a Fourier series they are invariant in the sector to establish plus one, but they also have this other property that if you take F from a to plus B over C Z plus D, we allow for something a little more general in case we want it to be C for the Z plus d squared times F of Z, that's as long as the matrix ABCD is an element from SL to Z with C divisible by some integer n, so for all that ABCD with C divisible by n and as long as you can pick an endpoint where it works, that's what matter, then this is a generalization, the non-abelian generalization of exponential functions, we get modular functions like this, so everything moves from abelian situations to non-abelian situations, so tonyamarie shamora's modularity conjecture was that there is some correspondence whenever you have an elliptic curve over the rationals, you can actually make a form of waiting module; those are the two here, so I have the final weight, but it just means this factor here and some level, that's the end, so the shape means here at the level here.

It doesn't really matter in detail right now what they are and what this correspondence is supposed to do, it's supposed to say that when you take the gawe group of the gawe action at division points p to n, see that in CL 2 of 0 or P T&Z because that's what the action looks like at the p TN division points under the Frobenius of L, as I described to you earlier, there is a notion of Frobenius, so I described it in the abelian environment of the roots of unity. carries data P 2 Zeta P to L, it is actually a little more complicated to describe it in this non-abelian setting, but there is a generalization because frobenius has its name, but for each prime L you have a Frobenius of L and under this correspondence goes to an element with trace al al is the ail from here then the Fourier expansion of this function tells you about the arithmetic of the elliptic curve then this is the arithmetic of the elliptic curve the action of gawe in these p to the n points of division this is the modular form, you look at its Fourier expansion and there is a magical relationship between the Al here and the Frobenius in arithmetic, so this conjecture seemed vague when first proposed by taniyama and became more precise with shamora , but it was a very article from 1967 where he identified the n level and proved a very beautiful theorem about it that allowed people to prove it to be able to prove this conjecture numerically and they got overwhelming evidence, people were convinced that this was true, so now that Ribbit had shown that the modularity conjecture, this implied our guys' last theorem, now we can be really sure that FAMAS's last theorem was true, it would be proven and it could be demonstrated in this way because this modularity conjecture is not something that mathematics can leave aside.

Fama's Last Theorem could have been like the perfect numbers, they are put aside and we can continue with mathematics without them, but we cannot continue without this. This is the core of the transition from number theory to mathematics.non-Abelian beam in mathematics to non-Abelian mathematics. The famous Last Theorem returned to center stage. Well, what is the result of this? Well, as I said, as soon as I heard that Ribbit had demonstrated this connection, I was fascinated and spent my time working on this, so why the case? P equals three, why is it so valuable? Well the point is what do we have so we have fry and then we prove that the modularity conjecture involves pheromones laughing theorem so now the problem just proved that the modularity conjecture is right and as I said when I started proving this, first after the first step and make the right decision make the right decision to get into arithmetic I think the first conceptually important step is that if we look at the three points of division, our website puts gl3 through BG L as well said mod three said p GL 2 said that mod three said that it is isomorphic to s 4, this is a group with a solution and when Langlands proposed this general program to understand extensions without a solution, there was a case that he could really understand and this is precisely when you have gower groups of the rationals in two or even any in the field in a solvable subgroup of GL 2 and s 4 is a subgroup of p GL a s bar is a subgroup, then what Langlands and actually this particular case is an extension of the theorem Langdon through Gerry's tunnel This is actually connected to the modular form, it's not the one we want to associate with an elliptic curve, it doesn't solve the problem immediately, but at least it gives you a tenuous link between the theory of elliptic curves and this theory of modular forms. he used this based on his change of base theory and uses the fact, as I said, that s4 is a group with solution, so this is still the first step, there is no way to avoid this step at the moment, we have no way to jump directly.

For extensions that have no solution, the only entry point we have is why these extensions with a solution in the middle, from what we understood a billion extensions in the first half of the 20th century, we have taken them a little further to these extensions with solution like s4 and we can use them. to get a little bit more, but we haven't really fully mastered them in solvable extensions, these simple groups of any kind, well, let me say a few words about this Langlands generalization that a lot of people work on in number theory. Actually, like I said, it's the main drive. from modern number theory because even working on other problems like elliptic curves and any kind of generalization of elliptic curves you want to solve equations, you're going to have to use this ultimately, so it's a generalization of class field theory.

Remember that class field theory describes a billion extensions here the problem is to describe all the Galois extensions of rationals or any number field and, in general terms, and I'm being a little dishonest, but in general terms, it generalizes modular forms to automatic representations, generally speaking, we use invariant functions instead of subgroups of sl2. but under subgroups of GL n to classify n dimensional representations of the Galois group, so it is a very beautiful idea that you can do this now, it is a bit dishonest because you really have to use a path and Del representation theory, but this returns to classical number theory. where analysis and now of course also representation theory are used to study number fields, we have been very lucky to have the help of algebraic geometry for 50 years and have sometimes been blinded by the fact that the Langlands test did not use algebraic geometry, it is just analysis. and Lang representation theory was a proof for the s4 case and this general case is still completely baffling to us, so let me mention that there is a little progress, so one of the simplest ones you can try to understand is the group of Gower, Gallagher's representation of Q.

BBQ Tucci L also said that the model said that this group has no solution, but if the determinant is assumed to be odd and the rows are reducible, then the row is associated to a modular form and The kind of way I've been describing the test uses similar techniques again. As an entry point, you have to enter using these solvable extensions, so you enter this theory using Langlands theorem using also the reducible case, but then with some very, very clever induction on primes, they managed to push these techniques to prove the modularity of the association to modular forms of these. types of extensions, so this corresponds to extensions of the rationals with certain types of Galois groups located within gl2 of said set of models, but this is still a very, very special case of the language program, unfortunately the method does not seems to work in the general case. the general case would be gln or even any other group and any number field instead of the rational ones, so we are at a stage where we feel we have mastered the abelian configuration.

We'd really love to understand what doesn't have a solution, even a simple group setup. we are just putting together some results on them, the solvable case, but we have some isolated results like this on the genuinely unsolvable case, but we are still at the beginning of this great undertaking, the Langlands programme. Thank you so much. It's time for some questions. Thank you very much for the exciting conference. You mentioned that one of the key ideas and the entire presentation was Fry's idea of ​​constructing an elliptic curve that is now called a Phi curve, so at the time I told myself that I thought this was some kind of joke to reduce an unsolvable problem to an even harder and more complicated one to solve a problem, it was obviously proven wrong, but I would like to hear your idea and how much you perceived it as a joke or if you immediately felt that this was realistic. and there was a chance, sorry, who said it was a joke, he did it right, he felt it was a joke to reduce something unsolvable to something even more difficult, yes, I didn't know he said that, but actually there was an even more story painful I think Aguar some years earlier had studied similar equations and had wondered if they were related to hunger and had shown them to a very eminent mathematician who said it was ridiculous that he could do much more that way and that's why, according to him, there was stopped working on it, sir, yes, yes, I think that happens.

I must admit that when Fry made this proposal, he was actually very skeptical, not about the modularity conjecture, but about what Ribbit proved, so I was very surprised that we were very happy when Ribbit proved it. I thought it was just an illusion, but yes, I would like to go back to primary school and to the beginning. Yes, you said that Thermo himself solved the problems for N equals three and four and I suppose after that there was a third one to solve it for individual numbers or classes of numbers greater than four five, for example, yes, could you say a few words ?

Then how? Yes, the first to realize it was Euler because the proof of N equals three had not been written, so Euler made the case N equals three. I think he made a mistake the first time, but I think he corrected it so that the next one was N equals five and that was very complicated, you can try to do it the same way. but it is much more complicated and I think the story is that I quite like this story with my age that dares to be in my early 20s and did a part of what is called the first case.

I think about the problem, since I mean, it is now. He called it so he made a case but he didn't complete it but somehow he was talking about it and a meeting, I think it was at 70 years old, I went in and he beat him to it, so I think it's but those two then I think the case which is equal to five the case was made N is equal to seven I think this precipitated the problems of the French Academy because I think LeMay claimed a very complicated proof and was announcing it to the French Academy and to some of the other mathematicians like Co , she got really excited and said yeah, I can do that too and I can do this and unfortunately I think that was where the mistake was the single factorization problem, he hadn't solved it, he had made some assumptions so I think finally case seven was done with practical methods, but I forget the details, but anyway there was an attempt with several non-prime numbers, but then when the participants' work became known, the participants' methods made all the numbers prime less than 100, except 37 59. and 67 unfortunately commas Methos could not even be used to prove that it worked for an infinite number of primes.

I don't believe it. It's stagnant. I mean, it looked very promising for a while, but uninstall it.