 Thank you Alina. So welcome everybody and I welcome especially those who are in the middle of the night and who listen to my talk. Hello Mike. And so I thank the organizers for this great idea of organizing this web seminar and also for the great idea of inviting me. So thank you very much. So I am going to speak on the representation of integers by cyclotomic binary form. You have seen the abstract which is that I will start with speaking of the representation of integers as some of two squares. It's a classical problem. And there is a constant of London and Ramanujan which comes there. And generalized by Bernays for positive definite binary quadratic forms. And what I am going to explain is some joint work with Etienne Fouvery and Claude Levec, where we consider the representation of integers by the binary forms which are deduced from the cyclotomic polynomials. And our work, which is published in the actoritmetica with Etienne Fouvery and Claude Levec and in the bulletin that Société Mathématique de France, it will appear in June with Etienne Fouvery. It is based on some result which are due to Stuart and Xiao. Okay, so I start with the sum of two squares. It's a classical problem which goes back to at least Pierre de Fermat. And the result for the prime numbers is that the prime number is a sum of two squares, if and only if it is either two or else congruent to one module of four. So we have the list of prime numbers, which are some of two squares, which starts with two, five, 13, 13. I will give very often the link to the online encyclopedia of integer sequences, which is a really beautiful database, which will be very useful for this lecture and which I recommend. So this was for prime numbers. Integers in general, there is a result which is the identity of Brahma Gupta, which is that the product of a sum of two squares is a sum of two squares. And we have in fact two identities depending on the sign that we have here minus plus or plus and minus. Using these two results, we deduce it's an exercise that a positive integer is a sum of two squares, if and only if each prime divisor which is congruent to three module of four occurs with an even exponent. So we have here the list of the sum of two squares and the list of the numbers which are not sums of two squares. And the goal is to look at these numbers below N and to ask how many of them are there, what is the density. The first result is that if you take the x y with x square plus y square less than N, you do not count the x square plus y square you count the x y, then you find a number which is greater than N. And therefore, there is some repetition. And it's a fact that if an integer is a sum of two squares, usually there are many solutions to the equation x square plus y square equal M. Well, if M is prime, there is essentially one solution, well, with the symmetries x minus x or you permute x and y. But if M is as the S prime devices which are congruent to one module of four, there are at least two power S minus one solutions. So in fact, there are many repetition when we write a number as a sum of two squares. And the result is that the number of sum of two squares has density zero. The number is asymptotically N divided by the square root of log N times the constant. And this constant is the so-called Landau-Ramanujan constant. It is divided by an infinite product, Eulerian product. And you see that this Eulerian product is known with a very high precision. I give only the first digits, which are on the encyclopedia. But to compute such an Eulerian product using, well, using the product itself is not efficient. In some other ways, there is a nice paper by Flageoli and Vardy that you find a reference on the encyclopedia of integer sequences, which enables you to get many decimals of this Landau-Ramanujan constant. This result of a Landau-Ramanujan can be made a little bit more precise. We have an expansion, which is in our paper with Etienne Fouvery and Claude Levec in Acta Arithmetica. We have a power expansion with many terms, which is given by this formula, alpha 1, alpha m are just real numbers. This work on the form x square plus y square has been generalized by Bernays to positive definite quadratic forms. So you take a form a x square plus b xy plus cy square, the quadratic form. You assume that the discriminant b square minus 4ac is not a square and is positive. And the result of Bernays is that the number of integers, which are represented, which means that they are of the form f of xy, where xy are in z. This number is asymptotically c, a constant depending on f, times n divided by square root of log n. So this is really the generalization of the previous result for x square plus y square. This was published in 1912. It was the PhD dissertation of Bernays done under Edmund Landau. So to a certain extent we can say that this is the conclusion of a lot of work which was started with Fermat and pursued with Lagrange, Le Gendre, Gauss. Bernays is an interesting character. He is mentioned with Landau. After that, the year after he got his habilitation with Zermolo, he worked with Polia, Einstein, Hermann-Weil, Hilbert, Neuter, van der Waarden, Erglot. And he's known for the Hilbert-Bernays paradox, which was published in the book with David Hilbert, von Lagern der Mathematik, and also for the von Neumann-Bernay-Gödel set theory. Misha, just one question please from Fabian Patsuki, the alpha i are algebraic or transcendental. Ah, this is a good question, which we don't know. You mean this constant c454? Ah, this alpha we don't know, even for c454, we do not know. There are just c454, we have a good expression, but we do not know whether it's transcendental or not. It is expected that it is transcendental. So after the quadratic form, I would like to go to the forms of a higher degree. The main difference between degree 2 and degree at least 3 is that if a positive integer is represented by a quadratic form, there are many such representations, and the reason is that a quadratic form has infinitely many automorphisms. And Stanley Xiao explains in his paper, which is here, that this is exactly the reason for which there are many representations. But if you take an irreducible binary form of higher degree, the group of automorphisms is finite. So we do not expect to have so many repetitions if we have a form of higher degree. I will expand a little bit the case of the form xk plus yk, where k is an integer at least 3. So for k equals 3, we look at the numbers, which are some of two cubes. And you got some information in the last lecture by Andrew Sutherland. And so we know that it's rather rare that the number is of the form xk plus yk for several values of xy, that there are several representations. For k equals 4, it occurs, but it's very rare that there are several representations of the same number as a sum of two fourths power. It is expected that for k at least five, when there is a representation, this representation is unique, but a two symmetry, but this is far from being known. So let me start with you know from the lecture of Sutherland that 1729 is the smallest positive integer, which is a sum of two cubes in two essentially different ways. 10 to the cube plus nine to the cube and 12 to the cube plus one to the cube. The story is written by Hardy, who say that in 1917 he visited Ramanujan who was in the hospital, and he told him that his taxi number was 1725 and that it was a rather dull number. It's difficult to believe that Hardy really considered that it is a dull number because it was already known by Freny Claude Bessie that this is the smallest integer, which is a sum of two cubes. But anyway, this is the story of Hardy. This gave the opportunity to read a little bit to say that every positive integer was one of Ramanujan personal friends. There are other properties of 1729, which I want to mention some transcendental number. If you look at the transcendental number E, the next 10 successive digits of this number if you start with 1729 decimal digits are just the 10 digits from zero to nine consecutively. And this is the first time that they appear consecutively. Okay, anyway, there is a sequence of taxi cab numbers. These are the numbers, which are the sum of two cubes in more than one way. And the first one is our friend 1729. The next one is 4104. And we have a sequence. This sequence is infinite for trivial reason. If n is in this is the sequence you multiply by a cube and you have another number in the sequence. But this is not the single the only sequence of taxi can't numbers. There is another one, which goes back to Fermat. You look at the smallest number that is the sum of two positive integral cubes in n ways. And in the Wolfram website, it is written T of n. So T of one is two because one is two is one square plus one square. There is one way to do it. T of two is of course 1729. And then you have T of three, which is the number with the sum of two cubes in three different ways. Then you have T of four, T of five. We know one number, which is the sum of two cubes in six ways. We don't know whether it is the smallest one or not, but it is very likely that it is the smallest one. This sequence is known to be infinite. The proof is given in the book by Hardy and Wright, and this is a result which was proved by Fermat. The numbers expressible as a sum of two positive integral cubes in n different ways exist for any n. But this is not the end. There is another sequence, which is very interesting. It is a cube free taxi cab numbers. If you look at the number which is written on the top here, you see that it is a sum of two cubes in three ways. There is a number which is cube free, and this is the smallest cube free taxi cab number, which was found by Vojta in 1981. There is another one, which is known, and which is a sum of two cubes in four ways, and it is cube free. So you have the list here on the website of the Anticlopedia of Integer Sequences. And the question is to know whether this list is finite or not. And it's interesting because there is a result by Silverman that if this sequence of cube free taxi cab numbers with n representation is infinite, then the model of a rank of the elliptical of x cube plus y cube equal a n tends to infinity with n. And this may be considered as an evident that the sequence should be finite, but we do not know. After the cubes, I would like to continue the story of Hardy and the taxi cab, because Hardy said that he asked Ramanujan whether he knew what a number which was the sum of two fourth power in two different ways. The answer of Ramanujan was that he does not know, but very likely this is a very large number. As a matter of fact, such a number was known by Euler, which is the number which is given here. And the infinite family of one parameter solutions for this equation is known by L keys, and it's not trivial, it's not possible by multiplying by a fourth power. So we know that the sequence is infinite. And for the sum of two higher powers. When we looked at the sum of two squares we saw that it was given by a congruence, but this is definitely not the case for higher power. And one reason is that there are not enough prime of this form, the known the primes of the form x cube plus y cube. We know the finite number of them, but we do not know whether the list is infinite. As was explained by Andrew in May 7, a number is of this form if and only if it is represented by a polynomial in one variable, 3x square minus 3x plus one. So we understand rather well the numbers which are represented by quadratic binary forms, but by polynomials in one variable of degree two. This is something which is not well known. It's the same if you look at the fourth power. If another integer is a sum of two fourth power, each prime number which is not congruent to one module eight has an even exponent, but this is not sufficient. You have here the least list of the sum of at most two non zero fourth power, which starts with zero, which is an empty sum. If you look at the list of primes of the form x four plus y four, we have this list, but we do not know whether it is finite or not. The largest which is known is a generalized for my prime, with more than one million digits. So let's get some further information on the prime numbers, which are represented by x to the k plus y to the k where k is the power of two up to 32. So I repeated that there are many things we do not know. I just want to quote some results, which are known and which are very impressive. The form x square plus y four, which is not the homogenous form. But if you take also the form x cube plus two y cube, which is a cubic form, then it is known that there are infinitely many primes of this form. And it asymptotic order is known by a Finlander and advantage for x square plus y four, and by his brown for x cube plus two y cube. So we know for x cube plus two y cube, but we do not know for x cube plus y cube. So, I am going to investigate the representation of integers by binary form of degree at least three. What do we expect? You take a binary form of degree at least three with non zero discriminant. And you want to count the number of integers of absolute value at most n, which are represented by this form. So there are f of x y where x y are in Z. What do we expect. I can give you the answer. It's a constant times n with the exponent two over D. And the reason is the following. You look at the integers x and y, which are of absolute value less than n with the exponent one over D. So the number of these integers is with the exponent two over D. And each of them will give value, which is bounded essentially by n. And we expect that these numbers will be essentially distinct. So this would give the answer. In the first way, if you have a number which is of the form f x y, if you have a pair x y with f x y less than n. Most often x and y will be bounded by n to the exponent one over D. It will be always the case if the form is definite positive. But if it is not definite positive, you have few x y for which the maximum is not less than n to the power one over D. So this is what we expect a constant times n with the exponent two over D. And as I explained, we want first to count the x y, which has the property that f of x y is bounded. And the expected result was proved by Mahler in 1933. You fix the integer real number z and you count the number of x y, not the number of f x y, but only the number of x y, so that f x y is z and z. This number is essentially the area of the fundamental domain times z to the power two over D plus an error term, which is one over D if there is no linear factor which will be the case interesting for us. So this constant f is the area, the big measure of the fundamental domain, which is defined by this. So this paper of Mahler like all the papers of Mahler is available on this website, which gives all the papers. Now, I want to look at the representation of integers by forms of degrees three and four. I repeat here that what we expect in general. Oh, excuse me, I have a small problem. What I don't know what you see. I'm afraid that. No, it's all good. I mean, it's all looking for me. I do anything. I lost. I lost to the my file. So, okay, it's okay. It's okay. Okay, for me, it came back. It disappeared, but now it is back. So I know that this number is a constant time and to the power two of D and this was proved in three papers to for the cubic binary forms and one for the quadratic form by Christopher who lay for the cubic forms. He first proved it in 1967 in some cases and the other cases in 2000. And for graphic form in 1986. So the case of degree three and four is sold by who lay. There are a few other cases where the problem has been sold. These are the forms. So, for example, XD plus YD, where these at least three, they are asymptotic estimate, and also more generally, hello Mike, but for a XD plus BYD when AB are non zero integers. So the whole case will be very recently by Stuart and the Seattle. And the result is the one which is expected. You take a binary form of degree D at least three with non zero discriminant. And the result is that there exists a constant CF such that for N going to infinity the number are N are F of N of integers bounded by N, which are represented by F is equivalent to CF and with the exponent to over D. I will come back to the error term which is quite sharp in their paper. So the constant CF as we expect is related with the area of the fundamental domain. And it is this area multiplied by a number which is less than one, which depends on the group of automorphism of F and AF is the area of the fundamental domain. So this was published in 2019. Okay, this is more or less the historical survey that I wanted to give. And now I come to the main topic of my talk, which is to consider cyclotomic binary forms. I start with the cyclotomic polynomials in one variable. I will define the cyclotomic polynomial Phi N by induction. We start with Phi one is T minus one and Phi N of T is T N minus one divided by the product of the Phi D for D dividing N and D different from N. So if we take this definition, we get the decomposition of T N minus one and we can compute the Phi N by induction. So the prime number, we have T P minus one plus T P minus two plus T plus one. For example, Phi two, Phi three, Phi five are given by this formulae. You have other formulae for four, six, eight and 12. And you see that the Phi four and Phi six are polynomials of degree two. Phi six is just Phi three of minus T and this is a general property that Phi two M of T is Phi M of minus T when M is odd. And then you have these two polynomials which have degree four, you have Phi five, which has degree four and also Phi ten because of this formula. So the summary of this cyclotomic polynomial is the so-called Euler torsion function, small Phi of N. There are three Phi in my talk. This is Phi for the Euler torsion function, this one for the cyclotomic polynomial and the capital one for the cyclotomic form, cyclotomic binary form. Just by making homogenous the polynomial Phi N. So this is a binary for degree N. You have the two first one Phi one and Phi two, which have degree one. I will just mention them once later, but otherwise they will not be so interesting for us. We have the forms, the cyclotomic forms of degree three, of degree two, which are Phi three, Phi four and Phi six. But Phi six is essentially the same as Phi three, they represent the same integers. And then you have the forms of degree four, which are Phi five, Phi eight, Phi 12 and Phi ten, which is Phi five of X and minus Y. So these are the objects that we are interested in. We want to investigate the number of integers which are represented by these binary forms. We can use the result of Stuart and Xiao if we fix one binary form. We know that the number of integers which are represented by Phi N is, so I denote it by Phi N of N, it is a constant depending on Phi N, N to the two of D, where D is the degree of Phi N. And there is an error term. And this constant C Phi N is the area times a small number WN. So the error term, I did not tell you before what it was, but I give you only for when D is even. And you see that for D, at least 10, it is one of it. So you can consider that this is essentially one of it. If I end this number here is the area of the fundamental domain. And this small WN is one over eight or one over four, depending on whether four divides N or not. And this depends on the group of automorphism of Phi N, which is isomorphic either to the deodorant group D2 with four elements, which is here, or to the deodorant group D4 with eight elements in case four divides N. So we have a good knowledge of what is this constant C Phi N. Maybe I can say a little bit more about the cyclotomic fundamental domain. You have a picture which was given to me by Francesco Pappalardi, which is the image of the curve Phi N of X, Y for N from one to 40. So you see that here. Do you see the my pointer? Yeah. Okay, so here you have Phi one, which is X minus one. Here you have Phi two. I told you that I will mention them only once. So it's here. And otherwise, you start with Phi three, Phi four, Phi five and so on. And so these are the fundamental domain. You have the pictures here for three, four, seven, nine, 12, 15 and 20. It's not too difficult to prove that this domain tends to the square centered at the origin and with size two. We have a more precise result, which is given here. But when N is large, it's very close to the square, as you see here. There's another exercise that the cyclotomic fundamental domain is convex if and only if N is either prime or twice a prime or power of two. If you look at this one, which is the N is nine. If you look carefully, you will see that it is not convex. It is not a prime, not twice a prime. Okay, so first result I want to explain is about the number of integers which are represented by at least one of the binary cyclotomic forms where N is the index is at least three. And this number is asymptotically a constant alpha times N divided by the root of log N minus a constant beta times N divided by log N to the three quarter plus a big O of N over log N. And I will explain the successive terms. The main term is this one with alpha and alpha is just the sum of C five four and C five three. So we have the value which is here. It occurs from the contribution of the two quadratic forms five three and five five four five six is the same is the same contribution. The next term, which is this one with beta comes from the contribution of the numbers which are represented by the form five four and also by the form five three, because here we counted them twice. So we have to subtract truth that we counted twice. And so this gives the number beta. I will explain that the error term is sharp. And this error term takes into account all the binary forms cyclotomic forms of degree at least four. So, here, we have the contribution of the forms of degree two, and there is some error term, and then we have the contribution of all the other cyclotomic forms. We already saw what is five four. It is the sum of two squares and the story is very similar. If we look at the form x square plus x y plus y square, the numbers which are represented by this quadratic form, the prime numbers are those which are either or which are congruent to one module or three. And it turns out that the form x square plus three y square represent the same numbers. Next for the product of two numbers, which are represented by this form, we have a number which is again represented by the same form, because of the formula which is here, which is just a formula for the norm of a product of element in the cyclotomic field q of square root of three, minus three, so q of zeta three. So, thanks to that, we can give a criterion for number to be represented by this form. And it is called the Lushian numbers on the encyclopedia of integer sequences. These numbers which can be written like this are those for which the prime devices of M, congruent to two modules three, occur with an even exponent. So, we are the same thing, we have the list of numbers which are represented by this form, the list of numbers which are not represented by this form. And what is the density of each? The result is that the density of this one is zero, and the density of this one is one. The number of positive integers which are represented by this form, x square plus xy plus y square, is a constant times n over log n square root. And this constant, again, is given by a Euler product. And this constant occurs with many decimal digits. You will not find more than six digits in the encyclopedia, but you will find so many digits in this paper of Etari, Ramari, and Surel, which is about a fast multi-precision computation. And again, like for the sum of two squares, we have an asymptotic formula for the number of integers which are represented with some other real numbers, alpha one prime, alpha two prime. And we do not know whether they are transcendental or not, even for C5 three. Now, this takes into account the main term, which was called alpha, which is C5 four plus C5 three. The next term in my formula was the numbers which are represented by the two forms. So they are both of the form x, y, four of xy and five three of uv. So we have to count them. And we know because of the previous criteria that the one which are of this form are the one for which the prime divisors not congruent to one module 12 occur with an even exponent. And this enables us to give another asymptotic formula, which is, again, like the previous one with Claude Levesque and Etienne Fouffry. And we have an asymptotic expansion with the number beta, which is given by a nice formula involving some oil products. And again, we know many digits thanks to the paper of Etari, Ramari, and Surel. So this is for the two main terms. I want to investigate the error term. In the main theorem that I mentioned before, we had an asymptotic estimate for the number of integers which are represented by one at least of the form with n at least three. So we have to look at the forms of higher degree. And when we look at these forms, we have all the prime numbers which come in because the prime numbers are represented by many cyclotomy binary form, all the phi with the index p to the r, where p is an odd prime and r is at least one. And of course p equals two occurs as a value of phi two, for example, phi four. So to count the number of integers which are represented by one at least of the binary form, we have always all the primes which come into the picture. And this gives the error term, it gives, at the same time, the fact that the error term is optimal, it cannot do better. But this is not exactly what we expect, we would like to get rid of the prime numbers which disturb the situation. So what we will do now is to omit the value one one, and we ask that the maximum of XY is at least two. So we will count the number of integers, which are of the form phi n of XY, m is less than n, but XY are restricted to the maximum of X and Y at least two. And what is changed here, compared with zero and one, is only the error term, which is a little bit more precise, it is n divided by log n to the power three half. So this is the result for quadratic forms. I have to explain the error term, why do we have this error term which is so small. So we have to take into account all the forms with phi of n greater than t, we have infinitely many forms. So I use this notation, ad of n, when d is fixed is the number of m, which are of the form phi n of XY. And for a greater than or equal to d, I introduce the restriction that the maximum of XY is at least two. So the theorem that I just mentioned, state that the number of integers, which are in a greater than or equal to two of n, is given by this formula alpha n of a square root of log n minus beta n of the root, the root three over four of log n, and then this error term which is here. So I will explain that and we have to look at infinitely many n. If we fix one form phi n, the contribution to the error term is very small, it's n to the power two over d, but it's much smaller than that. But there are infinitely many forms. So we need a uniform estimate. And we have a uniform estimate, which is not exactly what we expect because we have n two over d, but what is good is that we have an explicit constant 29, and this will be good enough. So this is the number of integers in a greater than or equal to d of n, phi of n is at least d. And these numbers are represented by one of the four. So we will see where this 1.161 comes from. So for that we need a low bound for the secretive polynomials. There is a result which goes back to the work of Gurian Lovac in 1970, and during 1977, which gives such an estimate, which is phi n of t greater than two to the minus phi of n times the maximum of one and phi n t, the power phi of n. And this result in the general case of the norm form for a CM field. And this is the case for the cyclotomic field. In their case, which is more general, the estimate that they get is best possible. And they give an example for that. In the case of cyclotomic forms, this is not enough and we need an improvement. And the improvement comes from a careful study of the cyclotomic polynomial. The cyclotomic polynomial takes only positive values. It has a minimum and infimum and this infimum we can estimate it and the estimate that we have here again is true. And so this gives the low bound for phi n of x y, which enables us, thanks to the assumption max x y greater than equal to two, to give an upper bound for n when phi n of x y is equal to m. This n cannot be large phi of n is bounded by log m. And this gives an upper bound for n. And this is where you see the constant one point one six one, which occurs in the previous results. And this is because of this upper bound that we get the explicit estimate that I mentioned and which is what what we need. So, I told you that this estimate for the number of integers, which are represented by a form with phi of n greater than or equal to D was not optimal we would like to get something more precise. And for that, we need some auxiliary result which may be of independent interest on the number of integers, which are represented by two non isomorphic binary forms of the same degree. So we I gave the definition of automorphism of form, I give you the definition of isomorphism of form to binary forms are isomorphic. If there exists a matrix, which is invertible with rational coefficients, so that f one of x one x two is f two of you one x one plus you two x two, you three x one plus x, you four x two. And now I would like to look at two binary form which are not isomorphic. And I would like to look at taken by both forms at the same. So you have x one x two x three x four, but the values are the same. And we will need an upper bound for the number of x one x two x three x four. And we will bound the number of f one x one x two but now we we found the number of x one x two x three x four. And there is this bound. You have two non isomorphic binary form of the same degree. The discriminants are non zero. And there is an assumption that we cannot avoid that one at least is not divisible by a linear form with rational coefficients. So the number of x, so that the two forms take the same value is bounded by be with an exponent, which is essentially one for the at least sufficiently large. So we have an explicit upper bound for this. I will just say a few words. So the proof basically that time is passing. So I will be brief for the sketch of the proof. The proof is based on results of is round pool a salberger. And in fact, this result is very close to the result some results oxidative result in the paper by Stuart and show that I mentioned before. So we went to an upper bound for the number of integral points on the hypersurface f one equal f two. So the x the work in two parts. The first part is to use a result of per salberger on the number of points for which the projective point does not lie on a complex projective line, which is containing x. And there are some assumptions in the work of salberger that we have to check, but it works and we get the main result, the main term. And this is of course to estimate the number of points which do not lie on a projective line containing x. And we have to make some work to check how many there are on each line. And this is where we use the fact that the forms are not isomorphic. And then one uses an upper bound for the number of these slides. And this gives the error term which is big O of B. I will use this for estimating the number of numbers of integers which are represented by six atomic binary forms of the same degree. So I start with two binary forms six atomic binary form fire and one and fire and two. And I want to count how many m can be represented by the two numbers fire and one for two forms fire and one and fire and two. And the result is that there are not so many of them. And the exponent here is a gamma d over D. This is explained just by the fact that now we bound M. And before we were bounding in B. So more or less in B occurs with the exponent D, which is the degree of the form. And so this is why we have to bound that so essentially we say that the number is bounded by N with the exponent one of the D plus epsilon. So there are not so many repetition for the forms of the same degree, assuming that they are not isomorphic. So this raises the question, how do we know whether two forms are isomorphic or not. And the answer is that we use this corollary to get the answer. So isomorphic. We take two integers and one and two and one is 11 and two. And the following condition are equivalent. The two forms are isomorphic, which imply that they have the same degree. The second condition is that the two binary form represent the same integers. And the third condition is that N1 is odd, N1 is a small one, and N2 is 2N1. So this gives a complete criterion. By the way, when am I supposed to complete in five minutes or something like this? Five minutes is perfect. Yes, Michel. Thank you. Okay, so I can give a few hints on the proof of this to difficult. First, we prove that if the two forms are isomorphic, then N1 is odd and N2 is 2N1. And the proof is as follows. You assume that they are isomorphic. So the primitive roots of unity of the corresponding indices N1 and N2 are related by this with a matrix, which is a regular matrix with rational coefficients. And therefore, the two cyclotomic fields q of zeta N1 and q of zeta N2 are the same. But now it's something which is not difficult to prove that the torsion subgroup of q of zeta N, the cyclotomic field of index N, is cyclic of order N if N is even and of order 2N if N is odd. So it just gives the conclusion three. The very easy part of the corollary is that if N1 is odd and N2 is 2N1, the two binary forms represent the same integers just because of this formula, which we saw several times. Phi 2N is Phi M of minus t. And now the interesting part is two in place one. Assume that the two binary forms represent the same integers. First we prove that they have the same degree. And this is just by counting how many integers are represented. It is essentially N to the power two of a D. And it is the same for Phi N1 and Phi N2. So the D is the same. So we have Phi N1 equal Phi N2. Now that they have the same degree, we look at the integers which are represented by both forms. This is the same as the numbers which are represented by one or the other. So it's not so small. And because of the previous result, we deduce that the two forms are isomorphic. Okay, so I will complete by giving a result, which is valid for any integer D, which is the value of the Euler-Toschan function. We call such a number a torchant. A torchant is a positive integer, which is the value of the Euler-Toschan function Phi. So it's a D, so that there exists N with Phi of N equal D. And I will denote by D dagger the next torchant, which is greater than D. It is always at least D plus two. And it is bounded by 2D by the results on the prime numbers Bertrand postulates as it is called. So we have a list of even integers which are not values of Euler-Toschan function. And you see that sometimes there are two consecutive numbers which are not such values. And here you have four consecutive even integers, which means that if you take D equal 240, then D dagger is 250. So this is a list of non-toschan, even answered that Phi of N equal N has no solution. And this is not so, we do not know so well on this, but there are several results of four, which gives some explanation of what is going on there. And now I come to the main result, the last main result of my talk. We fixed an integer D at least four, because for D equal two, this was a theorem one. We take N which tends to infinity, and we count the number of integers M for which there exists an integer N and the X, Y with Phi N of X, Y equal M. Phi of N is at least D and the maximum of X, Y is at least two. And we have an asymptotic expansion for that. It is C D N to the power two of D plus a big O, essentially of two over D dagger. We did not succeed to get it for D equal four, but we have it at least for D at least six. What is this constant? It is what we expect. We look at the cyclotomic binary form of degree D, which are not isomorphic. And not isomorphic means that N is not congruent to two modulo four. So this is the main term. And now for the remainder, it is N to the power two of D dagger. And I would like to explain that if D is at least six, because we want to be in this case. And if D dagger is D plus two, which occurs sometimes, as we have seen in the previous example, then the error term is optimal. So to prove that it is optimal. We can prove that D is a torchant. So it's some fire of N and D plus two is also a torchant. This means D dagger is equal to D plus two. And we look at the numbers five M of UV with the degree of five M is the dagger, which is D plus two. And among these numbers, we prove that the positive proportion is not of the form five N of AB with five of N and equality. And this will show that there exists a positive constant, so that the number of integers which are represented by a form of degree at least D is at least a D of N. The degree is exactly D plus a number which is a constant times N power two D two over D plus two. The proof of this result uses a lemma, which is the final lemma, which is the following. If you take an integer N at least two and the prime number which divides N, then for all A and B and Z, five N of AB is congruent to zero or one module of P. There are similar results for the congruences, module of four and nine. So this is sufficient to prove the result which is here. The confinement was coined by Etienne Fouvery in our joint paper, which was accepted in September 2019, because the values of five N of AB are congruent to the classes module of P, only zero and one module of P. So when this paper was accepted, we did not expect that the word confinement would become so popular after that. I will stop my lecture. Thank you.