 organizers. Well, it's my first experience in living. I talk on Zoom. So hope it would work. Do you hear me well? Hello, do you hear me well? Yes, I hear you well. Okay, very well. So I continue. So I will talk on a joint recent very recent joint work with Florian Luca and Damalia Pizarro. And it's based on the conjecture of one more colleague, Antonia Riffo, who was a PhD student. So, so well, let me start by introducing my co-authors. So maybe now they're not like this, they will wear masks and so on. Hope this time would will be back. And well, let me start from reminding some definitions. Well, our main character of this talk will be singular modulus, singular moduli. What is a singular modulus? It's just by definitions, j invariant of an elliptic curve having complex multiplication. So remind what is a j invariant of an elliptic curve given in right transform. But in this talk, in this talk, we will use an alternative equivalent definition is the value of j of tau, where tau is in Poincare plane is a quadratic rationality. So it's just an algebraic number of degree two. And j is a function, analytic function defined on Poincare plane with a familiar function given with this familiar series. So I'm going to remind more of the definitions with standard material. And while the two definitions are equivalent, if the elliptic curve obtained by quotienting complex plane, where the letters generated by tau and one, then j invariant of this curve is just j of tau. And there are two main numerical characteristics of singular modulus. First is discriminant of a singular modulus, defined delta sub x, it can be defined in two equivalent ways. One is kind of more conceptual way. It is well, since we have an elliptic curve with complex multiplication, so they are reading of endomorphism is set in imaginary quadratic order. And this is just the discriminant of this order. And well, when we know the discriminant, we know the order. So every discriminant there is exactly one order of the discriminant, which is this order, or equivalently, well, more computational way. And this is, well, how in this talk, we will understand the definition of the discriminant. So our tau has a minimal polynomial of degree two. Now that I write minus b, because I like to put plus b here. So there will be minus b here. And delta will be just a discriminant of this minimal polynomial. So what we know about discriminants that while it's so close negative number, because it's imaginary quadratic, it's also obviously congruent to zero one or two. And while we know that every delta with these properties serves as discriminant of some model. Now, the fundamental fact, well, besides discriminant, while there is a degree of singular modulus, and the fundamental fact that the singular modulus is an algebraic integer. And its degree is equal to the class number of this discriminant. Well, again, I don't want to go to define what's the class number. Well, if discriminant is what is called the fundamental discriminant, so the order is the, the maximal order, just the ring of integers or the corresponding imaginary quadratic field is just the class number of the field. In the general case, one should be a bit careful with defining, but one can again define the class group and the class number and so on. And another important fact that while for a given discriminant, all singular modulus is given discriminant for megalo orbital. So we have a since degree is h, then we have exactly h singular modulus given discriminant and they all are conjugate over. These are two fundamental facts that we're going to use throughout talk. And now what some consequences, for instance, we know that there are 13 discriminants of class number one. This is the famous class number one problem solved by Hebbner-Starkin Baker. And while here, you see the list of all these 13 discriminants and the corresponding singular moduli, they must be now algebraic integers of degree one, which means they're just rational integers. Now, a singular moduli, well, when there are 29 discriminants of four degree of class number two, which means the corresponding singular moduli i pairs, they are of degree two and 29 bellow orbits, 29 such pairs of singular modular degree two, there are 25 discriminants of degree three and so on. At present, all discriminants up to degree 100 are known due to the work of Watkins. Okay. And now let me quickly state what is reforced conjecture. So before conjecture the following that a singular modulus of degree, at least three, cannot be root of a trinomial defined operations. Where by trinomial everywhere in this talk by trinomial, I just mean the following polynomial. So it has polymonic has one term of maximal degree, then a certain term of smaller degree, but positive and then the free term. This is what I mean by trinomial. Well, in the way, I do not want to formally assume that the middle coefficient a is not zero. But well, if it is zero, then this conjecture is really very, very easy. It's just an exercise. So, so the problem becomes interesting exactly when a is not zero. And also why H must be at least three, because well, H is one, then the singular modulus rational number, obviously, it is a root of trinomial. And when H is two, then again, it's a root of trinomial infinitely. So the problem starts being interesting for age greater than three. And well, this conjecture is open as of now. Now, one may ask for motivation, but it is motivated. Well, to be honest, I don't think I'm me personally, I don't need any motivation. We have two very classical objects, singular, modular, classical objects known since 19th century work of no chronic or whatever. And we have trinomials, another very classical, very interesting object and well, to know how they kind of their relationship is, is of interest for me. And it is, of course, a problem of what there's their fountain flavor. So any problem interesting, their fountain problem is of interest, I don't need a special motivation for this. But well, maybe not everybody would agree. So I will try to provide some kind of motivation. And motivation comes from the very hot topic as of now equations, singular modular. And it all started in 1998 by theorem proved by even Ray, who proves the following theorem, assume that f is a polynomial with unit irreducible and it should not be special. And then the equation f of x, y equal to zero has at most finitely many solutions and singular mode. It was the first no trivial case of what is called Andrea or conjecture. Now there is much more, this is much, much more advanced to the work on Andrea or conjecture. But well, let us pretend that we don't know about what was done later, I would just want to concentrate on this on this result. Maybe it was defining what are the special polynomials that we have to exclude. So these are polynomials, well, of course, this kind. In this case, well, I must probably say special times, non zero, non zero scalar. These two, it's clearly special. And there is one more series of special polynomials are the classical modular polynomials of level n, which are by definition the irreducible polynomials over z, which are defined will satisfy this equation, give the algebraic relation with the J invariant and the J invariant of n multiple. Well, let me give just example, well, phi one is clearly x minus y because it's just J and J, already phi two is is a bit somewhat big coefficients. And for phi three, I will know the separate slides. So I don't give it. Basically, while we seldom work with modular polynomials as them, though there is some recent work, for instance, bounding their coefficients with the Pazuki. But for me, well, the good thing about modular polynomials is that they exist. But really, I try to avoid working with them as such. And well, of course, for these three examples, we have infinitely many solutions. For modular polynomials, this is if z is tau imaginary quadratic, and tau is also imaginary quadratic. So we have to single a modular problem. So we have to exclude them. And when we exclude them and their multiples by a scalar, then we must have finite solutions. And this theorem, there were many proofs, the original proof of Andre. Then the same year independently, both edicts often gave a proof. He had to assume GRH, but otherwise, very nice and conceptual argument, and was further extended. And the recent famous work by Cling, the rule moya five and so on was kind of continuation of this approach. Then the totally different argument was given by Pila. And again, it was continued. It was using all minimality and it's nice stuff. Oh, well, the problem with Andre's and Pila's argument that they went on effective. At some moment, they had to use Ziggle Browler. And, well, about like eight years ago, first effective proofs were given, while independently by Lars Kuhne and my joint work with Maser and Zanya. And later, Lars Kuhne came with another argument which I would call very effective because it was that argument that could be then adapted to obtain really very explicit results. I'm not sure that there was any kind of adaptation with the initial this argument. So, Kuhne and Maser Zanya and myself. And while I say, well, when you have effective, we want to have explicit and by adapting this Kuhne's wonderful argument, certain rather explicit results were obtained. For instance, Kuhne himself proved the proof the following that if we consider the equation x plus y equal to one, which is of course the simplest, non-special algebraic equation, then there are no solutions in Singapore. Three years after, this was generalized in a joint work with Bilal Ember and Damali Epizarro. If we take three rational numbers, and well, let us assume just to avoid some trivialities that a and b are both nonzero, and consider this equation. Then this equation may have only obvious solutions. This is our result. Well, which are obvious. First of the solutions, they are the solutions with x equal to y. And of course, when a is minus b and c is zero, then we have the solution. So like diagonal line, we have plenty of solutions like this. Now, the rational case, it's when both x and y are rational, and we saw that there are 13 rational singular moduli. So we have quite a few pairs like this. Then of course, we can draw many straight lines defined over q passing through these points. And the third obvious case of the quadratic case. Well, clearly, if we have a solution, then the field q generated by x or q in between generated by y or what you must be seeing. And if this field is of degree two, then what we do, we take our point x, y, we take x prime, y prime, the conjugate point over q. And if we draw a straight line, it will be defined over q. So again, we have a solution in this case. And well, our result, there are no other solutions. For instance, there are no solutions when x or y are of degree, or at least, before generalized, partially, this result in his work, he considered the equation, a x to the m plus b y to the n, equal to c, a b c as as before, and m and n are integers. And they again are known. They're not fixed here in his work. So we have actual equation. Well, let us count basically one, two, three, four, five, and six variables. Well, formally seven, but a b c format two dimensional variables and solve solve this equation almost completely. So, before did the main part, but in some cases, his method failed. So when then Florian Luca was listening to his talk, and he had me him to fix the remaining few cases. And what he proved the was the following that there are only obvious solutions with x distinct from y. So if we return to the previous slide, we see we had three cases diagonal rational quadratic, the diagonal being the easier than trivial case. So before proved that if we throw away the diagonal case, the case x equal to y, then only rational case and quadratic case may happen. But in the case x equal to y, which was the easiest, when m and n are just one, in general case, it is the hardest. So his argument did not work in that case. And it's exactly that that case, it reduces to the following problem. Determine singular moduli, which are the roots of trinomics x equal to y, we're training. So, of course, in the case for singular modular degree one or two, and the before conjecture that there are no others, this is how this conjecture matched. And while, as he concedes in his article, that we know a lot about trinomials, but while it's not enough, our knowledge is not enough to prove his conjecture to rule out all the cases of degree, at least three. And now here is our results. Well, all them appeared recently, rather long archive article, like 24 pages or so. Well, our first result is just if we assume GRH, then the before conjecture is true. There is no trinomial, no trinomial or a rational, vanishes at a singular modular degree three. Well, so with GRH life is nice, but well, if we want to do things unconditionally, then well, it's more complicated. And let me introduce one definition that will be convenient to use. We call a discriminant delta trinomial discriminant, if it is class number at least three. And the modulus of this discriminant is a root of rational trinomial. And this is equivalent, of course, to saying all singular modulus of these discriminants are roots of some rational trinomial, because they form a gal orbit of a cube. And therefore, conjecture in these terms can be restated like that this discriminant do not exist. So the rest of this talk will be study will be I will be speaking on kind of potentially potentially non existent objects. And here are our results about this trinomial discriminants. So first of all, they cannot be too small, at least 10 to 11. Well, they cannot be too big either, but well, we should allow one exception. So with one exception, then can run run away. They're bounded by certain huge number. Well, here we were not very careful about writing about obtaining optimal bound. Well, maybe it can be reduced to some 10 to 140 or 120. But I don't think one can do substantially better with using our methods. So anyway, there is some total explicit bound for all that one. Now, these trinomial discriminants are not just arbitrary. They are very, very special. They are either minus prime number or minus product of two distinct prime numbers and distinct odd prime numbers, of course. In particular, they're all fundamental discriminants. And they are odd discriminants. They cannot be zero, not four. And our last results is about trinomial itself, which also must be kind of very special. So if trinomial vanishes it in singular modulus, and here we should impose certain conditions of singular modulus of sufficiently big discriminant. Well, this is kind of lazy assumption. So I think that it will work harder. Perhaps we could eliminate this assumption, but I'm not sure. So let's assume that the discriminant is sufficiently big. And if trinomial vanishes at singular modulus of such discriminant, then the two non-constant terms are of almost same degree. So the degrees, this can be so n, can be only m minus one or m minus two. It cannot be smaller. So these are our results on reforce conjecture. So in the rest of this talk, I'll try to give an idea how to prove, well, at least the first four theorems. If time permits, I will also say a few words how to prove for the last theorem about trinomials, which is probably the most sophisticated of them. So, but, well, I will start by giving some basic facts about trinomials and singular modulating, and putting special emphasis on this trinomial discriminance. And, well, then I will say a few words how we obtain this lower bound and how we prove this structural theorem. And after this, proving the conditional theorem and proving the upper bound will be just through very, very classical analytic number theorem. And, well, as I say, if time permits, I will also say a few words about the last one. Okay, so let's start from trinomials. So the basis of everything is the following innocently looking lemma. So if we have a trinomial just with complex coefficients, and we have, it has three roots, where w is the biggest and x and y are smaller, so and x bigger than y, then we have this inequality. So x over w is kind of small number, and it raised to the power m minus n. And since m is bigger than n, it's also just, you can also put here just one. And for most application just one is sufficient, but for the very last theorem about trinomials, m minus n would play a role. And what it means in formally? Formally it means the following, that if a trinomial has three roots and all these three roots, one is much, much bigger than the two others, then these two others must be of almost the same absolute value. So we see y over x must be very close to one and x over n. And the proof is very easy, so this discriminant must be vanish, must vanish. And now what we do? We take the two biggest, so we have six terms and this discriminant and the two biggest of them are those which involve w to the m, and then they must involve x to the n and y to the n. And we put them on the left and the absolute value is this, or these two terms together. And it must be bounded by the sum of absolute values, with the sum of absolute values of four other terms, and these terms are bounded by, well the biggest of them is like w and xm, all the other are even smaller. So it's bounded by four times this, so it's very loose bound. Actually the article will use something more exact, but here we'll talk about this loose bound. Now we divide, we obtain this equality, and note that this is, well just by triangle and quarter this, and since y or x is smaller than one, it's great to go this and write that. So proof is just totally trivial, but it is what we use. Okay, so basically we don't use very little beyond this about three elements. At one point we also use a trinomial, cannot have more than four real elements. I don't think that we use anything else about trinomials. We also obtain a similar result of a well-impaedic metric in non-accommodian case, which is also used in some cases, but well for the purposes of this talk I prefer not to talk. Yuri, please for a question. Fabian, would you please ask your question? Yes, please Fabian. You just answered the question. I answered, okay. Okay, good. Sorry. Now, now let us speak about singular modulate, and again I want to speak about some very classical fact, which is, well it's cool to go back to Gauss. Consider a set of triples, ABC of integers, which satisfies certain conditions. So just for every given delta. So delta is b square minus 4ac, so it should be. They're also co-prime, and they satisfying certain well inequalities, and the inequalities are equivalent to saying just that this number belongs to the standard fundamental domain, you know this nice picture. And well work of Gauss, of course Gauss did not use this terminology, but his work can be interpreted like this, that there is a bijection between the set and the singular modulate, you know, discriminant delta, which is given by this, so to each triple ABC, we associate j to evaluate it to this number. And in particular this set t delta has exactly h of delta elements, exactly h of delta triples like this. And what is crucial for more, more or less, well all results on this explicit Andrea Orto, this explicit solving equations of the singular modulate, is among these triples there is exactly one triple with a equal to one, one bc, which I give here explicitly, so there is one and there can be no more, but one there is. And the corresponding singular modulate, well just the discriminant, well there will be no a here, just two. The corresponding singular modulus, we will call dominant singular modulus for the discriminant delta. Okay, now let us, let us see what is how, how the this triple ABC and the corresponding singular modulus relate. So we have for j, we have, we remember this expansion and you see there is the main term q minus one, which is kind of big term and the other terms are smaller. And well when they are small, when the imaginary part of z is separated from zero, in this case we just can bound all what is left, so it will be just q minus one and something bounded, bounded in terms of epsilon and well the same holds true also for absolute values and this is what we are going to use. Now if we have a triple ABC, what we have then b can be positive or negative, but an absolute value is bounded by a and then bounded by c and this implies, well by this simple argument that a is bounded by square root of delta, even square root of delta divided by square root of three and in particular the imaginary part is well separated from zero. Well then also observe it by just noting that this belongs to the fundamental domain. So it's well separated from zero and this means that what is the q, the absolute value of q minus one is just this number, exponent of pi square root of delta divided by a. So our singular modulus is this plus some bounded guy, so we have a very good idea about size of our singular modulus when we know this number a that corresponds it's just a through of this guy plus something and in particular for dominant a is one, so we have very big exponent of pi square root of delta, so really really very very big guy. If it is not dominant then a is at least two, so we have at most square roots, so dominant is much much bigger than all other singular modulus and this is crucial for us because well when we start we remember trinomials when there is one big roots then the other smaller roots we can say something very good look what happens. So if we just take singular modulus of given discriminant on a complex plane probably I had to write singular modulus of given discriminant delta on a complex plane. So in general we have one guy which is dominant and by the way it is real because it is only one if there are two then if it is not real then there must be also its complex conjugate but we cannot have two that's only one terminal. So it must be real it can be real positive like on this picture it can also be real negative but it is real and all the others are much smaller but kind of we don't have much information they're kind of chaotic if we don't impose any restriction of our discriminant. But if our discriminant is trinomial then again there is one bigger dominant and the others are much smaller and even really really very small so you see it's bounded by some power of delta this is what we prove and they have almost the same absolute value see it's kind of like almost on the circle yes I intentionally did it it's not a circle it's kind of good approximation to a circle on this picture so they you should think of them as like almost on the of same absolute value and this is what we use in our arguments okay so we saw that the number a plays an important role and well we probably should give it some name so let us call integer a suitable for a given discriminant if it emerges as a in some triple abc so there are some properties of this suitable so as we have already seen one is always suitable for every discriminant and the suitable it is bounded by square root of delta divided by square three and we'll also obtain some recipes for detecting these suitable integers for various kinds of discriminants for instance for even discriminants that are well not too small either two or four or eight must be suitable now very important if p is a prime number such that the chronicle is one and again it's not to be equal bound like this but one would expect here three but actually we can prove it on the floor then p must be suitable next one well well sorry in particular well also this holds for p equal to two what is the chronicle of delta two is just one if delta is one with eight delta is one eight and if not minus seven then two is suitable and here well is co-prime suitable with some co-prime guy but well if also divisor of delta can be suitable it must then satisfy which condition it must be really an exact divisor so can capture all the primes it touches so a and delta over a must be co-prime and again a should not be too big in this case a will be well we obtain some more sophisticated recipes for detecting suitable integers and we use them in our argument but for this talk it's enough these properties and well so far delta was more or less arbitrary now what happens when delta is a trinomial discrete well in this case well one is always suitable but beyond one we have very very strong constraints for suitable integers exactly they must be very big you remember that suitable integers cannot exceed square root of delta but they are almost like square root so square root divided by logarithm so they're really really almost as big as we want them to be well how we prove it well let's take the smallest of the suitable bigger than one and let's take one other suitable well it's a subtle question that this other exists because a priori they can be only one and one more suitable integer i prefer to not to discuss this question why it exists let us assume that it does and let x and x prime be two corresponding singular moduli they are not dominant which means that as we saw they must be of more or less same absolute value but what we know about the absolute values x it is it is this guy a throat of exponent of pi square root of delta and for x prime is a prime throat and a prime is bigger than a so is at least a plus one so and now when a is very small then this guy is much much bigger than this guy you see for instance when a is two then this will be square root and this will be cube cube root of of some very big number so they cannot be almost the same and so on and we stop getting contradiction only if a is enormous and this is how we prove well the actual prove is like about three pages so but the idea is is here and questions are sorry no so now let us see how we can use it so this is already more or less sufficient to prove the lower bound well as one can expect we prove this lower bound by running several pyres scripts and for deltas in the range from the previous theorem so you remember that delta should be a bit too big here actually one can put something like 10 to the three by compromising or this constant three like two but well let's put it like like here and well for all deltas in this range we show that there is a prime rather small small prime with a positive chronicle and such prime must be suitable but suitable for a trinomial delta must be bigger than this and this prime is always small so there can be no suitable in this range well we don't just check every discriminant this would be too long instead we use certain sieving procedure we start just seeing we see the way all primes with chronicle well all deltas with chronicle or two so we take away all primes which are one or eight then with chronicle three and so on and after some moment we just get an empty list the problem with sieving procedure it goes rather well the processor time is quite good about five six minutes but it requires a lot of memory because we have to make big lists well for smaller deltas we just indeed check everyone we just check our inequality for trinomials we simply detect three roots w x and y detect three singular moduli such that this inequality is not satisfied and we did it for all smaller delta with class number bigger than three with class number three does not work because we have just three guys the dominant one and two other which are complex conjugate x and y i hope like complex conjugate so y over x is one so this inequality is trivially satisfied so the discriminance with class number equal to three they require some special treatment and it was this is where we needed our pediatric considerations and well how we dealt with them would require one another talk so i just do this with this point out and well this was about five six minutes here about four five minutes for this step and well this step was theoretically hard but computationally it was very quick so about 10 minutes on kind of like and as i said the processor time was not the bottleneck the bottleneck was the memory because of sieving okay so now we know that they cannot be too small and this gives a certain room for maneuver so we know that we deal with very big discriminance in particular we can replace this three by four for these big discrepancies let's continue now the next result was about how about the structure the structure was of the form well let's say must be either minus p or minus pq and again we do this just in all other cases we detect a small suitable integer and the trinomial discriminant we know it cannot have a small so for instance for even the easiest way either thing is just to rule out all even discriminance because for them two four or eight must be suitable also it's easy to rule out discriminance having more than two prime or distinct prime devices because while the smallest it would take the smallest prime power of them it is it must be suitable and it is very smallest like you could have here we have square divided by law also for instance squares can be taken away for minus squares because for them or either five or thirteen or seventeen must be suitable so again we have a small suitable and so on but each step we have to be more and more sophisticated at the end we really need like to for instance to eliminate the case p square times q we need about two pages but the idea was the same just to detect a small suitable and now after this preparation well well we can just profit with analytic number 30 which was already done before us and this is how we prove the conditional result so the GRH in price reforces convention so we use the following we use a work recent work by Lamzuri Lee and Sundarajan while Kai is a primitive real character that can be modular some number m and if we assume GRH then there exists a very small or residue prime residue for this character so well actually the result is much more general but this is what what we need from the result now let delta be our trinomial discriminant and let m follow the absolute value then the chronicle is just a primitive real character mod m must be primitive because delta is fundamental and we obtain a contradiction if the right hand side is smaller than something like this but well this is true for m bigger than something like 10 to 21 and this is because of this 10 to the 9 term which is not enough for us because we know that m is bigger than 10 to 11 not 10 to 21 so we have to slightly adapt their argument little little work and we indeed managed to prove what we needed so take this character then we can put also this right hand side and so well so this is how we prove the conditional result and in a totally similar fashion we prove the we prove the upper bound for all guy but once again we use some classical analytic number theory again let kai be as before and then there is a linear kino grad of famous linear kino grad of theorem that there is there is a prime number p bounded by m to one fourth plus epsilon with k of p equal to one well perhaps one should impose i think linear kino grad of themselves prove this for prime m i think the proof extends for square three cube three m at least i don't maybe not in generality but well it's more than sufficient for us but well the good news is that this is smaller than square root one fourth and just one half smaller but the bad news that the implied constant is not effective why it is not effective because there are two ingredients in the linear kino grad of the argument the Burgess which is effective and Ziegel which is non-effective and as well people do there is this poor man's effective Ziegel which is Tatuzawa theorem that we have this totally effective lower bound for the L of one kai but for all m with at most one exception and this is what we use and after some well rather messy calculations we prove that with at most one exception every trinomial discriminant bigger than this we find the prime with this property and this gives a contradiction so this only one exception it can be such a trinomial discriminant all the others simply do not exist well i think that well do i have time for the proof of the last slide the organizers would say maybe two or three more minutes two three well maybe maybe four okay and well the last theorem that we prove is that the trinomial cannot be very must be very special can we just start let me indicate how we prove a slightly weaker result with m minus n at most four so first of all we know that we can see that the class number is at least six and well even it's at least 100 by the work of Watkins because Watkins listed all at least fundamental discriminant of class number at least 100 and they are much smaller than our 10 to 11 so plus number is big so we have a lot of freedom for choosing our roots and trinomial can have only four real roots so we can take two non-real singular moduli of our discriminant delta and such that they're not equal and so both x x bar y y bar are all distant and we consider the number z which is this number well it's not a very hard lemma to show that it cannot be zero x x y x bar cannot be equal to y y bar it's not hard so it is a non-zero real algebraic integer and we know that that this x and y are very close absolute value and the actual inequality is like this so it should be m minus where n if we forget about this guy 0.01 m minus n times pi delta minus pi square root of delta and this 001 is just some generic small number that takes care of certain noise and the hard stuff was to show that this same deep noise does not continue now let's consider this as i said a real algebraic number and what are its conjugates just we run over all galore stuff and among its conjugates well its conjugates are all of this guy of this kind they no longer x1 x2 no longer or complex conjugate they can be anything but among the conjugates of that there are exactly four where one of these four guys is the dominant guy so x1 can be dominant but they're all distinct so x2 y1 y2 are not dominant and and so on so we get four of them where one of these guys is dominant and these are big guys all the other guys are not big and again so what we obtain for the norm for the product of conjugates this upper bound again this is four 0.001 is just to take care of some noise and and this is z it also occurs in the norm and it takes care the norm is really much smaller than what we expect but this is algebraic integer the norm cannot be cannot be smaller than one and this means that m minus n m minus n cannot be smaller than like something like four four and zero two so most four and to get reduced to two we use again some pediatric arguments and we are instead of a trivial lower bound for the norm we obtain a non-trivial lower bound for the norm we showed simply that it's divisible by many many prime numbers and this gives this non-trivial so it will be like two two point two dot something yes well fortunately we cannot go further so this is all what we wanted to say so thank you very much and well these are my cats who tried hard helped me to prepare this so thank you