  ، నాబ్నిన్లన నారానినినితిత్ిని నిల౿నిిన నిహేలుఏం నిచాన్లన డిన్రంటా దినిసిత్ట్మ నిఖంటారియసినితిసిలేలస్సినిదినిరికవా. ギ ят dę coon  �  שא�  verme  majesty                                                                                                                                                                                                               quickest pitching bands andthree concept arm of a Psy long time nd one where N, as a product of the primes defining ABC. אני ഏികിക്ിലുൈോടൃമു്ന്്് കികൃുകൈടുകുുുകൈൃണ്ു്്നു്്ടൃണ്ന്ൎണ്ട simplicity tonnes ragout entity. C is a projection completion of the curve of equation y and Krishna අතා වටිල්රScript�්නාgage y andória converting to x ඔන෻නා එ කෙපලනී�anladım C is 2 කරරු. We can apply the effective modelpekrieste of the management that we является accessibility, tungaSabid canhaины y c y u x 5 x y ᵉ ᵃ ᵏᵍ different using a orage podeuded yezle grund ڭ z strong chris. ఎ technically you must use the elders if you use a point of view of the table ఒరి. ఒరి d � under the table కమ drauf. ఄారి ఉరి ఆdoo. We assume I have A plus B equals C. Then we can take Y equals C over A, therefore Y squared minus Y equals C squared over S squared minus C over A equals BC over S squared. Let's call this quantity M and the equation we have to solve is S5 equals M. So we introduce the number field F equals Q fifth root of M. And we can apply the effective model condition that we proved last time. Consider the point P equals fifth root of M, Y. So it's the point in C of F. We will define the height on C, namely HM to HM of P, the same as H by of the star over one of P. So the same as the height of P of P by the projection formula again. So this is H of Y, where H is the naïve height, that is the naïve height of rational number. It simply is the log of the supremum of the absolute, so this is the supremum of log of BC log of a square. We take log of the numerator, log of the denominator and we take the supremum of it. This is the naïve height of our rational number. So what we get here is something like 2 log C. So now we have the inequality HM P. Yes, I recall, 1 for the epsilon degree M log delta F F over Q plus O of 1. This is what we have shown last time. As a consequence of congestion B, we have this form of explicit effective model congestion for the curve. Well a little worried because of this 2 here. Ah, H of Y is supremum, Y is equal to C over A. So supremum of log C log A. We have to use Y equals C over A. So we get that log C is at most, while the degree of M is 5. So 5 over 4 plus epsilon log delta F divided by the degree plus O of 1. Now we can conclude this to number, namely the degree of F over Q is 5. F was obtained by extracting the fifth thought in Q. And there is a small m r, which is the delta F is at most 5 into the fourth. F was Q square of 5 of M. So the prime rhamifying in F are 5 because it is good. And the prime dividing M. And M was BC over S squared. So we get all the prime dividing AB of C. So this is N. And the exponent is 4 because it's only 10 divided by 5. That's the response. OK. So now we get the following. We like log C. That's what I call. 5 over 4 divided by 5 multiplied by 4. So we get exactly 1 plus epsilon log N plus O of 1. And this is what we wanted to prove. That C is at most N1 plus epsilon times the constant. Is there a question? Is there a question? OK. So if you wish, we look at the proof which goes from contract AB to ABC. For ABC, we know we want prime preceptilon. And this fixes the coefficient of log delta all along. This is really the way it was done by once, I mean, aiming at one preceptilon in the last estimate. And you have to control the coefficient of log delta in the theorem before. OK. So in fact, this theorem was announced by Mochizuki. It's already a long ago, 2012, a proof of a conjecture of VITAR. So the conclusion of this theorem is known according to this conjecture of VITAR. And as you know, people have a lot of difficulties to check the paper of Mochizuki. OK. So let's give some example. So we shall look at the few dial-fontane equations. The first is the Fermat equation. Extend plus y n equals 0. So it's of course of type A plus B equals C. And what do we get? C, which is z to the n. And that's how we call to A product of P dividing x, y, z to the power 1 preceptilon. So this is less than the constant A prime times z 1 preceptilon. So on the left hand side, we have z n. And on the right hand side, we have z 1 preceptilon. And z 4 n is bounded. So the A-B-C conjecture implies that for n large enough, Fermat equation has no solution. So it's less precise than the Fermat equation. But on the other hand, it's more general because we can look at a question which looks like Fermat equation. We'll then generalize Fermat equation. Assume we have 3 non-zero integers. And we look at the equation alpha x to the n plus beta y to the n equals gamma z to the n. So this is like A plus B equals C. And from the ABC conjecture we get gamma z to the n. It's less to recall to A product of the prime dividing alpha, beta, gamma x, y, z to the power 1 preceptilon. So there are really two cases. First case, assume alpha equals beta plus gamma. Then this equation has the opposite solution. X equals 1 equals z. Second case, alpha is not equal to beta plus gamma. And then we deduce from this as we did from the Fermat equation that gamma z is bounded. We get that n is bounded. By looking at gamma z less than gamma z to the n, being less than a multiple of gamma z. There exists n0, such that if n is bigger than the n0, either gamma equals alpha plus beta, or x equals y, x, y, z is 0. One of the variables is 0. Goldfeld stated something which is even more general. Let epsilon be between 0 and 1. Let ABC be between 0 natural integers. Then the equation alpha x to the r plus beta y to the s plus gamma z to the z equals 0 has finitely many solutions. Solution being x, y, z, r, s and t. So there are finitely many values of x, r, y, s, z and t, such that this equation holds and was first x, y, z is different from 0. Second, x, y, the greatest common divisor of x and y is equal to the greatest common divisor of x and z. And y and z is 1, so they are promised by pairs. And r, r, t should be positive numbers. And we have the following condition. 1 over r plus 1 over s plus 1 over t less than 1 minus epsilon. So r, r, s and t cannot be too big because they are all big, then the sum is less than 1 minus epsilon. So this is the generalization of what proceeds with the power of x, y and z not being the same. The proof is obtained by essentially the same method which I was giving for Frammer. OK. Now we talk about ABC for number fields. So we start with the number field k. We add r plus db and d of k for a log of the absolute discriminant. And now we have to choose the norm at this place. So uv is the place of k. We define xv equals the absolute value of x, uv is real. The square of the absolute value of v is complex. So per dig valuation of x, uv is given by a prime. But I have to tell you how I normalize the periodic absolute value. We do it as follows. We ask that the prime p has absolute value 1 over p 1 over r where p is simply the prime corresponding to p. OK. So we have normalized the absolute value. And we can define the height and the conductor as follows. If ABC has 3 non-zero element in k, we define the height of ABC to be the product for all pcsv of the maximum of norm of v, norm of c. That's why maybe I should put double bar here. So this is the height. This is the generalization of log of discriminant. Now for the conductor we do the following. We introduce a set of prime, a set of finite prime, which is those finite prime such that the maximum of ABC is simply bigger than the minimum of ABC. I mean if maximum equals minimum, this means that the absolute values are all equal. So there is nothing like we cannot say that the prime has something to do with ABC. We get the same norm for the three numbers of ABC. So to have the prime dividing ABC, we consider this set. The set of prime ideals such that the maximum of ABC is bigger than the minimum of ABC. Once we have made the definition, the conductor is defined as a product for all primes in r. The norm of the ideal p minus 1. Okay. And what is the ABC context of our number field? Well, with this notation, we have here a b prime with a sum context of bz whole. Then for every positive epsilon, there exists a constant a of epsilon such that if a prism is equal to c, where now ABC are elements in the number field, then the log of the discriminant for log of h ABC is at most one per epsilon log of the conductor, namely n ABC. This is the log of the conductor, but we also have some dependence on the number field. So this plus 5 for epsilon over 4 log of the discriminant of k divided by r plus a of epsilon. So there are two parts on the right hand side of this equation. There is the log of the conductor, which we can expect from the case of q, but also there is control on the dependence of the inequality on the number field k. It's not more than log of the discriminant times the constant. So of course, if k equals q, when we cover the ABC congression we mention above, we cover c and b. The other remark is that the proof is similar to the case of k equals q. We just have to be careful that someone will get log of the discriminant and we have to compute the discriminant of f in terms of the discriminant of k and the discriminant of the prime dividing ABC. Now what is interesting is the following. Is that the vector congression that Moshizuki has announced is weaker than what we have written. It says the following for every epsilon positive. Exist a constant which depends on r and n epsilon such that log of h a b c is at most 1 plus epsilon log of n a b c. So this starts the same way as the right hand side over there. But then we get this plus dk over r. So instead of 4 plus epsilon over 4 we get 1 plus epsilon. So this is better. But there is the dependence of the constant as the degree. Another way to say it is that you fix the degree r and you write an inequality like here. So the dependence on r is not explicit. So maybe I stop here for 5 minutes. So I will talk about another congression which follows from the discussion we had before. Namely the congression of the discriminant. So this is the congression on the elliptic curve which was formulated by Spiro at the end of the 70s. So some are before all the other congressions. They came after Spiro congression for discriminant. So it is as follows. Let e be an elliptic curve with semi-stable reduction. And let delta e be the minimal discriminant of e. And n e is the conductor. So to define the minimal discriminant you take a Weierstrass equation of the elliptic curve and you take the smallest power of p that comes in a Weierstrass model. So the minimal discriminant is the minimum you can get by looking at the Weierstrass model. And the conductor is as usual the product of prime of delta e. So there is a following theorem. Assume congression be whole. Then for every positive epsilon there exists a constant, maybe beta epsilon positive. Such that log of the discriminant is at most 6, 6 plus epsilon log of the conductor plus this constant. So several of the conjectures are really current. And here we discussed the conjecture of discriminant but we could say that conjecture ebc is a consequence of conjecture of discriminant. Because you look at the elliptic curve y squared equals x minus y x minus b and you apply conjecture of discriminant you will get essentially the ebc conjecture. So these different conjectures are very often equivalent. They are not equivalent to conjecture b. I don't know for instance how to deduce from ebc the conjecture b on omega 2. I don't see how one could get it. But the conjectures are very much equivalent to each other. So there is a proof of this theorem in Moribayi. One will take again k0 equals q and for c one takes the modular curve x0, 23. This modular curve has genus 2 and there is a formula in the book of Aug about the genus of the modular curve. There is a map from x0, 23 to p1 the J invariant which transits in forgetting the level of the elliptic curve to get the elliptic curve of z so that the g morphism and the degree of g is simply the index of g0, l, gamma 0, 23 and then c2, z. And we have 5, 24. So if m is the inverse image of the canonical line value on p1 the degree of m by projection formula will be 24. So we do as before, namely we call h the standard height, standard logarithmic height on p1 and hm is h composed with g is the height on c. So what happens is there will be a point corresponding to e in h0, 23. That is we can choose the level structure of the curve e and get a point in h0, 23. So we have this point such that the J invariant of the elliptic curve is the image of p by J. The point p is not defined over q it is defined in a number field of the degree 24 over q and one shows that f is an armified outside n of e. So g of e is governed by some formula with some constant g2 to the cube g3 to the square I see divided by delta is the denominator of g of e. So the log of delta e is less or equal to the nullified of g of e. And the nullified of g of e is the log of absolute value of the numerator divided by the supremum of log of the numerator and log of the denominator and we already have the denominator in the log of delta e. So this is the same as the h m of p and we can apply the effective model in equality. Log delta e less or equal to h m of p and by effective model value model this is the same. It is at most one of four plus epsilon times the degree which is 24 log of delta f divided by the degree of f of q plus some constant. So the actually here it's less or equal to so I get one of four plus epsilon times log of the discriminant log of the discriminant divided by the degree is at most one plus epsilon log of the conductor plus a constant. I mean this is the discussion on f and looking at which prime in f are unified and we find that prime divided by n of e. So the log of discriminant is bounded from above by this quantity. So if we put these two things together we get log of delta e is at most one plus epsilon one four plus epsilon 24 log of n e plus beta epsilon so this is 6 6 plus epsilon log of n e plus constant. So we get the condition of the discriminant with the exact coefficient 6 plus epsilon 6 plus epsilon was the constant predicted by Spiro when he formulated when he formulated the condition. We get exactly the right coefficient 6 plus epsilon. So I'm sorry I don't have much more to say today. Next time we will discuss a beautiful consequence of the conjecture which is that there is no zero zeroes. It's a result by Stark and Gonville relating effective model with the absence of zeroes. Or more in-ease ABC conjecture for nanmophiles. So what happens is that ABC conjecture for nanmophiles beta and Moshizuki is too weak to deduce some information on zeroes. But if we put in what we get from the conjecture we actually can prove that there is no zero zeroes. So I hope we will have some increasing argument next time about that. Thank you. Log data log data log data log�� log log data log data log data nanm for M Ḩᶦ� Avenue üzerinebum Ḩᶦᵍᵑᶜ� augment, ḣ��ᵥᶦᶃ �ヒᶸᵢ� buddy ḣᶔᶀᵉ ᶕ ᵜᶰᶸ ᶈᶼᶶ ᷴᵉ ᳄ᴀᶸ ʑἒᶦᶃ Ḹᶤᶸ ᶁᶦᶉ ʑᶥ, ᵏᵒᶜ x ᶲᶢᶸ ᶀᶼᶸ ᶃ ᵅᶥ ᵠᵍ ᶜᶻᶷ ᵔ� ᵒᶜ ᶔ�ᶆ ᵢᵇᶜ � iscernible ƒ. So, this was why this two numbers are the same. Are there other things which were not understandable so that I prepare for next time some kind of answer?