 So this is an undergraduate mathematics talk on the ABC conjecture. If anybody has come here wanting to know whether or not Mochizuki has proved the ABC conjecture, the answer is I don't know and I'm not qualified to say. What this talk is going to be about, I'm going to say what the ABC conjecture is. And then I'll give the proof of it for polynomials and then I'll make a few comments on the current state of it. So first of all, what the ABC conjecture is, it says if you have integers a, b and c with a plus b equals c, and here a, b and c should be co-prime, then c is at most the radical of a, b, c with a small fudge factor here. So what is the radical? Well, the radical of an integer n is the product of the distinct prime factors. For instance, the radical of 200 is 10 because 200 is prime factors two and five. And epsilon is some rather small positive number and this holds for all, but finite number of integers a, b, c where this finite number depends on what this number epsilon is. So let me give a few examples to explain this. So we have a plus b plus equals c and we want c is less than or equal to the radical of a, b, c to the one plus epsilon. Well, why do we need this number epsilon here? Well, things like one plus eight equals nine. And here the radical of a, b, c, so this is a, this is b, this is c, the radical of a, b, c is the radical of 72, which is six, and c is definitely bigger than six. So if we didn't put this number epsilon in here, the a, b, c conjecture would just be false. You can actually find quite a lot of examples like this. So the question is how big can epsilon be? And people try to find examples where epsilon is as large as possible. One of the biggest, for example, was found by Ray Satt and says that two plus three to the 10 times 109 equals 23 to the five. And the radical of a, b, c is two times three times 109 times 23, which is 15042. And this number c is 23 to the five, which is 6436343. So it's quite a bit bigger than this radical, but it's less than the square of the radical. So epsilon is less than one even in this example. So why are people interested in the a, b, c conjecture? Well, it implies a whole string of other very deep conjectures in number theory. I'm not gonna list them all. I'll just talk about one of them. So Fermat's last theorem says that x to the n plus y to the n equals z to the n has no non-trivial solutions for n greater than or equal to three, apart from one of these being zero or something like that. Well, if you write a is x to the n and y is b to the n and z is c to the n, then the a, b, c conjecture would say that c is less than or equal to the radical of a, b, c. To the one plus epsilon. Now the radical of a, b, c is going to be at most x, y, z. Here we're taking x, y and z to be co-prime to the one plus epsilon. So c is bigger than x and y. So we find that z to the n would be less than or equal would be less than or equal to z cubed to the one plus epsilon. And obviously if n is bigger than, if n is sufficiently large and epsilon is sufficiently small, then there only going to be a finite number of solutions to this. And if you had a precise version of the a, b, c conjecture, this would allow you to give a proof of Fermat's last theorem. Of course, the problem with this is that Fermat's last theorem was proved by Andrew Wiles by a different and very difficult method in the many years ago. But this just gives an example of how the a, b, c conjecture can be used to prove various other theorems. So what I'm going to do now is to show you how to prove the a, b, c conjecture in a slightly different version. This is the a, b, c for polynomials. And this was proved by Stother's and Mason round about 1980 or 81 or something. And what it says is that if a, b, c are polynomials, to be, let's make them have complex coefficients. No, this isn't really necessary. And suppose that a plus b equals c and a, b, c are co-prime. Then the degree of c is less than the degree of the radical of a, b, c. Here we assume that a, b, c are not constants. So what's the radical of a, b, c? Well, it's the product of the distinct linear factors since we're working over the complex numbers. If you're not working over complex numbers, the statement has to be modified very slightly. So the degree of the radical in this case is the number of distinct zeros of a, b, and c. And it has an amazingly short proof, which I think was due to Noah Snyder in about the year 2000, which possibly fits onto one piece of paper and uses nothing more than basic calculus. So if you need to know what the derivative of a polynomial and the derivative of a quotient is and if you know enough, that's enough to follow this proof. So suppose if a plus b equals c, then we know that a over b plus one is equal to c over b. Now we differentiate both sides and we find that a prime b minus b prime a is equal to c prime b minus b prime c. Here I've just differentiated both sides and then taken out a factor of b squared. Let's make f equal to this. And now you'll notice that f is not equal to zero, because if f was zero, then b would divide b prime times a, but b and a are co-prime by definition. So we're saying a, b, and c are pairwise co-prime. And we're working character zero, so b can't divide its derivative. If you want to do that, you can divide its derivative. If you want to do this over fields of characteristic p, you need to be a little bit more careful, but we won't worry about that. And now we also notice that, well, let's write a and a prime for the greatest common divisor of a and its derivative. And we notice that these all divide f and a co-prime so a, a prime divides f because any common divisor of a and a prime divides both that factor and that factor, and so divides f. And similarly, these all divide f. Since they're co-prime, we see the degree of a, a prime plus the degree of b, b prime plus the degree of c, c prime. This lets note the degree of f, which is the degree of a plus the degree of b minus one. And now we notice the degree of a and a prime is just the degree of a minus the number of distinct zeros of a. And this follows because if we write a is equal to t minus alpha to the n one, t minus beta to the n two and so on, then the highest common factor of a and a prime is going to be t minus alpha to the n one minus one, t minus beta to the n two minus one, and so on. So similarly, the degree of the highest common factor of b and b prime is the degree of b minus the number of distinct zeros and same for c. So the degree of a plus the degree of b might plus the degree of c minus the number of distinct zeros of a, b and c is less than the degree of a plus the degree of b minus one. And now we notice that we've got this factor degree of a on the left and the right and this factor degree of b on the left on the right. And so we find the degree of c is less than the number of distinct zeros of a and b and c, which is what we wanted to prove. So we can give an application of that. We can prove Fermat's last theorem for polynomials. So suppose x, y and z are polynomials in some variable t by, I mean, complex polynomials. And suppose they're co-prime and suppose that x to the n plus y to the n equals z to the n with n greater than or equal to three. So this is like Fermat's last theorem except that instead of x, y and z being integers, we take them to be polynomials. And now we let a be this and b be this and c be this. And then we find that the degree of c is less than the number of distinct zeros, which is at most the degree of x, y, z. So n times the degree of z is less than three times the degree of z. Here we're taking, we may assume that the degree of z is bigger as at least equal to the degree of x and y by swapping them around if necessary. So this gives n less than three, which is what we wanted to prove. If n equals two, or indeed solutions, for instance, we can take one minus t squared, squared plus two t squared equals one plus t squared or squared. So we can't do better than this. By the way, if you're working over a field of characteristic p, then you can get formulas like one to the p plus t to the p equals one plus t to the p. So this is working mod p. So you need to be a little bit more careful if you're not working over the complex numbers. So we've got a proof that works over polynomials. What happens if you try and modify that proof to work over the integers? Well, no one's been able to think of a way of doing that. The trouble is this proof very much relies on the fact that you can differentiate polynomials. And the trouble is nobody knows a really good way to differentiate an integer. I mean, the derivative of an integer is just zero, which doesn't really help very much. So I'll finish by just saying a little bit about the current state of the proof for the integers. So Mochizuki in 2012 put out several papers presenting a possible proof of the ABC conjecture. And the reaction of mathematicians, this was a bit odd. The problem was the proof was incredibly difficult. It was several hundred pages long and relied on thousands of pages of previous work by Mochizuki that most mathematicians knew very little about. And it wasn't very easy to read. I mean, I think most mathematicians did roughly what I did that spent a few hours trying to read it and then gave up and hoped somebody else would tell them what was going on. If you've ever seen one of these wildlife moves of penguins, say emperor penguins all want to jump into the sea and go swimming. But none of them want to be first in because there might be a leopard seal there. So all the emperor penguins are trying to push each other into the sea because they don't want to be first in. And mathematicians were behaving a bit like that with Mochizuki's proof. Everybody wanted someone else to be the first to read it and tell them whether it was safe to read the proof or not. Anyway, several people have read Mochizuki's proof and some of them say it's correct and some of them say it isn't. So we're kind of stuck. So in particular, Schultzer and Sticks have put out a pre-print saying something they say is a problem. And I'll try and describe roughly what the problem is. I'm going to have to simplify rather grossly. Let me, I'll explain what they say is a problem by a sort of analogy. Suppose you've got three countries and there's maybe a lake in the middle. So there are three borders and each of these currencies has currency as some current. Each of these countries has some currency. So this country uses dollars and at this border you can exchange the currency and say one dollar is say two euros. And at this border you can say maybe the exchange rate is one euro is three pounds. And at this border, if you say one dollar is six pounds we've got that the right way round two euros, just three pounds then there's no problem, everybody is happy. But if you said one dollar is equal to five pounds then there'd be rather interesting problems. People would be going round and round and round here making lots of money and things wouldn't be consistent. And Mochizuki's proof has a lot of one-dimensional real vector spaces in it and real vector spaces can be thought of as something like a unit of currency and an element of currency is in mathematicians terms an element of a one-dimensional real vector space. And you need to identify all these one-dimensional real vector spaces make them compatible and that means you have to choose it's like choosing an exchange rate between currencies. And furthermore, you need these exchange rates to be compatible that if you identify these two vector spaces and these two vector spaces and these two vector spaces you've got to make sure that all the three identifications are compatible. So are these compatible? So Schultz and Sticks isolated while there weren't three countries there were more like six or seven countries in a big circle with lots of identification. So Schultz and Sticks examined Mochizuki's proof and found all these vector spaces and studied the compatibility condition and said they thought these relations were not compatible and Mochizuki said, yes, they are compatible. So why don't we just check and work out what these exchange rates are and check where they're compatible. The trouble is Mochizuki's proof is so incredibly complicated that it's extremely difficult to figure out what these exchange rates are actually are and whether they're not, I mean, even if they're not compatible the inconsistency might be so small that it doesn't matter. So it's really unclear whether or not that there is this problem with the proof. With most deep mathematical theorems it's possible to give a very fairly short summary of what the key idea and the proof is. For instance, Perlman proved the Poincare conjecture a few years ago and the proof of this is really hard. It takes several hundred pages of hard mathematics to write out, but you can summarize the proof in a few lines. So what Perlman did is he applied something called the Richie flow where you take a three-sphere and you deform it according to certain equations and every now and then when you deform it something goes wrong so you have to cut a bad bit off. It's called doing surgery and then continue deforming the sphere. And you can sort of explain that in a few lines and mathematicians get a rough idea of what the proof is even if they don't understand the details. So you can explain the key idea of the proof even though the details take hundreds of pages. So what's the key idea in Mochizuki's proof and as far as I can figure out, that doesn't, no one seems to know. No one has been able to explain, give some simple summary of what is going on. So why don't you check it on a computer? We now have computer checked proofs of really very long and difficult theorems like the fight Thompson theorem is several hundred pages long and that has now been checked by computer. So why can't you do that with Mochizuki's theorem? Well, the problem is that you can only check theorems on a computer if the proof is really, really, really well understood by humans that the current state of the art for computer checking the computers are not very good at finding the proofs. They need to have the proof explained to them in extraordinary detail. And as far as I know, no one has managed to formalize the proof of Mochizuki's theorem yet. So the computer can understand it maybe in another 10 or 20 years computer proofs will reach that state. So at the moment we're in this very unfortunate state and have been so for eight years where it's simply unclear whether or not this theorem is proved. This is almost unprecedented. I don't know of any other cases when there's been this much controversy about whether a proof is correct or not. Anyway, I think that's enough about the ABC conjecture.