 Welcome back. In the last lecture we proved Lagrange's theorem which said that every natural number is a sum of at most 4 squares and then we saw 2 directions in which we could generalize this result. So those directions were the first one was given by Waring's conjecture which generalized squares to cubes by quadrates and so on and then the second direction was given by the geometrical figures. I must tell you that the explicit computations in both the directions are not yet done completely. The higher G k's in the Waring's conjecture which are mentioned in the last lecture those are not all computed and similarly for figure 8 numbers or the polygonal numbers not much is known. However there is something interesting that I must mention which is that Fermat the French mathematician had mentioned in a later written to Pascal that he could prove that every natural number is a sum of at most k k gonal numbers. So this works well because Gauss proved that every natural number is a sum of at most 3 triangular numbers no equal to delta plus delta plus delta. Lagrange also had proved already that every natural number is a sum of at most 4 squares and a square is nothing but a square. So that represents to 4 edges or 4 points. So it is a 4 gone and similarly Fermat said that he has a proof that every number is a sum of k k gonal numbers but no one has seen this proof. Fermat is perhaps famous more for his marginal note on what is now known as the Fermat's last theorem but this another thing about Fermat is not so well known. So you should know about that also. Now we want to go to another direction by which we can generalize Lagrange's theorem. Remember Lagrange's theorem says that every natural number is represented by the form which we refer to as the form in the last lecture which was x square plus y square plus z square plus w square. We are now going to allow some coefficients here. So let me make the notations clear. We consider a form a quadratic form in R variables a1 x1 square plus a2 x2 square plus dot dot dot plus aR xR square. So this is a form in R variables quadratic form in R variables and we also take aR to be natural numbers. And we call such a form to be universal if it represents every positive integer. So we would have Lagrange has proved that x1 square plus x2 square plus x3 square plus x4 square is universal. This is what is Lagrange's result in another language. Once we have introduced the notion of universal positive definite forms but we will consider let us not go in positive definite is of course something that you know but you know about it for binary forms the notion of a positive definite form makes sense for a form in general R variables but let us not go into that definition right now. So Lagrange proved that every this form this particular form is universal. And now the question is what are the integers that you can have here a1 a2 aR such that the form that you get is universal. So there are some very basic constraints that we have on the integers. For instance if all the ai's were to be bigger than one then there is no way you can write one as represented by the form. So if you write a1 less equal a2 less equal a3 and so on then a1 has to be equal to 1. Then a2 has to be 2 or more if a2 is 3 and onwards then 2 is not written by the form. So there are some such natural constraints and therefore whenever you fix an R you have a finite number of these forms. So Lagrange it is also known that if you take a form in R variables for R less than 4 then such a form is never going to be universal such a form can never be universal. So you need at least 4 variables to have a form to be possibly to be universal. And our very own Ramanujan worked on this after studying Lagrange's theorem and he wanted to answer what are all universal forms. So in 1916 remember this was the same time Hilbert had proved Waring's problem, Waring's conjecture and Weefrich had proved it and so on. And around similar time Hardy-Littlewood also had given their proof of Hilbert's theorem. So Ramanujan in 1916 he actually listed a set of 55 forms in 4 variables and he claimed that they were all universal. So he had a list of such forms of 4 variables and he claimed that they were all universal forms. His list needed a small correction there was one form which had to be deleted form from the list but otherwise it was right. However when a person like Ramanujan makes a mistake like this there is a natural question that how are you going to determine that some form is universal. Already Lagrange's theorem took some considerable effort to be proved. So one form is universal that form given by Lagrange is universal that took some quite some effort to be proved. So we will have to prove that for all these remaining 54 forms how do you prove that they are universal that is not easy and one is prone to making mistakes in this. So this is where brilliant mathematician by name J. H. Conway came up with an idea. So J. H. Conway this is a mathematician who worked in Princeton. He unfortunately passed away not long time back because of the Chinese virus COVID-19 but this is the story about some more than 20 years back. So when he was teaching his graduate course, graduate course spinning it is a course for PhD students. In Princeton he suggested that let us try to find a number K with the property that if you have a form if you have a quadratic form which represents all those K integers then it is going to be universal. He suggested this as a research problem to his PhD class in 1993 in Princeton. Once again the question that he proposed is that he said that let us prove this result. What is the result? He proved that there is an he conjectured or he formulated this question that there is an integer K such that any positive definite quadratic form represents every integer up to K then it is universal. So if this result was proved then checking Ramanujan's list would be a very easy thing because suppose that number K is some number say it is 100 for the sake of discussion. Then you have to check that all those 55 forms that Ramanujan had already listed or the 54 of them which turned out to be correct. All you have to do is that those 54 forms represent each of the first 100 numbers and you do not have to check it for everything because once 1 is represented 4 is going to be represented, 9 is going to be represented because once you have a representation for 1 as a sum of squares as per that coefficients given in the form you simply multiply the x, y, z, w by appropriate multiple. So that square multiples of 1 get represented. If 2 is represented 8, 18 all these numbers are represented. So of course you do not have to check for all 100 integers. So the actual problem of determining whether some form is universal or not will turn out to be a finite problem once you proved this result. So this was a very interesting idea that Conway proposed in his class and remarkably in subsequent lectures he and his students proved that K is 15. They proved that if you can show that a form a positive definite form represents all integers from 1 to 15 then it is universal. So for this computation you do not even need computers. If the K was 100 then perhaps we would need computer to show that 97 is represented by the given form. But if you have to check only for 15 these are squares up to 15 and their multiples. So it is a very finite problem. It is a problem doable by pen and paper once you are given a form. It is a very remarkable theorem and this is called Conway- Schneeberger theorem. So this is a theorem which says that if you have a positive definite form which represents every integer from 1 to 15 then it represents all integers then the form is universal. But there was a twist in the tail because the positive definite forms that we had taken had a certain constraint on the coefficients. The Ramanujan forms the forms that Ramanujan considered were what are called diagonal forms. So they were a 1 x 1 square plus a 2 x 2 square plus a 3 x 3 square plus so on the terms x 1 x 2 x 1 x 3 they were not there. So Conway said that let us introduce those terms also and now let us try to find the number K and Manjul Bhargava a recent Fields medalist was a student in Princeton then and Conway thought that if he could induce Manjul Bhargava to think about this more general problem then it would be a nice result to have. So Conway tried to induce Manjul Bhargava to study this slightly general problem where you are having positive definite forms but those are not only diagonal you have the cross terms also x 1 x 2 x 1 x 3 x 1 x 4 x 3 x 4 and these also then try to determine the corresponding K. It is as it happens Manjul was not a member of Conway's class when this theorem was proved K equal to 15 was proved so Manjul was not there and Manjul thought to himself that if he has to prove the slightly generalized result he should at least come up with a proof for this result. And so Manjul re-proved the theorem for himself and his proof was so simple and so elegant that Conway and the student Schneeberger ultimately decided to not publish the proof. So there is a paper by Manjul Bhargava which is called on the Conway Schneeberger 15 theorem and before that there is an introduction to this result. So ultimately this result was not published in a journal but in a conference proceedings on quadratic forms and the introduction to the proceedings is given by Conway where he lists this whole event of introducing this problem to the class and subsequently the solution by his students and him and then inducing Manjul Bhargava to solve this problem and then ultimately Manjul Bhargava solving this problem. So if you want to read more on this then you should go and search for Bhargava 15 theorem Bhargava 15 theorem the number for the generalized forms is 290. So if you allow the cross terms like X1, X2, X3, X4 and so on then you have to notice that everything up to 290 is represented then the form is universal. So this is the third direction by which we can generalize Lagrange's theorem and here we have a very satisfactory answer and we also see that there is a very important contribution from an Indian born Indian origin mathematician Manjul Bhargava. So we now come to the end of our fourth theme which is on binary quadratic forms and we now go to our last theme our next and final theme is continued fractions. So what are continued what are continued fractions clearly as the name suggests what were binary quadratic forms these were quadratic forms which were binary and what is a quadratic form it is a form which is quadratic. What is a form? Form is a homogeneous polynomial. So binary quadratic form where forms meaning homogeneous polynomials which are quadratic that means of degree 2 which are binary. So in 2 variables continued fractions are fractions which are continued but this is not a very well known term continued and so it is better to see an example here is an example. This is a continued fraction expansion you notice that there is a dot dot dot in the end. So this dot dot dot say that this fraction expands this fraction goes on towards infinity these kind of representations for a real number are called continued fractions expansion for the real number. Let me give you the explicit definition for continued fractions. So in general a continued fraction is an expression of the form a 0 plus 1 upon a 1 plus 1 upon a 2 plus 1 upon dot dot dot 1 upon a n. So we stop at a finite stage after n such things we stop this is what is called a continued fraction. We see here that each of these is a fraction and we are continuing these fractions in some way. This is why this is called continued fraction. You may ask what are ai's? So the numbers ai for i bigger than 0 that means from 1 onwards that means these number onwards we assume that these are positive integers and this first integer a 0 this is also an integer but this is allowed to be positive, negative or even 0. So the first one a 0 that can be negative 0 or positive if you are in good luck but all the others are assumed to be positive. This is called the continued fractions expansion. We have these ones here in the denominator. You may actually take some another non-negative integers also. Some people have the convention that when you are considering the continued fractions in the numerator you take any non-negative integers somebody may take 2 someone may take 20, 25, 317 any such integer but we are going to restrict ourselves to taking it to be 1. So by our note convention our continued fractions will look like what I have shown you in the last slide. But in the literature when you have to deal with the continued fractions which have these general numerators our continued fractions the one which have only one on the numerator they are called simple continued fractions and the other ones where you allow any non-negative integer they are called generalized continued fractions. But since we are not going to talk about those generalized continued fractions we are not going to introduce the word simple for our term we will just call them continued fractions. So these are the continued fractions for us you understand that you have a 0 plus 1 upon some bunch which is a 1 plus 1 upon some another bunch and so on. So it turns out that every real number in fact has a continued fractions expansion which means that every real number is a limit of such numbers. You remember that for the expansion for pi had the dot dot dot. So that means that you are not going to stop at a finite level but you are going to continue after that after every n and that is important because if you stop at any level then what you get is in fact a rational number. So we prove this result continued a rational number. So of course if you have a 0 plus 1 upon a 1 then this is nothing but a 0 a 1 plus 1 upon a 1 this is a rational number. So if your n is 1 then you get it to be a rational number. Now we go to the general case where we have a 0 plus 1 upon dot dot dot and finally we have 1 upon a n then we would look at the number which is here. This has one less element than our earlier expansion. So by induction argument if we assume that this is a rational number it follows that the whole thing is also a rational number. So we prove whenever you have a 0 and a 1 then it is a rational number that is proved when you have just a 0 of course it is an integer every integer is a rational number. If you have a 0 a 1 up to a n you ignore a 0 for the moment and write a 1 a 2 a n as b 0 b 1 up to b n minus 1. Now you have an expression of length 1 less and by induction hypothesis we will assume that for such an expression you have what the number you get is a rational number then by simple arithmetic it will follow that the whole expression you started with is also a rational number. So every continued fraction is a rational number and now we want to say whether every rational number has a continued fractions expansion. So let us do an example suppose we want to write 15 by 11 as a continued fraction. So we start with writing it as a 0 plus 1 upon theta 1 where theta 1 is bigger than 1. When theta 1 is bigger than 1 its reciprocal is going to be less than 1. So a 0 has the property that a 0 is an integer and the difference between 15 upon 11 and a 0 is less than 1. We have to find an integer whose difference with 15 by 11 is a positive number we want this to be bigger than 1. So this fraction is also positive less than 1 but it is a positive number. So a 0 is the largest integer smaller than or equal to 15 by 11. This is a notation that we use to denote the largest integer smaller than or equal to a given real number. So if I call 15 by 11 as theta 0 then we have defined a 0 by the notation square bracket theta 0. We note that 15 by 11 is 1 plus 4 by 11 and therefore its integral part the largest integer smaller than or equal to 15 by 11 is 1. So a 0 is 1 therefore we have written 15 by 11 as 1 plus 1 upon 11 by 4. It was 1 plus 4 by 11 but after taking the reciprocal it has become 1 by 11 by 4. We repeat the same procedure. Note that 11 by 4 has integral part to be equal to 2. So this is 2 plus 1 upon 4 by 3. Notice that our denominators are decreasing. Earlier we had denominator to be 11 then we got our denominator to be 4 and now the denominator is 3. So we get it to be 1, 2, 1 and then simply 1 by 3. The 1 by 3 is the reciprocal of 3. So you actually just have it to be 1 by 3 or you may further expand it adding a 1 in the end. But that is where the fraction, continued fraction expansion for 15 by 11 stops. So this example tells us that possibly every rational number has a continued fraction representation. We have that every continued fraction is a rational number necessarily. And now we will prove in the next lecture that every rational number is also a continued fraction. And we will ultimately prove that every real number is approximated by the continued fractions. This is our last and final theme in this course on basic number theory. This theme is going to be very interesting. We are going to see many applications of this theme. We will see some things related to transcendental number, approximations to real numbers by continued fractions and so on. Stick around. See you in the next lecture. Thank you very much.