 Welcome back, it is great to see you back. We proved the fundamental theorem of arithmetic in our last lecture, which said that if you take any natural number n bigger than 1, then it can be written as a product of primes in a unique way. If you are agreeing with me to write the primes in the increasing order and then when you have the same prime appearing many times, you just combine the whole bunch of those same primes and put a power to the head of the prime. So the decomposition is n equal to p1 power n1, p2 power n2 dot dot dot pk power nk, where p1 is less than p2 which is less than p3 dot dot dot less than pk and ni are natural numbers. So this is a very unique decomposition and which says that primes work like atoms if you are looking at natural numbers. The concept of primes actually goes beyond the realm of natural numbers and this is the study of what is called algebraic number theory. But we will restrict ourselves with the notion of natural numbers and so we will be looking at only the natural primes, the primes that we all know. And one question would be how are these primes distributed in the set of natural numbers? We know already that the number of primes is infinite. So it is not a finite set, the set of primes is infinite. So you have a prime after any given natural number. On the other hand as I told you there are sets of any given length which do not contain primes. So you can actually construct sequences of consecutive integers of any length you want such that there is not even a single prime in that sequence of integers, natural numbers. On the other hand I have been talking about this twin prime conjecture which says that there are infinitely many primes such that the very next odd number is also a prime. So the question is how are primes distributed and this is the question which has been troubling which has been bothering mathematicians since quite some time. Two major mathematicians, two well known mathematicians by name Legendre and Gauss they studied this question in the 19th century. So to explain this let me build a small notation for any real number x positive I will denote by pi x the number of primes up to x. So pi of 2 will be 1 because there is only 1 prime up to 2 which is 2 itself. Pi of 3.5 will be 2, pi of 4 will be 2, pi of 4.5 will be 2, pi of 5.1 will be 3 because now you have 2, 3 and 5, 3 primes up to 5.1. Pi of 1.5 will be 0 because there is no prime up to 1.5. So this is a function it is a step function if you draw the graph of this function. The question is how do we understand this function pi x? So this is a very irregular function because it is a step function it will be constant for a length and then it will suddenly jump to one more high level up then again it will be constant for some time then it will jump and so on. So in number theory what one does is that one averages the function. What we would do is that you would divide pi x by x which is some sort of average and then try to understand the function pi x upon x. The understanding is that you know the function x quite well and if you also know pi x upon x then you will know the function pi x that is the hope. So this is how one starts thinking about it and as I told you these two mathematician Legendre and Gauss in 1798 the end of 18th century that is when they thought about it independently and both of them conjectured a formula. What was the formula? They both conjectured that pi x is equivalent to x upon log x. Here this equivalence means that the quotient of these two functions goes to one as x goes to infinity. So we will divide pi x by x upon log x and then we say that this quotient goes to one as x goes to infinity which is the same thing of looking at pi x upon x and dividing that by one upon log x. That quotient goes to one so we want to understand the function pi x and we would do it by understanding pi x upon x the averaging function and this averaging function as x goes to infinity behaves like one upon log x. So these two functions are what are called asymptotic to each other. We learn about asymptotes, vertical asymptotes, horizontal asymptotes, slanted asymptotes. There the asymptotic function is taken to be a straight line so it is a linear function but here we allow the other function to be any other function just that we hope that the function be well understood. So here pi x upon x is asymptotic to one upon log x or which is the same thing as saying that pi x is asymptotic to x by log x. So this was still a conjecture it was not proved neither by Legendre nor by Gauss and meaning this result is very peculiar in the sense that every time there is something new happens to the result there are always two people working independently about it. So here we have Legendre and Gauss working independently and getting the result getting the conjecture that pi x is equivalent to or asymptotic to x upon log x. It took 50 years to prove this result which was proved by Hadamard and Dalawaleh Pousson. This was proved in 1896 no it took more than 50 years it took about 100 years 1798 was the time when it was conjectured and 1896 100 years later this was proved and the proof used this function called Riemann zeta function Riemann had introduced this function this zeta is a Greek symbol which is written like this. So here you will have S to be a complex number and normally what we assume is that real part of S is bigger than 1 this is when the function has very nice properties. But of course we would like to extend it to the whole complex plane what is called a meromorphic function C and then there is this famous line real part of S equal to 1 by 2 which is where we are supposed to have lots of non-trivial zeros and in fact all the so called non-trivial zeros are supposed to lie only on this line. This is the part of the famous Riemann hypothesis. But anyway so this function zeta S was used to prove the result conjectured by Legendre and Gauss 100 years back and then once a result is proved and then you quote the result several times how do you quote it do you call it Hadamard do you call it Hadamard's result or the Dalavalli-Poussot theorem or Legendre conjecture or Gauss conjecture how do you call it since there were several people involved in it mathematicians simply decided to call it prime number theorem. So this result later came to be known as the prime number theorem. So this was proved in 1896 and in the early part of the 19th century Hardy and others these were the prominent mathematicians both in number theory as well as analysis. So they started wondering so you know a philosophy of mathematics was also being developed then and so people started comparing different proofs as per their difficulty level. And the prime number theorem because it used complex analysis which is supposed to be a higher level. So you have the theory you have the system of natural numbers which is the set of integers this is contained in the real numbers which will give you real analysis. This is further contained in complex numbers giving you complex analysis. So if there was a result which can prove using only integers then one would say that such a result is elementary compared to a result which uses real analysis or calculus. And the result which uses real analysis will be called elementary compared to the result which uses complex analysis. And something which uses properties of the Riemann's data function which is a very difficult function actually defined only on the upper half plane on the right half plane then one would say that this problem is very difficult. So Hardy and others were wondering where whether there can be an elementary proof of the prime number theorem. And people had different opinions and they were all stunned say in 50 years after the result what was first proved in 1948 when Selberg and Erdos gave an elementary proof. Elementary proof means that it did not use complex analysis it did not use any contour integration or Cauchy's formula or any such thing. However, the proof is quite intricate it is quite involved and there is a small story involved here which proves that mathematicians are also humans. And you will notice that again there are two people. So the result was conjectured by two people Legendre and Gauss in 1798 it was proved first by Hadamard and Dalavalli Poussin in 1896 independently and again Selberg and Erdos in some sense their work was also independent they proved it without using any complex analysis. So Selberg was at working Erdos is a mathematician Erdos was a mathematician who worked on a plethora of problems he worked on several problems he was a problem solver whereas Selberg was the person who was a developer of a theory. He would really think about a concept and develop the whole theory Erdos would try to prove problems would try to solve problems and the moment there was any new tool which was made available to him he had the knack of solving it solving a solving new problems with solving problems with the help of the new tools which were made available to him. So the story goes as follows that in 1946 or 1947 Selberg proved one very fundamental asymptotic equality which is really an inequality and he proved this inequality he was at University of Princeton that time and he had to travel to Montreal for a few days and before going he told one of his assistants spinning a person who has completed his PhD but was visiting Princeton University for a while he told this assistant about the inequality or the asymptotic equality and then he went to Montreal. Meanwhile Erdos arrived in Princeton and he got to know about the inequality from this assistant whose name was Turan by the way. So Turan and Erdos knew each other from Hungary days and the moment Erdos heard about this he thought let me use it to prove prime number theorem when Selberg returned from Montreal to Princeton Erdos told him that he thinks prime number theorem can be proved without with the help of the fundamental inequality that Selberg had proved but Selberg said I proved the inequality because I wanted to prove for prime number theorem meaning it would not be fair if somebody else did it with his inequality he had the very much aim he was going to prove prime number theorem using the inequality. However Erdos had almost completed his proof of the prime number theorem by then and so it turned out that Selberg heard Erdos's proof he observed that there could be some more improvements that could be done and finally Selberg understood that he can prove prime number theorem without using any of results of the Erdos any of the results that Erdos had proved. So it was settled in some sense by Selberg agreeing to let Erdos prove his proof first and then Selberg would prove his proof but mathematical community was mostly on Selberg side there were some mathematician on Erdos's side also however it is always called Selberg Erdos proof of the prime number theorem. So this is the interesting story about the prime number theorem which says that pi x the function which calculates primes up to x behaves like x by log x when you go to infinity in the sense that the quotient becomes 1. There are more functions which were developed which were supposed to have given you a better asymptotic result but we will not go into all that we will go to a another problem regarding primes once again. Now that we have established that there are infinitely many primes out there one would ask how many of them are even and how many of them are odd. So we know that there is only one even prime the oddest of them that prime is 2 and once you take 2 all its multiples which are the even numbers cannot be primes because 2 divides them so all other primes have to be odd. So because we have infinitely many primes it follows very easily that there are infinitely many odd primes. So odd numbers which are of the form 2n plus 1 contain infinitely many primes in fact they contain almost all primes except the prime 2. Again you can ask further every odd number can be written as 4n plus 1 and 4n plus 3. So the infinitely many primes which are in the odd numbers where do they go and sit? Do these go and sit in the 4n plus 1 or do these go and sit in the 4n plus 3? You may imagine that there are two lines one is of the type 4n plus 1 other is of the type 4n plus 3 and the moment you observe a new prime you check whether it is of the form 4n plus 1 or 4n plus 3 and write it in the corresponding line. Which line grows faster? That is the question. But do we at least have that there are infinitely many primes in each line? That is the question first we should ask. Are there infinitely many primes of the form 4n plus 1 or are there infinitely many primes of the form 4n plus 3? And we would like to have a proof of this result whichever went is mathematicians are not unhappy about the result which is proved they just want to have a proof about it. So we will first prove that there are infinitely many primes of the form 4n plus 3. Let us see how the proof goes. One can prove easily that there are infinitely many primes of the form 4n plus 3. The proof will follow on the same line we will assume that there are only finitely many construct a number which should have a prime of the form 4n plus 3 dividing it and this prime cannot be part of the set of primes that you have already considered and thus you have a new prime this is the idea of the proof. How do we do it? If there are only finitely many primes of the form 4n plus 3 let us write them p1, p2, p3 up to pk these are those primes. Now we construct a new number from these primes. When we worked with Euclid's proof we had constructed a new number simply by taking the product and adding one here we do a similar construction the construction is take the product of these primes multiply to it by 4 and then subtract 1. So what did we do? We constructed a new number capital N which is of the form 4 times p1 times p2 times dot dot dot up to pk and we subtracted 1 from this. So we have that our N is of the form 4 times a number minus 1 or which is of the form 4 times n plus 3 for some n. So we multiplied the primes that we had p1, p2, pk by 4 and subtracted 1 and now this number is of the form 4n plus 3 but more importantly it is an odd integer. So all its prime factors will be odd. Since n is odd all of its prime factors are odd 2 cannot be a prime factor here. Now since the prime factors are odd the prime factors can be of the form 4n plus 1 or they can be of the form 4n minus 1 or 4n plus 3. Can all the prime factors all the prime factors of this capital N be of the form 4n plus 1? No they cannot be. All of them cannot be of the form 4n plus 1 for the following reason. If you take any two numbers of the form 4n plus 1 say 4a plus 1 into 4b plus 1 the product will be of the form 4c plus 1. Because 4a plus 1 into 4b plus 1 will give you 16ab plus 4a plus 4b plus 1. So there is a multiple of 4 and you have plus 1. So if you take any elements any number of integers of the form 4n plus 1 and take their product the product is again going to be of the form 4n plus 1 whereas our number n that we constructed here is of the form 4n minus 1 or 4n plus 3. Therefore all its prime factors cannot be of the form 4n plus 1. So it should have one more odd prime factor and that has to be of the form 4n plus 3. So this way we get one more prime factor of this number which is of the form 4n plus 3 and this new prime factor cannot be equal to p1, p2, pk any of these because if it was any of the p1, p2, pk it would divide the product 2 square into p1, p2, pk it also divides n and then it will also divide 1 because 1 is obtained as 2 square p1, p2, pk minus n. So we have a new prime of the form 4n plus 3. If you add this call this as pk plus 1 I will repeat the process and get yet another prime of the form 4n plus 3 and so on. So this contradicts that the assumption that there were only finitely many primes of the form 4n plus 3. Therefore there have to be infinitely many primes of the form 4n plus 3. This completes the proof. Now that we have proved that there are infinitely many primes of the form 4n plus 3 what about the primes which are of the form 4n plus 1. You see we began by achieving that there are infinitely many primes out there. Out of them 2 is a singleton even prime. So all the remaining primes which is really an infinite set has to go and sit in the odd numbers and among odd numbers we have now found that 4n plus 3 type of primes are infinitely many. What about 4n plus 1 then? Are there only finitely many such primes of the form 4n plus 1 or are there infinitely many such primes? So we have to just check can you get 5 such primes of the form 4n plus 3, 4n plus 1, 4n plus 1. Remember 4n plus 3 are infinitely many we are now wondering what about 4n plus 1. So we have to type multiples of 4 and add 1. So 5 that is the first one that should come to your mind 4 plus 1, 4 into 2 is 8. 8 plus 1 is 9 which is not a prime, 4 into 3 is 12 plus 1 is 13 that is again a prime. So we got 2 primes of the form 4n plus 1, 5 and 13, third one is 17 which is 4 into 4 plus 1. So 5, 13, 17 after that the next one we get is all the way 29 because 21 and 25 are not primes of the form 4n plus 1. So 29, 3 we got 5, we got 13, we got 17 and 29 and after 29 the next one we get is 37. So there are 5 primes of the form 4n plus 1. Given if you ask me to list out 100 such primes I may be able to do it given sufficient amount of time. If you ask me to do, write 1000 such primes I may again be able to do it given sufficient amount of time. But that does not prove that there are infinitely many primes of the form 4n plus 1. So we have to give a proof to show that there are 4n plus 1 type primes and that set is also infinite. We give a similar proof to what we have done now except that we have to construct the number capital N in a slightly different way. So one can also prove that there are infinitely many primes of the form 4n plus 1. What we do is replace the earlier n by this new n which is 2p1p2pk whole square plus 1. In the proof for 4n plus 3 we had 4 into p1p2pk minus 1. Now we have n to be 2 into p1p2pk square plus 1. So it is a square plus 1. It is an odd number still but it is a square of an even quantity plus 1. This should have a prime factor which is odd. It cannot be among the p1p2pk which are your earlier assumed finitely many primes of the form 4n plus 1. The only thing you should prove is that any odd factor, any factor of capital N has to be of the form 4n plus 1. So think about this proof and let me know what you think about it. Let me know your comments. We will stop here now. I hope to see you in the next lecture. Thank you very much.