 So tonight, I'm really delighted that we have Kathy O'Neill here to introduce our speaker. Kathy is a data scientist, and she is the author of a book called Weapons of Math Destruction. And any of you who have ever looked at the exhibit names here at MoMath know how we feel about names like that, and so it gives me particular pleasure to welcome Kathy. Hello, everyone. I'm so happy to be here tonight to introduce Professor Carl Pomerance. I got a PhD in mathematics myself in number theory, and he's a number theorist and he's an amazing guy, so you guys are all very lucky. Right now, he's a Dartmouth professor, but he's also a professor emeritus at the University of Georgia, and he got his PhD at Harvard in 1972 after an undergrad at Brown. He has 120 publications in mathematics or more. And by the way, I just found out also that he has Airdish number one. I don't know how many of you know what that means. Okay, that means that he's written a paper with Airdish, but actually, he wrote 23 papers with Airdish, and I think that means he has Airdish number 123rd. What do you guys think? Yes? Or maybe one over two to the 23rd. We can argue about that later. He is a celebrated teacher and researcher. He's got numerous rewards in both teaching and in research. He gave the George Polia Lectures for the Mathematics Association of America, and he was also an invited speaker at the International Congress of Mathematicians in Switzerland in 1994. He also, in 1994, had a big part in factoring a 129 digit number, which was the biggest number that had been factored at that date, using his quadratic sieve algorithm, which he'll talk about a little bit. Very exciting. Here's one of the things that you might not know about Professor Pomerance. He's also an activist, and I really appreciate that because I am as well. So he told me that in 1964, which is called the Freedom Summer, he went down to Mississippi to help people register to vote the blacks in Mississippi. Now, Mississippi claimed that blacks just didn't want to register to vote, but they were going to raise one of the obstacles, and they were going to give them the option to register to vote with the caveat that their names would be published in the local newspapers, so they might be blacklisted, so there was still a threat there. In spite of that, he and some of his friends, civil rights workers, went down and knocked door-to-door to talk to people and tell them, you have people who are behind you, the rest of the country wants you to enjoy the freedoms of your civic responsibility. Just to give you a little bit of color on that, when he got there to Mississippi, three other activists had disappeared. Two white activists from New York named Goodman and Schwerner, and one black activist named Cheney from Mississippi had disappeared. They were gone that entire summer, and everybody was looking for them. They were found at the end of the summer, their bodies were found, I should say, by the FBI, the local authorities would not prosecute, and then the federal government had to step in eventually. So it was a real risk he was taking to go there. He also, by the way, if you guys are big mathematician fans, shared a jail cell, I might be exaggerating, but that's my thing, shared a jail cell with Richard Stanley later on in the Vietnam protests. So this is a guy who does great math, a great teacher, and he's also a great guy, so let's welcome him, Professor Pomerance. So thank you, Kathy, for the introduction, and thank you, Cindy, for the invitation here to speak at MoMath. It's a tremendous thrill. I'm very impressed that so many people should come out in the evening to hear about arithmetic. I mean, you'd think that arithmetic is boring, right? Well, I'm going to try to show you that it's not boring in some sense. In fact, it's not boring because there are unsolved problems. As you saw perhaps in the abstract for this talk, that you'd think that issues about addition and multiplication were sewed up in third grade. Well, we're going to learn some things that they didn't tell you back then. Here's something they did tell you. Here is a multiplication problem of a three-digit number by a three-digit number, and I'm sure you all remember how to do it. Here it has worked out, hopefully I didn't make any mistakes on this, but you multiply, you know, four times each one of these digits and you get an answer and then eight and there are carries and so on and you get this little parallelogram of numbers and then you add them up and you get the answer. And you can see that it's, you know, maybe it was the kind of thing where you came home and you had a page of, you know, ten or twenty of these that you had to do for homework for the next day just to make sure you knew how to do this skill. And it was boring and okay, good. So what if instead of giving you a three-digit by three-digit problem, suppose I gave you a much bigger problem, twenty-three digits by twenty-three digits. Now, you probably always thought that math teachers were a little bit sadistic and cruel. No, no. Just, you know, maybe a tiny bit, but not that much. I'm just doing it in theory. I actually gave you one to work out right now on those little pads that were at your seats. You probably wouldn't have, the pad is too small. You'd need a big sheet of paper to do it. The question is, how much work to do that if you did it? Let's have a thought experiment. Now, the first thing to notice is that it should depend on what the numbers are. So for example, ten to the twenty-two, which is one followed by twenty-two zeros, that has twenty-three digits. And if I were to square that number, in other words, multiply it by itself, the answer is ten to the forty-four, one followed by forty-four zeros. You can do that problem in your head. So it depends on the problem. But let's say we don't have a special problem like that. Suppose we just have some random digits. Well, you can argue, hey, wait a minute. If there are twenty-three digits, then some have to be repeats, like maybe two is repeated three times. So that means when you multiply two into the top number, you can copy that line down, lower down, maybe two more times. Okay, great. However, you still have to write down everything. So here's a diagram showing everything written down. So if you count it, there are twenty-three A's here. They don't all stand for the same digits, just placeholders. Here are twenty-three B's. They don't all stand for the same thing. And I couldn't really fit it on the slide. But there's supposed to be twenty-three rows here. And this parallelogram of numbers has to be written down to get the answer. So even if it's easy to write it down, you still have to write it down. And how many numbers are there? Well, at least five hundred and twenty-nine, which is twenty-three squared. So in general, if you multiply two N-digit numbers together, this method that we all learned in school takes about N squared steps. Just to write down everything in that multiplication parallelogram. Now you could argue, well, actually it takes a little bit more because you have to, you know, do the addition at the end and so on. But so maybe that changes it to two N squared or three N squared. But it's still, we say it's of order N squared. That's the complexity of school multiplication. Now you could ask, does any faster way to do it? And in fact, a Russian mathematician named Kolmogorov, who had a seminar in Moscow, asked this question in 1960 at a seminar. Is there a faster way to multiply two N-digit numbers than the method that Russians and also Americans and everybody else learns in school? And Karatsuba was one of the young people at this seminar. And a few weeks later, he came back with a method that took fewer than N squared steps. It took about N raised to the 1.6 steps. So Kolmogorov couldn't believe it. He thought that the answer should be its school multiplication is the best and he was looking for a proof of that. He wasn't looking for a faster method. But Karatsuba came up with the faster method. So let's see what his method is. It just involves high school algebra, period. Here's how it works. Suppose we're multiplying together A and B, and each have N digits. So let's let M be a half of N. So we could assume that N is even if you want. So we write the numbers A and B, the top part and the low part. The top part and the low part. So the A0 is the lowest M digits, lowest order M digits. And A1 is the top M digits. And same here. If you multiply A times B using high school algebra, you get this product. There's a 10 to the 2M term coming from the leading term here and the leading term there. And then there's the two middle terms and then the bottom term. Now let's see how we've helped ourselves or if we've helped ourselves. Instead of multiplying the long number A times the long number B, we have four shorter problems. The A1B1 problem, the A1B0 problem, the A0B1 problem, and the A0B0 problem. Each of these shorter problems should take about M squared steps because we're dealing with M digits. Now M is a half of N, so each of these four shorter problems takes about a quarter N squared. Four problems, each taking a quarter N squared. Kolmogorov must be right. You've not gained anything. You've just made it messier. Four times a quarter N squared is N squared. We're back to N squared again. However, there's a little trick. Addition is pretty easy. I didn't mention that before, but the school method for addition, how can you improve on that? It's just if you have two N digit numbers, adding them takes an order N. So I could easily add together the low-order digits and the high-order digits of the number A and the same for the number B. And let's look at this multiplication problem, which was not part of the problem, but let's look at it anyway. If I multiply that out with high school algebra, these are the same three coefficients that appeared on the previous slide. Here are the three coefficients for the things we want. And here they are here. So suppose I work out A1, B1, and I work out A0, B0, and I work out this thing. That's three multiplications, not four. And when I do that, I can, after I've worked out this one on the left and the first and the third on the right, I just subtract them and I get the middle coefficient that looks like it should have taken two multiplications, but I only did one. So I made a savings. Instead of N squared, it's now 3 quarters N squared. But wait a minute. Each of these three smaller problems can then you can cut them in half two and then cut those parts in half two. You get a fractal of multiplication problems and if you add it all up and figure out how much work you've done, when N is large, it takes about N to the 1.6 elementary steps. Okay, cool. But is that the best you can do? Well, Karatsuba's method was later improved by Tum Koch, Schoenhaage and Strassen. It's called the fast Fourier transform and how long does that take to multiply two N-digit numbers? It takes about N times L of N. L of N is just the number of digits of N, the length of N. L of N is a very small function. It's smaller than any fixed power of N. So this is quite fast and in fact there's even been some improvements on this in the last decade by Fur, Di, Kur, Saha and Satarishi. But they've been small improvements and we don't know if we've reached the limit of the method. So here's an unsolved problem in mathematics. What is the fastest way to multiply? Isn't that incredible that we don't have the answer to that? We have some very good methods but we don't have the fastest method. Let's play a game. It's called Jeopardy multiplication. Now I trust you've seen the TV show Jeopardy. If not, I'll explain the idea to you. Alex Trebek, who's the emcee, will give an answer. For example, he'll say 1492. And then the contestants have to say what year did Columbus discover America? That's an example of how it's done. So you're supposed to find the question that goes with that answer. Jeopardy multiplication is like that. If I say 15, then you say what is 3 times 5? Remember, they kick you off the show or you don't get any prize. If you don't ask it, it's a question. You have to say what is 3 times 5? So let's try playing this. So I'm going to give you the answer to a question. Now in this particular game, my rules are that the numbers you use have to be whole numbers and bigger than 1. So no negatives, no fractions, whole numbers bigger than 1. Beautiful. That's it. What is 3 times 7? Okay, so let's up the ante. Let's do a harder problem. 13 by 7? You're correct. Terrific. Okay, this is a strong audience here, so I have to go to some harder problems. How about this one? Yes? Okay, but you asked it as a question, which I'm very happy you did. However, you violated another rule. You're not allowed to use the number 1. The numbers have to be larger than 1. Excellent. How did you do that? I observed that 347 is 13 less than 1. That's an excellent observation. So what he said is that he added 13 onto this number and he got 260 and he recognized that instantly as a multiple of 13. So this number had to be 13 times 19 instead of 13 times 20. So in fact, you might ask why 13? Where did 13 come from? Was it a bolt of inspiration to do 13? Well, I'm guessing you tried earlier things before 13, which didn't work. So it's like 2, that doesn't work. 3, that doesn't work. 5, that doesn't work. 7, the last digit is 7. Take it away. 24 is not divisible by 7. 11, there's a test for that. It doesn't work. So 13 is really only the first try and it worked. Okay, cool. So we have a lot of experts here in the audience. So I have to zoom up to a much harder problem when I have a story about. So while you're thinking about the answer to this or the question for this, I have to tell you a story. So I went to high school not far from here in Nassau County and I was a math elite, which was a math contest, and the way it was organized back then, I'm not sure if it still is, is that it's not like you went in and you took a test and had 10 questions. You went in and there was just one question and it was timed like maybe five or six minutes. And so the question that was posed to me was this. It said factor 8051. I don't know if jeopardy existed back then, but it wasn't done in that way. It just said factor 8051. That was the question. Now I was pretty good at arithmetic. I had five or six minutes to do this and I figured, okay, this method that we looked at 247 with where you just try small things one at a time to see if they go. Well, how far would I have to go to do it? Well, the square root of this number is about 90, so I'd just have to try prime numbers up to 90 to see if any one of them went. And I figured, well, I could do that. I could get this problem right, but I'm good at arithmetic. So however I thought like a student in a contest, that would be a very stupid contest question. If all you're supposed to do is arithmetic like crazy for five minutes and just barely beat the buzzer and get it. There has to be a trick. So I spent about maybe half of my time looking for the trick and not finding it, and then I said, oh, I better go with the way that I know. And I wasn't that good and I got it wrong. So the question is now to you. Now, this is a story that I remember from a long time ago. What is the trick? Yes? It's 93 squared minus 7 squared. So 83 times 97. What is 83 times 97? Excellent. What is the trick? 90 squared minus 7 squared. Right. So the idea is to say 8,051. I gave you a hint when I said that square root was near 90. So in fact, 90 squared is 8,100. So this is 90 squared, which is 8,100 minus 7 squared. 8,100 minus 49. And so now you can use algebra to factor it. So this is 90 minus 7 times 90 plus 7. Or as was said, 83 times 97. Okay, so now that you know this trick, let's look at some other numbers where it works nicely. Here's another one. Okay, now that you know the trick, this should be easy to factor, right? Who has the question to ask me? Yes? What is 29 times 31? Excellent. That's it. What is 29 times 31? Because you wrote it as 30 squared minus 1 squared. That's correct. 30 squared minus 1 squared. So it's 30 minus 1 times 30 plus 1, or 29 times 31. Here's one more. That's not too bad. Here's a big number to factor. 9,991. Yeah, what is 97 times 103? Because you can write it as 100 squared minus 3 squared. Okay. Here's another number. I'm going to go back to the slides now. This one's also a good number for jeopardy multiplication. However, it's not so easy to do. While you're thinking about 1649, let me just tell you that the problem we don't know, because this is what we still don't know, is we don't know the fastest way to play this game. We don't know the fastest way to factor. If you use that method of trial dividing up to the square root, then doing an n-digit number could take, in the worst case, when you have to actually go up to the square root, it would take maybe about 10 to the n over 2 steps. Now, we know a much faster way known as the number field sieve, which takes, instead of 10 to the n over 2, it takes 10 to the Hubert of n steps. We don't know if this is the fastest we can factor. However, it is the fastest we can factor right now. We don't know if it's the champion method forever. And it's interesting, it's fun, that the fact that we cannot, at present, factor any faster, because this is the state of the art, the number field sieve, that has an application in cryptography, which we'll see in a second. Now, this factorization of 8051 and 9991, is a method due to Fermat. Try to write the number as a difference of squares. In other words, find the first square above the number, like if you have 8051, the first square above it is 90 squared, and then subtract the number and see if the difference is a square. So 90 squared minus 8051 is 49, and that is a square. But let's look at 1649. 1649 certainly looks like a square, because it starts with a square, it ends with a square, and it has a square in the middle. But it's not a square. In fact, the first square above 1649 is 41 squared, which is 1681, and if I subtract 1649 I get 32, and 32 is not a square. If I go to the next square, 42 squared, I get 115, and that's not a square. If I go to the next square, 1849 subtract, I get 200, and that's not a square. In fact, Fermat's method will eventually work. If I have enough energy, I could just go through this list of squares and eventually find one. But in the worst case, Fermat's method is worse than just trial dividing up to the square root. Numbers like 8051 and 899 and 9991, they were all very special cases of Fermat's method that worked on the first shot. Here we have a much more typical situation with 1649 where it's not working. However, we can use these numbers, here are these numbers, 32, 115, and 200 to make a square. If I forget about the middle one, I just look at the first and the third, the 32 and the 200, their product is 6400, and that's a square. So how can we use this? Let's see how it works. I want to introduce some notation where if you studied math, you probably know this notation and if you haven't, this is totally new and might be scary, but I want to tell you that it's not square, it's just something also that you learned in grade school. It's just the notation that's different. So what you learned in grade school is how to divide. And you learned that division and sometimes leaves a remainder. That's the radical thing, it leaves a remainder. So how might you write that two numbers leave the same remainder when they're divided by a third number? Well, that's the way you write it, right here. And we read this, x is congruent to y mod n. If you divide n into x and get a remainder, divide n into y, get a remainder, they're the same. That's what this notation means. So for example, if I divide 17 and 37 by 10, I get the same remainder. So it's legal to write 17 is congruent to 37 mod 10. And you can check that this is also right, that 43 is congruent to 98 mod 11. Now, we'll use this notation as follows. So the first equation we had a few slides back was here. I'm going to let n be this number 1649. So that's what the n is over here. And 41 squared is 1681. And if I divide 1649 into it, well, it goes one time with a remainder of 32. So it's legal to write that. And the third one was 43 squared after a remainder of 200. So we have this correct also. And now what I want to do is I want to multiply these two things as if they were equations. And it's very, very handy notation because that's perfectly legal. It might not be obvious to you that that is legal, but if you work out the details, it is. And if I multiply the two left sides, there they are, 41 squared by 43 squared, multiply the two right sides, 32 times 200, there they are. Well, this thing here on the left side is a square. It's the square of 41 times 43. And the thing on the right, we just noticed was a square. That's 80 squared. Now, 41 times 43 is a big number, but its remainder, when I divide by 1649, is 114. So what I have here is two squares that leave the same remainder when I divide by n. So before we had the number n was equal to the difference of two squares. Now we have the number n is a divisor of a difference of two squares. So I don't say that n, I don't use algebra to factor this like this, but I do use algebra to say, look at this number, 114 minus 80, which is 34, or I could look at the other one if I wanted to, but this is smaller. 114 minus 80 is 34. I can work out the greatest common divisor of that with my number 1649. There's an old method of Euclid from 2300 years ago that helps me do that quickly, and it's 17. So this means that 17 is a prime factor of 1649. And I can solve the jeopardy multiplication quiz by just dividing 17 into 1649, finding that the other factor is 97, and so the answer is what is 17 times 97? Okay, so we looked at three methods for factoring. We looked at trial division. We saw that that's pretty slow. We looked at Fermat's method, which in some cases is fast, but in other cases is even slower than trial division. And then we have this so-called method. It's not clear how all of these different steps actually would work if you tool it up, but this does work if you tool it up. And this is the core of the quadratic sieve algorithm that Kathy O'Neill mentioned when she introduced me. So this method actually works and can factor big numbers. The number field sieve is just a fancier version of the quadratic sieve. It also has the underlying flavor of assembling two squares that are congruent mod n, whose difference is divisible by n. So the record that we've now done is about another 100 digits above the 129 digit mark that Kathy mentioned in 1994. It was recently a success with a 220 digit number. But progress is slow and if you give me a 300 decimal digit number, that's hard. We don't have any way of dealing with that. Now the fact that we don't have a way of dealing with that, as I hinted at before, we can use this to make a secure cryptographic system. So let's see how this cryptographic system might work. I call this game hack the bank. So the bank has two public numbers. We'll call them n and e. And when I mean public, there's not a technical definition of the word public. All I mean is on the bank's website, you can look it up. What are your numbers? There they are. They're right there in the front page of the website, n and e. They're public. Everybody knows them. Now the bank is not telling you everything because the bank has a secret number, d, that it does not reveal to the public. Now say we have a message that we want to send to the bank. I'm going to call this message m. Now m could be written in English, for example, but we're going to digitize it and make it a number. And in addition to that, if it's a long number, we're going to break it into pieces so that each piece is smaller than m. And good. So let's assume that m is just a piece of the original message. I'll full call it m. And here's how we're going to encrypt our message that we're sending to the bank. What we're going to do is we're going to multiply this message m by the public number e and then we're going to divide that product by the public number n and get the remainder r. That's it. So r is some jumbled up version of the message m and that's the encrypted message. Now what does the bank do to decrypt this? Well, the bank has a secret number d and what the bank does is multiplies the encrypted message r by d divides by n gets the remainder that gets you back to m. Now you might not believe that this could actually work. So let's look at a numerical example. So just running through this quickly. I've chosen for the public number n to be 1001 which is not only a pretty number but it's easy to divide by that and get a remainder. So we can essentially do that part in our heads here and only the only hard part is the multiplication. And let's say the encryption, the public number is 96 and let's say our message is the number 561. So we have to encrypt the number 561. So this is the way we do it. We multiply, we do this multiplication problem. There it is. You can check me. I think it's right. Good. Then we subtract a multiple of 1001 from this to get a remainder. Now 53053. It's not an odd to the number 53053. No, that's not what I'm talking about. This number is a multiple of 1001. And we're going to subtract that from 53856. We get 803. So that's the encrypted number. There it is. Message started. This makes perfectly good sense, 561. 803 is just some random looking string of digits. It's the encrypted number. Now here the bank has to come back and find 561. How does it do it? It has a secret number D. Now in this case the secret number D is 73. And so the bank does the same kind of thing. It multiplies 73 by 803. Gets this product. Subtract 58058 from that and you get 561. Now it's not important that the message is 561. This would work for any message as long as it was smaller than 1001. Any message whatsoever from 0 up to 1000, this would work. Now let's see this from the bank's perspective. An encrypted message has come in and it is the number 591. We're using the same public numbers 96 and 1001. And the bank's secret number is 73. Their computer is down at the moment. So let's help them as if they need it. But let's help them decrypt this message. So 591 is encrypted. What do we have to do? I'm just repeating what I just said. What do we have to do to do this? Well, we have to multiply. Let's try it over here. So this is what the bank would do. And then we subtract from this and we find 100 is the message that was intended. Now this part's more fun. Suppose we don't know D. Suppose we don't have that secret number. We want to hack the bank. So let's try it with some, here this is. We're even dropping. Remember the public numbers are public. Everybody knows that even the black hats, even the people that even the people that want to hack the bank. And the encrypted message comes in 789. What could the real message be? We have to decrypt it. We don't know what D is. If we knew D we could do like I just did by hand on the previous example. Now remember what I said that it doesn't really matter what the message is. The D would work with any message. In particular, it would work for the message 1. If 1 were the message then E times 1 is just E. And so whatever D is, it should leave the remainder 1. D E should leave the remainder 1 when divided by N. So we're searching for a number D such that if you multiply it by 106 and divide by 1001 the remainder is 1. Now you've seen equations like this. You just divide both sides by 106. That's what they taught you in algebra. But that doesn't make sense in this case because dividing by 106 is not, would not give you a whole number for D. But what you could do is just start D equal 1, D equal 2, D equal 3 and up through D equal 1,000 perhaps. And there are just maybe a thousand things to try. And you'll find the right one. Well that's not so bad. However, instead of 1001 which is a painfully small example, suppose we tried a Google plus 1 which is 10 to the 100 plus 1. Then you wouldn't want to try everything up to this number. You'd be going way too long. So the question is how can we hack the bank without guessing? Now there is a method for doing this and I'm not going to show you the details. But the method is actually due to Euclid. I mentioned Euclid a few minutes ago for an algorithm that it came up with 2,300 years ago to find the greatest common divisor of two numbers. That same method can be used to find the number D. Anyway, so using Euclid's algorithm from 2,300 years ago, you can find the secret number very quickly and therefore hack the bank. Euclid could hack the bank. So that's not a good cryptographic system if it could be hacked so easily. But in fact, there's a very similar system which cannot be hacked so easily and is in use. And it works like this. Again, you have public numbers E and N, you have a message M. Instead of multiplying E times M, E take M and raise it to the power E and find the remainder when you divide by N and that's R. And then to decrypt there's a secret number D and you decrypt by taking R to the D power divide by N and that's the message M. Now why can't you use Euclid? The D and the E are related in the same way that they were in the previous slide. Except the thing we divide by to get a remainder is no longer N but a number N prime. Another number that's also part of the secret. And N prime can be computed easily by anybody who has the number N and it's prime factorization. So if you're good at playing jeopardy multiplication, really, really good then you can hack this system too. However, nobody's that good at it and if you've gone to other talks here at MoMath people, mathematicians proudly tell you of all the wonderful applications of their work. It's of course fun to do mathematics, fun and games but also there are nice real world applications. Well, here's a real world application of my subject tonight namely a secure cryptographic system but it's not something I'm personally very proud of because it's an application of something we do not know how to do. Namely factoring big numbers. I'd like to breeze by a few other unsolved problems that involve addition and multiplication. Both of these are very famous. Goldbach's conjecture says that every even number starting with 4 is the sum of 2 primes. So for example 4 is 2 plus 2, 6 is 3 plus 3, 8 is 3 plus 5, 10 is 3 plus 7 and so on. It's really, really easy to find pairs of primes. In fact, as you go up it gets easier and easier because the numbers have more than one representation as the sum of 2 primes. So for example, 14 is 3 plus 11 and also what is it, 7 plus 7. So you have a second way. When you get to 16, whatever. 3 and 13, 7 and no not that. 5 and 11 and so on. So it's obviously true nobody has ever proved it. It's twin prime conjecture. For infinitely many primes when you add 2 to them you get another prime. So for example, 3 plus 2 is 5. 5 plus 2 is 7. 17 plus 2 is 19. 197 plus 2 is 199. Those are both primes. This doesn't happen for every prime when you add 2 you get another prime. In fact, it starts really thinning out beyond where the primes thin out. However, it's not so hard to find large, large examples of twin primes. We don't know if there are infinitely many. Euclid proved that there are infinitely many primes. We don't have a proof that there are infinitely many pairs of twin primes. Here's another problem which I'll skip because I don't have the time. It's also a famous problem. You've probably heard of it even in the N plus 1 problem or the 3x plus 1 problem. Here's another problem. I encourage you to go to my website and find these slides and look at this problem because it just involves the idea of least common multiple and adding up some fractions. If you can prove this inequality that's here at the bottom of the screen then the Clay Math Institute will pay you a million dollars. There's no opportunity to make some money at this talk. The Clay Math Institute website won't have this version of the problem. They have another problem called the Riemann hypothesis and this one is equivalent to it. So check this out. I'd like to look at quickly at an unsolved problem concerning just addition. Here we have the addition table from 1 to 10 and the color code here is that the red numbers are new and the numbers that are not red are not new. So, for example, take one of these blue numbers, 13. It's not new because it appeared on the previous line and this 13 is also not new because it appeared on the previous line. Wait a minute. That's the very first time 13 appeared in this table so it's right. So if it appeared on a previous line, it's not red. If it's the first time, it is red. And if you count up the red numbers which are just the distinct numbers that appear in this table there are 19 of them. So this matrix, 10 by 10 matrix of 100 numbers there are really only 19 numbers in it. Now, in general, if we had an n by n addition table with the numbers from 1 to n there'd only be 2n minus 1 different numbers because the sums are just ranging from 2 to 2n. And it's not so hard to convince yourself about the answer to this question here. Can you find n numbers, not the numbers 1 to n but some, you have complete freedom. Choose the best n numbers you want to choose and have fewer than 2n minus 1 distinct sums. It's pretty easy to convince yourself that no, you cannot do that. There always have to be at least 2n minus 1. That's not hard to prove. Now here's the question. Sometimes if you take n numbers there are lots of sums. So here are the first 10 powers of 2 and there are lots of distinct sums. There are I guess maybe 55 red numbers here. So an Israeli mathematician, Gregory Freiman has a nice theorem that talks about the structure that's forced on the n numbers if there are very, very few distinct sums. The unsolved problem is what's the boundary for Freiman's theorem? So the structure that he finds when there are very few distinct sums is that the n numbers have to sort of almost be in order like the numbers 1, 2, 3 up to nr. For example, they could be the even numbers 2, 4, 6, 8 and so on or they could be a slightly more complicated things than just an arithmetic progression. Those are the forms that the n numbers have to be if there are very, very few distinct sums. Now this distinction between few and very few is a little bit subjective but it's carrying me too far afield to reason that out, to say specifically what's happening there. Okay, well let's move on to the multiplication table. Now if I look at the multiplication table I can ask how many distinct numbers appear there. So I'm looking at just the numbers 1 to n. I'm not being fancy. It's a very basic question. How many distinct numbers are in the n by n multiplication table? The code in this talk is if the question is in red we don't know the answer. Okay, so well we do know the answer to some extent. Well for example, we know it for n equal 10. It's not hard to work out. Here's the 10 by 10 multiplication table and there are 42 numbers in red. These are the numbers. There are 42 distinct numbers out of it. The numbers of course below the main diagonal have to all be seen before because of the commutative law of multiplication. So for example, 6 times 5 is 30 but that appeared on the previous line as 5 times 6 but in fact that appeared on the previous line for another reason that was 3 times 10. Anyway, 42. The question is how does this function of n grow as n goes to infinity? How does this number grow? Now if you look close up you'll see that if you take m of n and divide it by n squared those fractions tend to zero as n grows larger and larger but it's a very lazy approach to zero. In fact, if you look from a distance it's about n squared. So it's a small constant times n squared where the small constant is growing to zero very slowly. But how slowly? This is, it's n squared divided by something. Remember l of n, it's the length of n, the number of digits of n. So that's already very small compared to n and I'll raise that to some exponent e. That's not a public, this is another e. That's the wonder of math now, e is something else. Okay, e is a number that's about 0.086. So l of n is pretty small and you're raising a pretty small thing to about 0.086. That's even tinier. Here is something that's even smaller. It's raised to a higher power but it's the number of digits or the number of digits of n. So it's even smaller. We don't know the exact answer. This theorem that involves e, I called it e in honor of Erdisch who was mentioned in the introduction and that he and I had a bunch of papers. I'd like to take a break in the technical stuff in this talk and tell you a story about how I first met Paul Erdisch. It's sort of an interesting story. I guess the year was 1974 and at that time I was an assistant professor at the University of Georgia and I used to watch occasionally baseball games on television and they had the Atlanta Braves on, being in Georgia. This was an exciting time because Hank Aaron was gunning for the record home runs, career home runs. So the record had been held since the 1920s by Babe Ruth and the record of Babe Ruth was, as I think many of you might know, 714. Now this record of 714 was sort of like Steph Curry's three-point record this year in the NBA, totally beyond what anyone else had ever done. I mean, maybe this was like 50% larger than the runner-up. I don't remember who that is. Anyway, Hank Aaron from the Atlanta Braves was threatening to break this record. He had already tied it and would he hit 715? So the question was about 715. Would he get it? And I was watching TV the night that he broke the record. I later learned that this was more than just a baseball feat. He had received death threats for trying to break Babe Ruth's record. In fact, Hank Aaron's mother came down to the field and gave him a hug. That's nice, but she wanted to take a bullet in case that was going to happen. It was unreal time that that should be a worry, but it was. Anyway, I didn't know that at that time and I was watching the game and I thought, 714 and 715, can I play Jeopardy! multiplication with them? Can I factor those numbers? Well, they're not very hard to factor, are they? I mean, one of them is, even the other one ends in five, so they're pretty easy to factor. In fact, 714 is 7 times 102, so it's 2 times 3 times 7 times 17. And 715 is 5 times 143, so it's 5 times 11 times 13. And then I noticed, wait a minute, not only were these numbers easy to factor, but they involved the first seven primes. Now, that's sort of interesting, isn't it? I have two consecutive numbers and their prime factorization involves all of the primes up to some point. Does that ever happen again? You know, that's the way mathematicians think, I'm sorry. Well, you try it and it's pretty easy to convince yourself that it probably doesn't happen again. It happens for smaller numbers, so for example, 14 times 15 involves the primes 2, 3, 5, 7. Even smaller than that, 5 times 6 is 2 times 3 times 5. But anyway, bigger numbers, probably not. So the next day, I went into work at the University of Georgia and I saw a colleague of mine, David Penny, and I asked him, can you find an interesting property of 714 and 715? Well, he found the same thing, but he then asked the question to a class he was teaching and one of the students in the class came up with another property. And this was the other property. If I add 714 and 715 together, I get 1429. Now, what's the interesting property about that? If I add up the primes in 714, 2 plus 3 plus 7 plus 17, that's 29. If I add up the primes in 715, I get 5 plus 11 plus 13, and that's 29. So forget about the fact that I added 714 and 715. Just look at the last two lines there. If I add up the primes in the two numbers, I come to the same sum. So Penny and I, and a grad student in Carol Nelson, wrote a paper together about the numbers 714 and 715. It wasn't meant to be a serious math paper. We had jokes in the paper, even in the first paragraph. We talked about how all of the certain interesting properties about the numbers 714 and 715 had been lost in all of the hubbub, and we were going to set the record straight. And this last property seems to occur infinitely often. In fact, we found a bunch more examples. We have two consecutive numbers, and if you add up the primes in one, you get the same answer as when you add up the primes in the other. And so we called these things Ruth-Aaron Pairs. So this was a lot of fun. In fact, if you Google this, you can find that this, you know, there's a little cottage industry now on Ruth-Aaron. Anyway, fast forward about 20 years. So, no, before I fast forward, let me tell you what happened. This appeared in the Journal of Recreational Math, which is a defunct journal right now. I don't think we are the cause of why it's defunct, but anyway, it's now defunct. And guess who read our article? It was Paul Airdish. And we had guests in our humorous article that this was an unusual property that the sum of the primes in n should be equal to the sum of the primes in n plus 1. And he wrote to me a letter, and he said he knows how to prove that. And he'd like to come and visit me, and we can discuss it. So I was totally dumbfounded, and so that's how our collaboration began. And that was the subject of our first joint paper. Now fast forward 20 years, and Airdish was being given an honorary degree at Emory University in Atlanta. And a professor there who had arranged it named Ron Gould, well, he knew that Airdish was going to be there. He invited me and my wife to come as well. He knew about another honorary degree recipient who was going to be at the same party for the recipients of the honorary degree. And that was Hank Aaron. So I introduced myself to Aaron, and I said, you're not going to believe this, but your baseball feet changed my mathematical life. He looked like he had just met a very strange person. And so anyway, I introduced him to Airdish, and then Ron Gould had come with some baseballs for Aaron to sign. He knew that Aaron was going to be there, but he didn't know about the connection to Airdish. And he gave me one of the baseballs. I had Aaron sign it, and I had Airdish sign it. And so I figured that Aaron now has Airdish number one. Also, doesn't have a joint paper, but he has a joint baseball. Okay, back to serious business here. I want to close out with one unified problem that involves both addition and multiplication tables. And the problem is due to Airdish and another Hungarian mathematician named Samarit. And the idea is to have N carefully chosen numbers and look at both the addition and multiplication tables. Now we've seen that if we take the first N numbers, we get close to N squared distinct entries in the multiplication table. It's actually a small thing times N squared, but from a distance it looks like N squared. But very few in the addition table. At the other extreme, if we take for our N numbers the powers of two, then the addition table sums are essentially all different except for A plus B equals B plus A. So we have about a half N squared numbers in that table. But now the multiplication table looks like the addition table on the previous case, because you're just adding X bonus. So when we choose our N numbers this way, one is big and the other is small, we choose our N numbers this way, it's the reverse. Now you might have wondered, well we'll get to that, what the poster for this math encounter is. It had seesaws on pictures of kids on seesaws, what that has to do with this talk. So we'll see that in a moment. I just mentioned here that if we take N random numbers, then it's likely both tables have close to N squared distinct entries. But here's the question and then it's unsolved. If we choose our numbers so that the number of distinct entries in one table is small, must the other always be large? So here are the poster from this evening. We can see children on seesaws and that's my illustration of making one is large, the other is small. So here's our example with 1 through 10. Multiplication table has many distinct entries, addition table few. So we have many products and few sums. And here's an example with the powers of two. This is the multiplication table on the left. There are very few distinct products and there are many distinct sums. So we have the reverse, few products and many sums. So put more precisely if we have N distinct numbers, must at least one of these two quantities, number of distinct pairwise sums, number of distinct pairwise products, be greater than almost N squared into the, say, 1.999 for all large values of N. We don't know the answer to this question. And it's not for lack of trying. There are some people that have worked on this problem. Erdisch, Semmerer, they came up with the problem. Nathan Sinu is a professor here in New York City. He's worked on it. Chen, Elikesh, Borghain, Chang, Green, Tao, Solimoshi. Solimoshi has the record. He didn't get N to the 1.999. He got N to the 1.333. He got N to the four thirds. That's the record result. Now, this is a very illustrious group of mathematicians. I don't know if you know all of the names here, but this list contains two fields medalists, namely Borghain and Tao. Also two Crawford Prize winners, the same two people. In fact, let's skip them now because they... I mean, if you look at Tao's awards, you must have about 30 of them. We have an Abel Prize winner here. That's Semmerer. We have a Wolf Prize winner. That's Erdisch. We have four Salem Prize winners in American Math Society. That's Borghain, Konyagin, Green and Tao. We have an Eisenstadt Prize winner that's given in Canada. It was given to Solimoshi. And still, with all these heavy hitters, the problem is still not solved. So I'm not trying to intimidate you. Well, actually, it intimidates me to think of all these people not solving this problem. So my message is that we could use a little help. In fact, I have one prop that I wanted to share with you here at the end of my talk. I mentioned Steph Curry and the Golden State Warriors, so I have this t-shirt, which is sort of apropos of the theme of this talk, strengthen numbers. We could use some more numbers, numbers of people working on these interesting problems. Thank you very much. Well, fortunately, we've left a little time here for a special homework help session with the Master of Arithmetic and Multiplication, Multiplication in Edition. So if you're having a little bit of trouble with one of those problems, or if you have anything else to ask, Professor Pomerance, please raise your hand high. Over with the mic. Stand up and ask your question in the mic. So any questions this evening? Surely someone has something that has vexed them about arithmetic over the many years. You could also ask if there were any answers. Oh, true. He asked for your help. So if somebody has any insight to offer on Multiplication in Edition tables, please step forward now. The rewards will be wonderful, I assure you. Okay, we have a question. It's almost a, it's a personal question. What is the chance to get number, Erdos number two? Well, you can write a joint paper with somebody who has Erdos number one. That's the usual way it's done. There are quite a few people around who have Erdos number one. Yeah, roughly how many folks are there living with Erdos number one? I don't know the answer to that. There's a guy at Oakland University which is near to Detroit. I think his name is Gerald Grossman who maintains a website about Erdos numbers. You can see that. I don't know the question of how many of them are living. Okay, well, there's where you find your hot prospects for lowering your Erdos numbers and for lowering your Aaron number. I'm not quite sure how you find out how to do that. Any other questions this evening? Oh, I see one. Same deal. Wait for the mic. Please stand up. Maybe you said as much as you remember about this, but I'm just totally curious what Hank Aaron made of your story about the impact of his accomplishment on your accomplishment. Well, you know, he's a very famous person and I personally found it very difficult for me to go up and introduce myself to him, you know, but I screwed up my courage and did that. I mean, it's a strange story even if you're in mathematics. He was very polite and he smiled nicely to me. I mean, what do you say? Going back to the quadratic, you tried three times to find, like you said 41 squared, you subtracted n, you got a remainder which wasn't a square, then you'd 42 squared, then you'd 43 squared. I was curious, in the more general case, do you need to do it three times or five times? Is there some kind of limit to how many times you have to try before the product of two of them is a square? No, obviously this was a toy problem and in the real world when you do this and you do the quadratic sieve, you have a very long string of big numbers coming at you and you have to find some subset of them whose product is a square and so this is the way it's done. You take the numbers that are coming at you and you try to factor them. Now, I just said factoring is hard but some numbers are easy and so sometimes these numbers coming at you and this long string of numbers are easy to factor so that's good. Keep on collecting these ones that are easy to factor and so like 32 in the example, that's two to the fifth and 200, that's two cubed times five squared, maybe, I think so. The key thing is that they just involve the primes two and five and you just look at the exponents that appear. So the exponents for two to the fifth, that's five twos, zero fives, so you have five zero. For 200, three twos, two fives, that's three two. You think of them as vectors but now it's not just twos and fives but long, you know, maybe 10,000, 30,000 of these exponents and so you have these lots of vectors and you use linear algebra to find a dependency among them, mod two, which means you have a square. So that's how you make the toy problem into a real algorithm. Now, I've heard that somehow quantum computers are really good at this multiplication jeopardy. Is that the case and is that going to mean that we have to find totally new cryptosystems as soon as we get those quantum computers working right? Well, yeah, it's a big F, isn't it? Yes, it's been shown theoretically that a quantum computer could play multiplication jeopardy and win in quote, polynomial time and in fact crack the RSA cryptosystem. However, this is just theoretical. We don't really have quantum computers that can... I mean, when we played multiplication jeopardy, the first couple of examples we had were 15 and 21. I think they built an MIT, a quantum computer that can do those numbers. But beyond there, I mean... So I guess it all depends on whether there's a Moore's law of quantum computing. Let's thank our speaker one more time.