Primes and Equations | Richard Taylor

today our speaker is Richard Taylor Richard joined the Institute as a professor in our school of mathematics in January before that he was in residence here as a visiting professor for the previous year and a half Richard obtained his bachelor's degree in mathematics from Cambridge University in 1983 and his PhD from Princeton in 1988 where was a student Andrew Wiles who is a trustee and former member of the Institute Richard was a lecturer and then reader in the Faculty of mathematics of Cambridge University from 1988 to 1995 when he became civilian professor of geometry in the University of Oxford a year later he moved to Harvard as professor of mathematics becoming Herschel Smith professor from 2002 until his move here last month Richard's work has been concerned with number theory with Andrew Wiles he found a fundamental and unexpected method to demonstrate the correspondence between two different mathematical objects certain Galois representations and elliptic modular forms this provided a crucial step needed for the completion in 1995 of Andrew Wiles proof of Fermat's Last Theorem he's responsible for many other important developments with collaborators he completely resolved the shamira taniyama vague conjecture in the theory of electric curves that Michael Harris he proved the local Langlands conjecture and he also proved the Soto Tait conjecture a long-standing problem in the theory of elliptic curves Richards work has been recognized by very very many prestigious awards and prizes including the 2007 Shaw prize which he shared with Robert Langlands the emeritus professor of the Institute today Rich's title as you can see is primes and equations thank you Peter as Peter says my title today is primes and equations but what I would really like to report on is the remarkable progress in recent years on the theory of polynomial congruences a topic in which the Institute for Advanced Study has played a really outsized role reporting on developments in pure mathematics is at least for me a real challenge mathematicians have developed and are still developing an extremely powerful language in which to express their ideas but this language is not widely understood to express the same ideas in English or any other natural language it's often such a cumbersome process that they ideas lose their natural beauty I've tried hard to make this talk elementary while still conveying some cutting edge research in an hour you'll be able to judge for yourselves how far I've succeeded I must start at the beginning by describing some of the actors in the story I wish to tell and at the very beginning we should start with prime numbers so prime numbers are whole numbers integers bigger than 1 which are only divisible by themselves and by one so the smallest examples are 2 3 5 7 11 13 17 each of these take seven for example the only whole numbers that divide 7 are 1 & 7 itself other numbers like 6 or 9 or 51a divisible by more number so 6 is divisible not only by 1 and itself but is also divisible by 2 and by 3 a 9 is divisible not only by itself and by 1 but also by 3 so these are not prime numbers prime numbers derive their importance in mathematics because they are in a sense the building blocks for all whole numbers any positive work any positive whole number can be written uniquely as a product of prime numbers so for instance 2012 is two times two times five hundred and three each of which are prime and it's not it's pretty obvious that any number can be written as a product of prime numbers but in fact it's not very obvious that there is only one way of doing that or only a one way up to reordering them I think many people believe it's it's obvious that there's the only one way of doing this probably because we were taught it at a very young age but in fact if you ask yourself why 2012 can't be written as any other product of prime numbers it's not such an easy thing to see nonetheless it's true the second actor I need to describe is congruence --is so technically what's that the definition of a congruence two whole numbers are said to be congruent modulo another number P which most of this talk will be a prime but for now can be any number if exactly when P divides the difference of those two numbers and when that's the case I'll write a a sort of equal sign with three lines B modulo P so as this is going to play such a big role in this talk I'm going to spend a little time trying to reinforce the concept for those of you who are new to it so there's a first example 2 is congruent to 17 modulo 3 what does that mean means the difference which is 15 is divisible by 3 the most basic example of congruence is is congruence modular congruence is modulo 2 with which we're all familiar an even number is exactly 1 divisible by 2 which is exactly a number congruent to 0 modulo 2 and an odd number is a number that leaves remainder 1 when divided by 2 or something that's congruent to 1 mod no to so the concepts of odd and even numbers are exactly the same as the concepts of numbers modulo 2 the first prime another case of congruence is with which most people are familiar is congruence is modulo 10 two numbers a congruent modulo 10 exactly when their difference but when they have the same last digit because they have the same last digit exactly when their difference has last digit 0 exactly when their difference is divisible by 10 another example of congruence is with which people are familiar congruence is modulo 12 are clocks at least the hours marked on our clocks based on congruence is modulo 12 1300 hours is the same as 1:00 p.m. this means that as you turn around the clock when you get up to 12 you go back to the beginning as it were and start counting again so what was 13 becomes 1 and this is I think a good way to to think about congruence is for those of you who are not familiar with them before there's no particular reason why our clocks should have 12 hours mark round they could have any number of hours for instance 7 hours and the study of arithmetic modulo 7 congruence is modulo 7 sometimes could kroc arithmetic you imagine the integers equally spaced around a circle with seven points mark seven hours marked and as you just as when you whole numbers are introduced in alimentary school you describe them laying out on an infinite line starting at the origin zero one two three and so on we're going to do exactly the same thing but on a circle and when we've moved seven around we come back to the beginning just as on a clock when you move around twelve hours you come back to the beginning and we can do all the arithmetic we can do with normal whole numbers in fact often a bit more with these numbers module seven so for instance our one can add one could add what was the example I was going to do five plus six so I can use this thing if we start here we add five and six we first of all count round five one two three four five and then we want to add six so we count round another six one two three four five six and we see that in this world of arithmetic modulo 7 5 plus 6 is 4 we can do a subtraction 3 minus 5 I count round in the positive direction 1 2 3 and then go back 5 1 2 3 4 5 and I find that 3 minus 5 is again 5 in this world of arithmetic modulo 7 just as I would have done if I was adding or subtracting on the number line in elementary school except then I wouldn't have gone back round the origin I would have just stayed on the infinite line I can do multiplication I can look at 2 squared for instance that's two lots well let's do 3 squared 3 lots of 3 I count round 3 3 times 1 2 3 1 2 3 1 2 3 3 squared is 2 and equally 4 squared is 2 if I take that's for lots of 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 4 squared in this world or arithmetic modulo 7 is again - so I've recorded in this example of arithmetic modulo the prime 7 the examples I've just talked about but I've also because we will need them later recorded all the squares there are only as we saw 7 possible positions for the our hand on this clock and if I square each of them I get 0 1 for 2 and then I repeat 2 4 & 1 so only four numbers modulo 7 are squares 0 1 2 & 4 okay well what I want to say is mostly about these congruence ease but I thought having introduced them I would digress for a bit to give some motivation of how why they were studied historically and after I've done that I will come back to talk about modern results about what new things we know about these congruence ease so a very old so called daya phantom problem daya phantom problem means problem I want to solve in fractions rational numbers not another way of putting it decimals that either terminate or recur not are infinite decimals that never repeat themselves that not numbers like the square root of 2 or e or pi is to describe all right angle triangles with hypotenuse one and the other two sides rational numbers so by Pythagoras theorem two numbers X and y are the short sides of a right-angled triangle with hypotenuse one exactly when the sum of the squares x squared plus y squared is equal to one so we're asking to find rational numbers fractions x and y such as x squared plus y squared is 1 so this at least appears to be a problem that was considered nearly 4,000 years ago by the Babylonians this is a famous cuneiform tablet and this column and this column seem to be a list of 15 solutions to the equation I just wrote down in fact they're they describe just one side this is the numerator this is the denominator the ratio of this over this is one side of a right-angled triangle with hypotenuse one whose third side is also a rational number I've written those Babylonian solutions to this equation in in modern notation here see though some of them are really quite complicated it's actually interesting that so it's the first fraction in every case that's recorded on the tablet you look at the second fraction in every case the numerator tends to be divisible by 60 or a power of 60 or at least something has a high common factor with with some power of 60 and the babylonians of course worked with numbers to base 60 so these would have looked like particularly simple numbers to them so we don't know why the Babylonians made this list of numbers but it's at least tempting to believe that they were interested in this question and had at least this list of solutions that's one example of such a triangle but it was certainly by the time of the Greeks they certainly knew how to completely solve this equation if I think of it there is a simple procedure that actually parameterize --is all rational solutions to this equation if I look at all real solutions to this equation and plot them in the XY plane I get a circle a circle of points that are distance 1 from the origin of course an uncountable number of real numbers and what the Greeks understood was that if you looked at the points on this circle which happen to have rational coefficients both their coordinates were rational numbers then their densely spaced round this circle leaving no gaps at all and as I say they knew very clearly how to parameterize them and the process is not very difficult but to save time I think I won't describe it now ok but now let's vary the problem a little bit let's ask ourselves what about right-angled triangles with rational sides whose hypotenuse is just slightly shorter let's say naught point 9 9 9 9 9 9 1 million less than before if I draw the graph of the real solutions to this equation I get another circle which is just ever so slightly smaller than the one I wrote it for what about which rational numbers points with rational coordinates lie on this circle I like the rather surprising fact is that none at all these circles from a geometric point of view look to be almost exactly the same but one had many many ah rational points on it in a dense way make it slightly smaller you can't really from a geometric point of view see the difference but suddenly you miss every rational point and the explanation for this comes from the theory of congruence --is that I was just describing so Hendrik Leinster is fond of saying a mathematics talk without a proof is like a movie without a love scene and so this is going to be the proof for this mathematics talk I'm going to explain to you why it is that this there is no right-angled triangle with rational sides and hypotenuse nor point nine nine nine nine nine nine so if there were such a thing I could write the two sides as rational numbers with a with a common denominator and I could always arrange that the denominator is divisible by a thousand just by multiplying numerator and denominator by whatever I need to be convenient later and I can also arrange that a B and C have no factor in common if some number divided bigger than one divided by and C I could simply cancel it in both fractions and I would have a simpler way of writing the fraction so I can assume that nothing divides them all and now substituting it into the equation we had before and clearing the denominators I now have to look at the equation a squared plus B squared is 999,999 times C squared where now I've gained the fact that a B and C have to be whole numbers not necessary fractions now whenever I have whole numbers that satisfy such a relation when I do the addition and multiplication on the infinite number that will still be true that relation when I rap the number line round a circle with any number of hours marked I like I look at the prime factorization of 999,999 here it's written out I'm going to focus on one of the prime factors 7 7 divides this 999,999 I could equally focus on 3 or 11 but for definiteness let me focus on 7 and I'm going to consider this as equation not now in the whole numbers but as an equation in the whole numbers modulo 7 as I say take the number line wrap it round the circle with seven hours marked and I get another equation now when I wrap it round the number line this last number is divisible by seven so this equation when I consider it more if that one square number plus another square number is zero but if you remember I listed the squares modulo seven and there were only four of them zero one two and four on the seven mod 7 o'clock arithmetic these are the squares now there's not one obvious solution we could have zero plus zero equals zero that would certainly work but that can't have come from our original problem because if zero plus zero equals zero it would mean that a was divisible by seven and that B was divisible by seven so this side would be divisible by 49 this is only divisible by 1 7 so 7 would have to divide C squared as well so 7 would have to divide C and that would mean that we hadn't written a B and C in the simplest possible way we could factor cancel a factor of 7 and then repeat the process ok so this solution although a solution to the equation modulo 7 doesn't provide a solution to the original problem we were interested in and then the key fact is there are no other solutions to this equation modulo 7 C plus one is one that's not divisible by 7 0 plus 2 is 2 0 plus 4 is 4 none of those are divisible by 7 1 plus 1 is 2 snow good 1 plus 2 is 3 that's no good 1 plus 4 is 5 no good 2 plus 2 is 4 2 plus 4 is 6 4 plus 4 is 8 when I divide by 7 I get a remainder 1 none of the other possibilities work there's only one solution to this equation modulo 7 with a and B and C all being zero and that as I say doesn't lead to any rational points on the original circle ok well that's an application of the congruence is and I think the sort of reason that congruence is were first considered to to solve problems daya phantom problems such as this they have many other applications they have applications in everyday life the many of the error correcting codes that allow information to be efficiently packed onto a DVD and not when it gets lightly scratched or something you don't lose all the information or information sent through communication with a spacecraft or something uses the theory of congruence --is and in fact even the theory of polynomial congruence is that we're coming to but I I'm not an expert on that and I'm not going to talk about these real world applications I'm going to stick to the the pure mathematics involved so we just in connection with this dive and higher equation we looked at this particular congruence we looked at the equation x squared plus y squared congruent to 0 modulo 7 x squared plus y squared adding up to 0 in this clock with 7 hours and we saw that there was only one solution the only possibility is that both x and y is 0 but one can consider any other sorts of polynomial equations one could consider a one variable equation does to have a square root we saw that it does it has two square roots three and four I showed you that one could consider higher degree equation one can look at numbers that cube to give one indeed there are three such numbers one two and four or cube to give one for instance four cubed on the usual number line is 64 when I divide by 7 7 goes into 63 the remained as one that tells me that when I wrap the number line around the 7 art clock X 4 cubed becomes 1 so in a way this this world is better than the world of usual integers because certainly 2 has no square root in the usual integers 1 has only one cube root in the usual integers namely well either ya namely 1 and I can look at more complicated equations I can look here's in a cubic equation in two variables this particular equation has 9 solutions which I've listed here for instance saw this one ah so if I put x equals 5 I get 125 minus 25 is 100 when I divide by 7 7 goes into 98 so the remainder is 2 so this side becomes 2 the other hand if I put Y equals 1 I get 1 squared plus 1 is 2 so this is indeed a solution of this equation modulo 7 and these are all of them a feature of these which I hope you've noticed is that the number of solutions is always finite because there are only a finite number of possibilities for each variable they're only a finite number of hours on the clock well the first are a big topic I want to talk about its so-called reciprocity laws ask the question if I fix the equation but vary the prime number P how does the number of solutions vary as I vary the prime number so let me again start with a simple example the equation x squared equal to 5 so for instance we saw that modulo 7 no five had no square roots the squares mod seven with zero one two and four mod three it's the same story mod three X can be 0 1 or 2 the squares are 0 1 or 4 which mod 3 are just 0 and 1 we never get 5 which is the same as 2 not 3 on the other hand mod 11 there are two square roots 4 squared is 16 take away 11 I get 5 7 squared is 49 take away 44 I get 5 and if you check there are no other possibilities so here's a table for the thus giving the solutions for the first few primes I've excluded 2 of 2 and 5 because they occur in the equation so they behave slightly differently and you'll see immediately that there are always either no solutions or exactly two solutions that's actually not so hard to prove but can anybody is there a pattern as for which when you find no solutions and when you find two solutions can anybody who doesn't know Gauss's law see the pattern these which numbers have I written none against in which ones have I written two other numbers against well if the number ends in a 3 or a 7 there are no solutions 3 7 3 7 3 on the other hand if the number ends in a 1 or a 9 there are two solutions one nine nine one one nine one one nine and this is a general phenomenon what it means is that the number of solutions of this equation is governed by the value of P modulo 10 we saw that the last digit of a number is like looking at congruence is modulo 10 and this pattern as I say was a is a general one as was first realized towards the end of the 18th century first of all by linear Euler who didn't phrase it quite like this but was more interested in the sort of Diophantine equations I was talking about a minute ago and as you saw congruence ease came into their study and so Euler needed to think about these sorts of questions and basically saw this pattern it was formulated in the in the way I've just described first by Legendre and I apologize for Legendre for this picture of him but according to Wikipedia it's the only surviving portrait of him in many books you will discover a much more picture of a much more distinguished looking man but again according to Wikipedia it's now been discovered that that gentleman was his cousin and not the mathematician so the law they discovered now called Gauss's law a quadratic reciprocity is that for any whole number n if I asked where the N has a square root when I do arithmetic on a clock with P hours PR I'm well first of all how many square roots there either 0 1 or 2 square root sum 1 square root hardly occurs only for a finite number of values of P that are bad in some sense so they're basically either has no square root or it has two and crucially you can determine which by looking simply at P mod P where P lands on a clock with 4 n hours so this may be on the face of it doesn't seem so surprising we've got this croc arithmetic occurring twice maybe they've got something to do with each other but at least to me it is an extraordinary surprising fact the solving an equation with clock arithmetic modulo P should have anything to do with P modulo something else it's something that in a sense I still don't understand many statements in mathematics when you've really understood a proof you feel you know when it's true there's no other proof you could write down although there are many proofs now known for this fact seemed to me to be to really explain why it's true Gauss found this as a teenager found this proof as a teenager or his first proof as a teenager and it's supposed to have been why it supposed to have been his favorite theorem yeah it said that he called it his theorem herim his golden theorem that I don't know whether there's really evidence that he did so he certainly came back to it again and again throughout his life providing at least eight different proofs during his lifetime gasps let me just do one numerical example to try and illustrate the power of the result suppose we asked ourselves wherever three has a square root modulo 20 million 130 2011 or 2013 2011 which is a prime number well with a calculator or a computer it would be easy to run through the 20 million possibilities but if you wanted to do it by hand that would take you an awfully long time on the other hand Gauss's theorem tells us that whether this equation has solutions only depends on this prime modulo 4 times 3 modulo 12 so let's find what this is on the usual 12 odd clock if I divide it by 12 the remainder is 7 so Gauss's theorem tells us that the this is has a solution exactly when this has a solution but we've already calculated the squares modulo 7 it didn't take us a moment to run through those there are just four of them 0 1 2 and 4 not 3 is not a square modulo 7 so we immediately conclude the 3 doesn't have a square root modulo this large prime either ok well that's the very simplest example the first example of the so-called reciprocity law law that provides some way of us count a number of solutions to a fixed congruence modulo aver for a fixed equation modulo a variable prime what about more complicated equations well there was a lot of work that went into generalizing Gauss's law for the next 150 years culminating in class field theory which was one of the great achievements of mathematics in the first half of the 20th century particular art ins emails art ins reciprocity law but they were all restricted to a very small class of equations it wasn't until the last 60 years that people realize that these reciprocity laws were much more general or should be much more general so then we look at a more complicated example let me go back to the case of a cubic equation in two variables so here we want to think of a B and C and D as some fixed integers I'll take a concrete example in a minute and I want to consider x and y as variables so we're going to fix a b c and d and we're going to then let the prime p vary and count the number of solutions x and y to this equation modulo P how does this vary with P and some notation I'm going to write n P for this number of solutions so let me as I said I was going to take a fixed example we already looked at this example modulo 7 we saw that it had exactly nine solutions which I've listed here but I can do this modular any prime I can look modulo 2 modulo 2 X gives 0 and 1 so this will always be 0 and Y can be 0 or 1 and this will always be 0 in the case Y is 1 I get 1 plus 1 which is 2 which is equivalent to 0 so there are 4 solutions modulo 2 and in fact there are 4 solutions modulo 3 and 5 this repetitiveness is is because this equation actually has some symmetry that's not immediately apparent it's not an important phenomenon way we can easily calculate these numbers modulo 11 on the 11-yard clock you get 10 solutions on a 13-yard clock you find 9 solutions which I've listed here modulo 17 there are 19 solutions modulo 19 there are 19 solutions so I've written this R actually relatively small amount of data as a table for the Prime and the number of solutions to this equation modulo that prime so I think the first thing that's immediately should be immediately obvious is that it's not like the quadratic equation in one variable that Gauss considered there the solution was always like the number of solutions with always either 0 or 2 here at least it appears that the number of solutions is growing as the prime grows that's true was proved by Helmut Hassler in 1933 but the number of solutions to first order is the same as the size of the Prime the number of solutions minus the prime is a order twice the square root of or is bounded by twice the square root of the prime so the leading term this is this is just behaves like P with some sort of relatively small error this result R has an extremely important enormous generalization that was proved by Pierre Deline professor here at the Institute which was originally conjectured by Andrey Vey who is also a professor here at the Institute which is really the basis for for everything that we'll be doing the rest of today but I I'm not going to talk about anymore because it's not what I wanted to focus on anyway in this case the number of solutions is approximately P and so may be better to take off this easy leading term and to find out what's really going on we want to focus on the error the difference so let me instead give the table of the differences so here's the prime and here's the difference between the number of solutions and P and now these numbers are obviously much smaller with this amount of data it may be not obvious wherever they're bounded for instance wherever they state they stay in a finite range that's not the case in fact houses theorem is best possible you there are certainly examples where you get as big as this bound but can we predict do we have some way of predicting what these numbers are and the answer is yes this was discovered by Eichler about 60 years ago so I'm going to look at the following infinite product I'm going to mount use I'm going to multiply so Q is just some variable I could have called it X or T it's conventional to call it Q so I've called it Q so I'm going to consider the infinite product Q times 1 minus Q all squared times 1 minus Q 2 xi or squared times 1 minus Q squared or squared times 1 minus Q to the 22 or squared and so on I hope you see the pattern although this is an infinite product so you could never finish multiplying it out if you're interested in the coefficient of any particular power of Q you only need to consider a finite number of terms if I'm interested in the coefficient of Q cubed or even Q to the fourth I don't need to go beyond this point I only have to look at the terms up to here because everything in here will involve more than a q cubed so it'll just affect the higher order terms it's a pretty laborious thing to multiply this product out but I did I calculated the first 20 terms here and to my mind completely remarkable thing happens if I look at where a Q is raised to a prime PACU squared Q cubed Q to the fifth and so on and look at the coefficient if I look at the coefficient of Q squared minus 2 which was exactly the error in the point count modulo 2 I look at the coefficient of Q cubed it's minus 1 which is exactly the error in the point count modulo 3 I look at Q to the 11th the coefficient is 1 which was exactly the error in the point count modulo 11 I look at Q to the 17 the coefficient is minus 2 which was exactly the error in the point count when I counted solutions on the clock with 17 hours Q to the 19 doesn't occur at all this coefficient is 0 which was and there was zero error when I made the point count with a 19-yard clock and again I would like to stress how surprising this is to me why multiplying out this extraordinary expression should have anything to do with calculate count counting solutions to equations modular equations are completely different sorts of mathematics one can ask if this was just the coincidence of it and the answer is no as I say martinique approved form a member of the Institute martynuk approved in 1954 that for all Prime's P the error was exactly the coefficient of Q to the P in this infinite product expansion and this was really the the birth of a whole new chapter in reciprocity laws it's the first where people call non abelian reciprocity law everything that had happened to date based on Gauss's work somewhere in the background there was a group that was a billion here there's no such thing well within aa very shortly after I class work people began to suspect that this wasn't an isolated phenomenon first taniyama suggested that something similar to two actus things should work in general without being terribly precise what goro shioma a former member of the Institute and professor at the University made a really pretty precise conjecture as to how you work to do this and Andre Vey who who initially didn't appear to believe she Maura's conjecture when he eventually paying became convinced he completely tied down one or two loose ends in jhamora z-- formulation so there should be some similar effective algorithm whenever I have a cubic equation in two variables I'm sorry I should have said that this may seem a particular special cubic equation I've got a quadratic in y equals a cubic in X but in fact any cubic equation in these two variables can essentially be put in this form by some change of coordinates so I don't lose much generality or any generality by thinking of equations in this particular form they congestion there should be some similar algorithm well you have to be a bit care for what similar meant it doesn't mean multiplying out some enormous product of the sort we had there that was rather accidental and I chose that example because it's easy to explain but underlying that product there's the study of certain symmetries certain symmetries of a two-dimensional geometry two-dimensional geometry code a hyperbolic plane geometry in which are two lines going out from a point at a given angle somehow move apart faster then you would expect them to if you think about the usual flat two-dimensional spurt you could in two dimensional space a tabletop a hyperbolic plane is often described as the interior of a circle but it's you should think as the distance here so first of all here is the hyperbolic plane being tired by triangles first of all you should think in this hyperbolic world all the edges of these triangles are straight lines although they look to us as if they're curved in the hyperbolic world they would be straight lines and also the triangles in this picture in the hyperbolic wad all the same size so although they look to us much bigger here than they do out by the edge in the hyperbolic world as you go out towards the edge things are much sort of you should think of them as expanded enormous Li and so one can consider symmetries in this world just as one can consider symmetries in the real world just as a triangle has a rotational symmetry of order three in this world there are many symmetries one can study and the thing that's predicting the error in the point count when you're try and solve this modular equation is something to do with the symmetries in this strange geometry there also another way of thinking of them is as 2 by 2 matrices with whole number entries in determinant 1 this group acts on this space and this is a group of symmetries here so again to me it's an extraordinary thing that why we doing this algebraic problem counting solutions to equations modulo some prime should have anything to do with sort of continuous geometric questions this hyperbolic plane was made brought to a general audience by Asher who when Center ah drawings like this one by the mathematician HMS coxeter was uh inspired to to make various woodcut prints based on these symmetries so here is one where he's got fish tessellating the hyperbolic plane well this conjectures no longer a conjecture it was finally proved in 2001 based on the revolutionary idea is introduced by Andrew Wiles another former member of the Institute and professor at Princeton University and this received a lot of attention because prior to Andrews work Gerhard Frye and Ken ribbet had shown that this reciprocity law implies Fermat's Last Theorem so just as the Gauss's reciprocity law has consequences for the study of quadratic equations fontaine equations the work of this shamora taniyama conjecture has dire Fantana coin applications to Fermat's Last Theorem but that's not something I want to talk about received a lot of attention so I'm going to talk about other things today well you can ask all right we've looked at this is it how general is this phenomenon we looked at the quadratic equations in one variable cubic equations in two variables and we now believe thanks to the extraordinary insights of Bob lineman's professor at the Institute that this should apply to any system of equations any number of equations in any number of unknowns of any degree they if you try and count their solutions modulo a varying prime number P you should be able to predict the answer you get in terms of the geometry of some of some space or equally I can think of all the symmetries of some space non Euclidean space not in the usual flat space but some more complicated space or you can think of it to do with n by n matrices with a whole number entries in determinant one that's another way of thinking of the same thing ah as I said the case N equals two was related to the hyperbolic plane which you can sort of draw pictures of the next simplest case N equals three is related to the symmetries of a five dimensional space which I can't illustrate but I do want to give you some idea of what's going on I can oops that's not what I wanted to do this is a simulation made by people at the University of Minnesota of hyperbolic 3-space which is the geometry that would be related to so-called if one was doing G l2 over the gaussian integers would come up they've tessellated a hyperbolic 3-space by dodecahedra and i think what you might get from this is a sense that as you move through this the size of things appears to change quicker things come in from the edges at a different rate from what you would expect if you were moving through usual Euclidean space ok well the second thing I want to talk about is distribution theorems so far we've asked for an algorithm that predicts allows you to predict how many solutions a given equation has modulo some varying prime but one could also are simply for statistical information one could ask for what fraction of primes does such an equate for instance li does n have a modular what fraction of Prime's does then have a square root or two square roots and what map modulo what fraction of Prime's does it not and if we looks for this so in the case considered by Gauss this quadratic equation in one variable the answer was found 40 years after Gauss's theorem by dear Ashley who proved this as long as n is not a perfect square which will behave differently then what you might expect is true modulo 1/2 the primes this has a solution a modular the other half it doesn't dearest's lays proof of this relied crucially on Gauss's law of quadratic reciprocity use somehow can't get at these density theorems distribution theorems unless you know reciprocity first having said that rather surprisingly our Frobenius 40 years later founder the similar distribution question for ah any equation in one variable no matter what the degree mode since the answer is less as less you wouldn't have guessed the answer in this case and also the reciprocity law in question is still not in still not known it was being conjectured by Langlands relatively recently but nonetheless Frobenius back in the 19th century was able to answer prove this distribution law because he realized you needed something slightly weaker than a reciprocity norm but I won't probably for lack of time let me leave that for now okay now let me move on to equation in more variables again cubic equation favorite cubic equation in two variables again just as I had to before assume that n wasn't a square in the case of the quadratic equipped with some n I have to make some slight assumptions that they're not extra symmetries present are generally satisfied in the cases when this is not known or easier and one can ask about the distribution of the number of solutions well we saw that in this case it grew like P so any given number of solutions is only good there are only four solutions for three possible Prime's number of solutions is four for only three Prime's P this you can't look at a statement of exactly the form that dear Ashley proved but if instead I look at the error this will be smaller and if I normalize it by what has approved was the order of magnitude of the error so I look at the normalized error term in the point count then it's going to be by hasis theorem it's going to be fixed in the range from minus 2 to 2 angkong can ask how is it distributed in the range minus 2 to 2 is P varies and the answer was conjectured by mikio Sato and John Tate about 50 years ago Miki Osato because he he did a series of computations here at the Institute and which he completed just after he left back up in Japan and John Tate for completely different reasons because he had heuristic reasons to believe something like this might be true and the answer is the distribution should be a sort of semi circular law this looks like a semicircle so this graph the red line is the distribution predicted by Sato and Tate and this bar graph blue bar graph drawn behind it is the actual distribution you get for this normalized error term by looking at all Prime's less than 1 million and you can see that already there's a reasonably good fit to the prediction this is now known it's true proof for most elliptic curves about five years ago completed even more recently and one needs more than the recipe turns out that the difficulty here is one needs more than the reciprocity law for a which the shamone taniyama conjecture he was sorry as Thea it's the cubic equation in question one needs a reciprocity law for self products which will be somehow have take the self product n times I'll get a equation in to n variables or something like that so you have to have reciprocity laws for equations in arbitrarily many variables to be able to answer this to use what was essentially Jewish ladies method then to to obtain this distribution okay well I wanted to finish ah I've said what I wanted to say about congruence is but I wanted to finish with a another application of congruence --is to Diophantine problems again an old die a phantom problem as far as we know this one was first posed by Al qiraji at the end of the first millennium qiraji was a mathematician working in Baghdad though maybe he was the first origin and he asked the following question for which whole numbers n is there a right-angled triangle such that all the sides of fractions all the sides of rational numbers again and has area n so we're looking for a right-angled triangle we have given n fixed areas of one for instance is there a right-angled triangle with area one or whose sides are rational numbers of fractions so this is not at all clear how you would go about answering this question I mean you could could run through sides x y&z with small denominator and see if the area happened to be one but I mean you could do all denominators up to a million and you wouldn't find any solution and you could ask yourself does that mean there isn't a solution or the just the first solution is much bigger and algebraically we're asked to solve the equations two simultaneous equation x squared plus y squared equals Z squared Pythagoras's theorem that's the condition for XY and z to be the sides of a right-angled triangle and the area is 1/2 base times height 1/2 X Y so we want that to be equal to n you can actually eliminate a variable and end up with a cubic equation in two variables one of these elliptic curves that we've been talking about well the remarkable answer we believe although it's not yet proved that there is a good answer to this question now believe let me give you two numbers that are much easier to calculate I'm going to take n and try and write it as a square of a whole number plus twice the square vu whole number and eight times the square of a whole number and I'm going to count the solutions to that where W here is even and the solutions to that where W here is odd I get count them both I get two numbers these are in a sense easy to calculate because it can certainly do it in a finite amount of time because each of you V and W have to be less than the square root of n so I just have to run through small number of possibilities and see whether they satisfy this equation and we have a W is odd or even they either tribute to this one or this one and then what was realized by Jerry Tunnel form remember the instituting professor at Rutgers was that the answer to this question should be yes there is such a right angle triangle exactly when the number of solutions to this equation with W even equals the number of solutions with W odd so for instance we were talking about triangles of area one if I take N equals one with N equals one V and W had better be zero because two already 2 times 1 squared is too large or 8 times 1 squared is too large so the only solutions are you is plus or minus 1 and V is 0 and W is here both of which are even so a 1 even is 2 a 1 odd is 0 so according to this conjecture there should be no right-angled triangle with area 1 this was in fact proved by firmer in the 17th century it was already known but in certainly for general n it was unknown in the 1980s Gerry tunnel dick gross Hans Agee and Victor khaliv Argan proved half of this they've shown that if these two numbers that are easy to calculate are not equal then there can never be a triangle with it rational sides right angle triangle with rational sides and area n so this reproves sorry this this reproof fern axiom for instance there's no such triangle aerial one I want to mention though not explain why that the proof is based on one of these reciprocity laws in this case our tins reciprocity laws from the 1920s counting solutions to congruence is in the other direction it's still certainly an open question but we would believe that we know an algorithm will always find a solution when these two are equal we believe we know an algorithm that will always find a solution but at the moment nobody knows that that algorithm will terminate no we certainly believe it so here are some I took some small values of n each case calculated a n well my son Jeremy calculated a and even an a an odd here they are so the as we saw they're not equal when N equals 1 there's no rational right triangle at area 1 similarly there's no rational right triangle of area 3 and no rational a right triangle of area 155 no rational and right triangle of area 163 but we believe if we believe the conjecture there should be one of area 5 1 a very 157 159 163 165 sometimes they're easy to find so in the case of 5 which we've already observed by Fibonacci in the 13th century there is a solution the triangle with sides 9 646 and 41 sixths not over in their lowest terms has area 5 and does form a right-angled triangle but sometimes it's far from obvious if there's any such solution richard fineman in his book you must be joking mr. Feynman says that as a graduate physics graduate student at Princeton he used to like to challenge the mathematicians say that he could find a solution to any problem they posed he wouldn't guarantee to give the proof but he could always figure out by a combination of guesswork or trying small examples the answer to any question they posed and he claims to have always been successful at this I wish one of the math graduate students who challenged had posed to him the problem given a list of numbers like this and asked him for which of those is there a right-angled triangle of the given area and rational sides because without knowing the theory I've partially described I think it would be very hard to guess the answer so for instance 157 here we believe there is a solution we believe we have an algorithm but trial and error would never find it for you here is the simplest right triangle with area 157 and rational slides so with that I'll stop thank you are there any questions yes for the other known examples for these yes ah I don't know off the top of my head what they are and none of them are quite as bad as 157 I chose this range exactly so I could finish with this slide it was I should say it was done zag yay but as you might have guessed if you know the number theorists who found this example yes maybe can you give an example of a diaphragm line problem for which congruence --is don't help so it depends what you mean by don't help I mean for almost every diakon time problem they help they sometimes don't give you the whole solution sorry sorry so if if what you mean by the question so sometimes there's a beautiful theorem if you took a quadratic equations then there's a there's a solution exactly when there's a solution modular wall congruence is that's a theorem of hassle in Minkowski and if you mean if you mean by congruence is helping whether this look so-called local to global principle holds then the answers certainly that there are examples and I don't I mean for instance already amongst the elliptic curves or curves of the genus one or example I don't remember the specific example but maybe I don't know is that the question you were asking but I'm tempted to say that even even when they don't give the answer they help your a lot further forward when you've ruled out anything that can be ruled out by congruence is than you were at the beginning any other questions was a bit perplexed between yeah alright Frobenius and shabbat Arif so Frobenius what did you have in mind a that Frobenius didn't before shamatari Frobenius given any equation in one variable with rational or integer coefficients Frobenius told you the fraction of primes for which it had a given number of solutions so that is actually easier than that sure if you figure figured out it's easier than the Chabot or of density theorem but if there no more questions I suggest we go to the reception enfold hall where we can continue discussion but before that let us thank Richard for a really beautiful talk

Channel: Institute for Advanced Study

Views: 27,760

Rating: 4.8120804 out of 5

Keywords: Richard, Taylor, Math, Lecture, Number Theory, Diophantine Equations, Prime, Numbers, Congruences, Reciprocity

Id: 5pTaZu3C--s

Length: 63min 59sec (3839 seconds)

Published: Fri Apr 27 2012

Reddit Comments

[u/reph]

Fantastic. Thanks.