 This is an introductory mathematics talk on diaphanetine equations Originally given to the Mathematics undergraduate student association at Berkeley So let's start by just giving some examples So the first one that everyone has come across is Pythagoras's Equation x squared plus y squared equals z squared And you all know the solutions three squared plus four squared equals five squared and There's 18 eight squared plus 15 squared equals 17 squared and so on and The point of diaphanetine equations is you've given it an equation and have to solve it in Integers if we're trying to solve this in real numbers, it wouldn't be terribly interesting We just take z to be the square root of x plus y And So Particularly notorious example is Fermat's last theorem So Fermat changed the exponent here from two to some high in number and asked whether x the n plus y to the n Equals z to the n as any solutions apart from the trivial ones where x or y or z is zero so here n greater than three and He said he had proved there are no further solutions although he probably hadn't and For a couple of centuries, this was the most notorious open problem in mathematics It's just Fermat's last theorem until it was finally solved by Andrew Wiles in the mid 90s So a special case of this is x cubed plus y cubed equals z cubed and Oil originally showed it at no solutions and What I'm going to do in today's lecture is not show that equations of no solutions, but show how to find solutions So this isn't a terribly interesting equation because it's got no solutions to find So instead we could look at some variation of it like x cubed plus y cubed equals 9z cubed and This certainly has a solution 1 cubed plus 2 cubed Equals 9 times 1 cubed and we can ask if it has any other solutions, and if so, how do we find them? So another example of an equation with a Curve actually has a singularity as we will see later is something like y squared equals x squared plus x cubed so this is a solution 6 squared equals 3 squared plus 3 cubed or 24 squared equals 8 squared plus 8 cubed And again, we can ask can we find all integer solutions of this? So next example is the number 1729 is fairly famous for a reason I'll explain a little bit later And it can be written as a sum of two cubes in two different ways So you can ask are there other numbers that can be written as a sum of two cubes in two different ways So we're trying to solve W cubed plus x cubed equals y cubed plus c cubed And that there are some trivial solutions where w is equal to y or w is minus x or whatever So we we want to find non-trivial solutions of this For example, we could have 2 cubed plus 16 cubed equals 9 cubed Plus 15 cubed and again, we can ask how can we find solutions of this? Well one way to find solutions of all these equations is just a kind of stupid one We could just get a big computer and have it checking every possible case and This would sort of very slowly find solutions like this but what we want is a is a sort of lazier way of finding solutions with less effort and What I'm going to do is show how to find solutions of Diophantine equations by using geometry and the geometry makes it much easier to find solutions so let's look at the first example Pythagoras's equation and We can divide it by z squared. So if I put x equals x over z Y equals y over z and this equation becomes x squared plus y squared equals 1 And you can recognize this as the equation of a circle So what we're trying to do is to find points on the circle with rational coordinates So how can we find those systematically? Well, what you can do is there's an obvious point with rational coordinates, which is this one and I guess These four here, which just as coordinates minus 1 0 and What I'm going to do is I'm going to just draw a line through these two I'm just going to draw the straight line through these and This straight line has a slope It's called the slope t and the slope t is easy to figure out It's just this distance here divided by this distance here, which is just y over x plus 1 and Now we can work out the equation of this line So this line is just the line y equals x plus 1 times t Now you notice t is obviously rational if x and y are rational So we can say x y rational implies t is rational And we can ask the converse if t is rational or x and y rational And you can see the answer is yes without doing any calculations as follows Because this is circle has a degree 2 equation So if you try and find the intersection points, suppose you try and find the x coordinates of the intersection points of a line on the circle There'll be two intersection points which has a solution of some quadratic equation this quadratic equation is integer has rational coefficients and We know one of the roots is rational So the other route must be rational because the sum of the roots of a quadratic equation with in the rational coefficients is rational so We know before we do any calculations that if we Pick a rational t then x and y will be rational. Well, now let's actually calculate them So we have y equals x plus 1 times t and we also know x squared plus y squared equals 1 And if we substitute y into here, we find x squared plus x plus 1 t all squared Is equal to 1 and now this is a quadratic equation in x So it is two roots and one of the roots is minus one So it must factorizes as x plus 1 times something and you can work out what the something else is as t squared plus 1 times x plus t squared minus 1 equals 0 so here's the root and x equals minus 1 which corresponds to this point here and Here is the other route which corresponds to this point here and You see we can work at what x is we have x is equal to 1 minus t squared over 1 plus t squared and Y is equal to x plus 1 times t which is 2 t over 1 plus t squared So you can check by algebra now that if t is rational than x and y are rational But these are slightly complicated expressions and we could have found the x as we said We don't need to do this calculation to see that x and y are rational Anyway, we can now find lots and lots of solutions to Pythagoras equation just by choosing t For instance, if we choose t equals a half We find x equals three fifths and y equals four fifths Which gives us the equation three squared plus four squared equals five squared If we choose t equals it's just choose any random rational number three over eleven We find x is equal to 56 over 65 and y is equal to 33 over 65 giving us the solution 56 squared plus 33 squared equals 65 squared and you can obviously generate effortlessly generate enormous numbers of solutions to this just by picking a rational number t and Working out x and y from that So this actually gives essentially all solutions of this equation because solutions of this equation more or less correspond to rational numbers t There's one slight exception this point here doesn't correspond to a rational number t it corresponds to The point t equals infinity, but apart from that we get a one-to-one correspondence between solutions of this And rational numbers, so we've got a complete description of the solutions Um, there's so that's one geometric way of finding solutions There's actually a second geometric way of finding solutions of this So we again draw a circle so and This time Let's write down a solution Yes, we've got some point x1 y1 And let's just look at this angle theta And suppose we take another solution And let's call this angle phi So here we're going to have a solution x2 y2 and you can see this is equal to cosine phi sine phi and this one is of course equal to cosine theta sine theta And now what i'm going to do is i'm just going to add up theta and phi So now i'm going to have this angle here is going to be theta plus phi and the That the coordinates of this point will be cosine theta plus phi sine theta plus um phi And now um, we can use the formulas for the sum of two angles. So, um, this route here will just be cosine theta cosine phi minus sine theta sine phi And then we have cosine theta sine phi plus sine theta cosine phi And now what you notice from this is that if all these Numbers here are rational then these numbers here are also rational So what this means is if we've got a solution x1 y1 and a solution x2 y2 of Pythagoras's equation Then we can find a new solution as follows. So so we've got um x1 y1 With x1 squared plus y1 squared equals one and we've got x2 y2 x2 squared plus y2 squared equals one and now we take this point here, which is the point x1 x2 minus y1 y2 x1 y2 plus x2 y1 and if we call this x And this y then we see that x squared plus y squared will also be one So this gives a way of generating new solutions from old solutions using this formula Notice that algebraically this formula is a bit mysterious. It's not something you would Immediately think of but if you think of it geometrically, it's obvious what's going on. You're just adding up two angles And for example, we had a solution three fifths four fifths And we had another solution 56 over 65 33 over 65 And if we take this to be x1 y1 and this to be x2 y2 We can work out this equation here and it turns out to be 36 over 325 323 over 325 So we found another solution of Pythagoras's equation 36 squared plus 323 squared equals 325 squared and we can obviously carry on doing this and produce more and more larger and larger solutions And incidentally this shows that the solutions of Pythagoras's equation form a group Which means we can sort of There's a sort of way of adding up two solutions to get a third solution So here we take this solution and this solution and sort of add them together to get that solution Um Okay, now let's look at uh another Some equation So let's look at the equation y squared equals x squared plus x cubed So we have solution six squared equals three squared plus three cubed for example, and we want to find all solutions And we're going to do it geometrically So let's draw a graph of y squared equals x squared plus x cubed And it looks something like this You sketch it And you notice there's a sort of funny singular point here This is called a singularity So singularity just means a point where things go wrong. So here we've got two The curve is sort of crossing itself And we also notice there's one obvious solution here minus one zero as well as the solution zero zero Um, and now suppose you've got any other solution Say we've got a solution x naught y naught Well, what can we do with it? Well, what I'm going to do is I'm going to draw a line through this and another point And the most obvious other point to draw the line through is this singularity because there's obviously something rather special about it And let's draw the straight line through these two points And let's try and find the equation of this line. Well, it will have slope Is equal to y naught over x naught which is equal to t And this is a cubic equation So we expect every line every straight line will generally intersect in three points Because if we've got a if we set y equals a x was being substituted in we'll get a cubic equation for x So if x naught and y naught are rational then obviously t is rational So, um, let's just note that down. So x naught y naught rational implies t is rational And the question is if t is rational does this mean x naught and y naught are rational? and the answer is yes because If we fix t um, we have the equation y equals t x And if we substitute this in we will get a cubic equation for x And two of the roots are already rational because they're zero so the third root is also rational So let's just carry this out. So we substitute y equals t x and we get t x squared Equals x squared plus x cubed. So we have x cubed plus one minus T squared x squared equals zero and here we've got a cubic equation And you can see the roots are pretty obvious. There is a root zero and zero and the other root is T squared minus one and the two zero roots Um, we can see them there. They're where this line intersects the The orange curve in this double point and you see where it's very useful as a double point here because it means we get two Roots that are both zero Um, so now we can solve this equation We find x naught is equal to t squared minus one and y naught is equal to t times x naught, which is t times t squared minus one so Here's a way of finding solutions of this equation just by writing down rational numbers And of course, we want integer solutions So we might take t to be an integer. For instance, we could take t equals four and we get the solution x equals 15 y equals 60 and Similarly, you can generate as many solutions as you like um Well, that that cubic equation was particularly easy because it had a double point Um, what happens if you get a cubic equation without a double point? So let's try The example I mentioned earlier x cubed plus y cubed equals nine times z cubed And that's before let's just sketch this to see what's going on And it looks something like this. It's got a sort of asymptote at y equals minus x and it ends round like this So this is x cubed plus y cubed equals nine z cubed. Sorry. I equals nine. Sorry. I should have said I'm changing um Everything by dividing by z cubed so let's put x equals x over z and y equals y over z And then we find x cube plus y cubed equals nine And we know two points on this Because we've got the point one two And the point two one And how can we find other points on this? Well, well one idea is we could just draw a straight line through these two points Um, but that wouldn't be very interesting because the straight line Would just go like that and it would intersect this orange curve at infinity But it wouldn't have a third intersection point with finite coordinates. So what else can we do? Well, if we've got a a point on this curve, let's call its coordinates a b for the moment what we can do Is we can draw the tangent Um To the orange curve at this point so it looks something like this So here we've got a straight line And it intersects its orange curve in three points and it sort of Intersects it twice at this point here because it's a tangent vector. You see a tangent vector sort of Intersects a curve twice at one point So as usual, this means the third point of intersection will also have rational coordinates So we've got this very simple geometric idea if we Take a tangent vector to a solution and find the third point of intersection it will give us another solution So let's do this first of all we need to work out the coordinates of the tangent line well We work out the slope of the tangent line which is Minus a squared over b squared and then we can complete the tangent line as x minus a Plus b so it's beginning to look a bit messy And if we substitute this into this equation here We find we get x cubed plus minus a squared over b squared x minus a plus b All cubed equals nine and this is a cubic in x and It's got three roots So two of the roots are a Because it intersects This orange curve twice at a and there must be a third root Um Well, if you've got a cubic equation Something like a x cubed plus b x squared plus c x plus d equals naught You know the product of the roots is going to be minus d over a Um, so we can work out the third root just by saying The product of a and a in the third root is minus the constant term of this divided by the coefficient of x cubed and if you work this out it turns out to be minus a cubed plus b cubed all cubed Plus nine b to the six divided by b to the six minus eight six And then we have to divide by the two roots we thought of You can simplify this a bit because we know a cube plus b cube is equal to Nine for example, but anyway, so we have an explicit formula for the third root x x is now equal to this and similarly you can find y um So for example, so if we found one root we can now find another so let's take a b To be the root we already know about one two Well, this gives us the root minus 17 over 7 20 over 7 So we find minus 17 cubed plus 20 cubed equals nine times seven cubed So we found another solution of this Um, well, there's another way of finding solutions What we could do instead of taking a tangent line Is we can take any two solutions say this one and this one and draw the line through these So this will intersect in a third point And just as before if these two points are rational the coefficients of this green line will be rational So it's Intersection with the orange curve will again have rational coordinates um, if you do this the algebra becomes really a bit of a mess um So i'm not actually going to carry it out explicitly Um, but i'll just give an example. So if you take This point and this point two comma one And draw the tangent line through them and work at its intersection point. It turns out to minus two seven one over four three eight nine one nine over four three eight And this gives us the solution minus two seven one cubed plus nine one nine cubed equals nine times four three eight Cube which as you see would be a bit tiresome to find by trial and Using this geometric argument. It's quite easy to find solutions like this And obviously you can just keep on doing this Um, as long as your patient holds out every time you find a point you can take tangent line Every time you find two points you can draw a line through them So you can generate endless numbers of integer solutions to this equation in a in a fairly systematic way Um So the Um, there's actually a famous theorem related to this due to more dell Which says suppose you take any reasonable cubic equation All right, so for instance y was x cubed minus x might look something like this What you can do Is I guess it should go through the origin never mind you have to pretend Um, what you can do is whenever you've got three points on this On this cubic whenever you've got two points You can generate a third point by drawing a line through your first two points Um, this is called the chord tangent process for finding solutions Because you can either draw a tangent to a point you found or you can draw a chord through two points um, and It turns out that you actually get um Something called a group. So I suppose I've got two points p and q here Then I get a third point here and if I change its y coordinate to minus the y coordinate I get another point that I'm going to call p Uh sort of plus q. It's it's a funny sort of addition and it turns out That this funny addition of points behaves obeys the usual rules of algebra So p plus q is obviously equal to q plus p Because the lines through p and q is the same as the lines through q and p um Slightly more difficult fact is that p plus q plus r is equal to p plus q plus r This is really a total mess to verify because the formula for the coordinates of p plus q um It are really rather complicated and trying to calculate this explicitly is really rather a headache Um, there is actually an easier way to prove this without doing explicit calculations, but whatever so um What we get is a sort of funny addition of points which obeys the usual rules of algebra And punk hurry ask this question. Can you find a finite number of rational points on any cubic? So that all rational points can be obtained from this finite number by by using this operation And mordell showed the answer is yes He showed the rational points Are finitely generated In other words, you can find a finite number of rational points Such that all the other rational points are generated from these by the chord tangent process um, so um So now we get so far. We've been looking at curves now. I'm going to look at an example of a surface. So this is Related to remanigans famous number So there's a famous mathematical anecdote about this. Um, this appears in hardy's book on Remanigans, so I have here a copy of hardy's book And on page 12 he has he has this anecdote that I'll just read out So hardy says I remember to go going to see him once when he was lying ill in chutney So he's he's talking about remanagement, of course I had written in taxi cab number 1729 and remarked that the number seemed to be a rather dull one and that I hoped it was not an unfavorable omen No, he replied it is a very interesting number It is the smallest number expressible as a sum of two cubes in two different ways I asked him naturally whether he would tell me the solution of the corresponding problem for fourth powers And he replied after a moment's thought that he knew no obvious examples and suppose the first such number must be Very large So as a result of this story this this particular number 1729 has become Really famous. It's even got its own wikipedia article if you want to read about it um so What i'm going to do is show how to find Other examples of numbers with this property So we want to solve a cube plus b cubed Equals c cube plus d cube and obviously we want Non-trivial solutions where a isn't equal to c or d and it's not equal to minus b and so on You can rearrange this by changing the sign of c and d so Uh, we so it becomes a bit more symmetrical It looks like that and now we can just divide by d cube. So let's put x equals a over d y equals b over d z equals c over d So we get x cubed plus y cubed plus c cubed equals one Um, so this is a bit like the earlier equation. We had x cube plus y cube equals nine except now We've got three variables and this is now the equation of a surface in Real three-dimensional space and we're trying to find points on the surface with rational coordinates um and now We can do a similar trick that we did with cubic curves. Let's take two points p and q And draw a line through them And this line will usually intersect this cubic surface in another point. So r is the intersection point of this line and the surface x cubed plus y cubed plus c cubed equals one assuming it line intersects this occasionally the line would just be asymptotic to it, but we'll ignore that problem And now just as before if p and q have rational coordinates Then the equation of this line will have rational coordinates So the third intersection point with this surface will also have rational coordinates So this will be a rational point Okay, well, let's try this first of all We notice this equation has lots of obvious solutions like it's got x cube plus minus x cubed plus one cube equals one cubed and it's got um y cubed plus one cubed plus minus y cubed equals one cubed and One cubed plus c cubed plus minus c cubed equals one cubed. So these are the um obvious not very interesting solutions Um and we can ask what happens if we take two of these boring solutions and do this process We draw the line through them and take the third intersection point and the answer is you get a third uninteresting solution So if you take a point on with satisfying this and a point satisfying this and carry out this process We just get a point here. So that's not terribly interesting However, we've got um a more interesting solution minus nine cubed plus minus 10 cubed plus 12 cubed um plus one equals zero um, so um I guess it should be a minus one there never mind um, so Uh, what happens if we do this? Well, let's just see. So let's take the point p to be one minus one minus one And we're going to let this be Correspond to a rational number t being naught and at t equals one We're going to let it be the point q which is equal to minus nine minus 10 12 And now we're going to draw the line through these points and it will be the set of points one minus 10 t Minus one minus nine t minus one plus 13 t so this is a line through p and q because you can see it's Line parameterized by t and it goes through p and q when t is not all one. So this is going to be our third point r And now we we want the sum of these three numbers cubed You want x cubed plus y cubed plus c cubed plus one to be equal to zero So we get one minus 10 t cubed plus minus one minus nine t cubed plus minus one plus 13 t cubed equals one So plus one Zero keep getting that the wrong way around um, and you can see that it has three roots It has roots t equals zero Or t equals one and the third root Is going to be rational and you can work it out explicitly by doing some calculation It turns out to be minus one over 26 in this case So we now find our point r is equal to 36 over 26 minus 17 over 26 minus 39 over 26 Um, so we get a solution 36 cubed plus 26 cubed Equals 17 cubed plus 39 cubed. So we've got another solution to ramanogen's equation um and um Um, obviously you can keep generating more and more solutions by just drawing a line Through any two solutions you've got um incident you can ask If you get a group like this and the answer is you don't because if the points p and q are equal You think you would take the tangent line to p and q But the trouble is if you've got a point of a surface it doesn't have a tangent line It has a tangent plane So you don't really know which Line tangent to it to take so you can't really define An addition operation on points because this breaks down if the two points are the same um so um in general what happens with playing curves um is If you've got a plane if you've got a curve then it might have degree two And if you've got a plane curve of degree two It acts rather like the equation x squared plus y squared equals one if we can find one rational point on it Then we can find all the other rational points very easily just by drawing a line through both points If we've got a degree three curve Such as x cubed plus y cubed equals nine Then um, this is an example of something called an elliptic curve Unless it happens to be singular such as the curve y squared equals x squared plus x cubed Um, so singular elliptic curves with singularities behave a bit like degree two curves elliptic curves without singularities behave like the degree three curve we had That's you more del showed you can find a finite number of points that generate all of them Um, what about degree curves of degree greater than three? so If you have something like this um Then um curves of degree greater than three Generally only have a finite number of rational points on them This is a very deep theorem proved by faultings a few decades ago. It was originally conjectured by moor del So anyway, we we turns out that curves split up into three different groups There are those for which the rational points are really easy to find An infinite number usually Um, then we get elliptic curves Where the rational points are at least finitely generated And then we get all the other curves where there are only a finite number of rational points and they seem to be very hard to find Um, these differ by something called the genus. We say these curves have genus zero these curves of genus one These curves of genus greater than one And the genus turns out to be what happens if you take the curve over the complex numbers Which gives you a surface and then the surface has a number of Sphere with a number of handles and the number of handles is the genus So we've got this really weird relation between the topology of a complex curve and the complexity of the number of points on it Okay, I think I'll just finish by giving a suggestion for further reading if you want to find out more about this Um, there's an introductory book called undergraduate algebraic geometry by miles reed and the early chapter on it explains um This method of generating um Rational points on cubics. So here chapter two is on cubics and the group law And he explains this method of generation new points from old points in more more detail Okay, I think that's all for this lecture