Transcription
I want to tell you about something very simple that leads to very weird behavior. Take four numbers, x, y, z, and w. And we want the sum of their squares to be equal to their product, x, y, z, w. And we want x, y, z, and w to be whole numbers. Let's say positive. Presumably, there are solutions to that equation.
Well, maybe if I tell you there's a very symmetrical solution. You can guess. Maybe the people watching can spot it. But it's 2, 2, 2, 2. On this side we have four * 4. Four fours. In this side we have four * 4. So this is 16 and this is 16. Got it. We solved it. All right. Well, maybe there's maybe there are other solutions. Are there other solutions? Um, well, yes, there are. Um, and so one way to find another solution would be just to maybe look at the equation and try to guess another one. But there's there's a more clever way to do it. Show me. Show me. What is it? What are we going to do?
Okay, so the clever way to do it is to notice that this equation has a has a very special property, which is even though it it it's degree four, which means that four things are multiplied together. If I view y, z, and w as constants, then it becomes degree 2 in x and therefore much easier to deal with. Even though there are four things multiplied together here, there are only ever two x's multiplied together. So let let's try to work with that and rearrange things to view this as a quadratic equation in x. So we'll write it as x^2 - yzw * x + y^2 + z^2 + w^2 = 0. So that's just a that's just a standard rearrangement of what we had before.
Right. Exactly. It's just a rearrangement, but I've written it in the form that makes it obviously a quadratic equation in x. This might be familiar to you if you know about the quadratic formula. This is in the form ax^2 + bx + c = 0.
Yep. Classic. Just to remind everyone, the quadratic formula says if ax^2 + bx + c = 0, then the solutions are x = (-b ± √(b^2 - 4ac)) / 2a.
Okay. And from this you get Vieta's rule. So if x and x' are the solutions, the plus and minuses cancel out. Plus so x and x' correspond to taking plus and minus here. Um, I'm assuming there are two solutions. They cancel out and we get that x + x' = -b / a. That's true for any quadratic equation.
Yeah. If there are two distinct roots, the sum is -b / a. And we have a quadratic equation here. Right. So if there are two roots, then one of the roots is the one we have here. And the other one will be related to x by x' = yzw - x. That's -b because there's a minus sign here. It cancels out with this minus sign. And then we have to divide by 2a. But a is one here. Or in other words, if I know one solution of x, I know the other one by x' = yzw - x.
All right. And we know that well, we've got one set where they're all twos, right? Yeah. And so, okay, so we have all the solution all twos. So we can start from 2, 2, 2, and we can change x by this formula which says we should replace 2 by 2 * 2 * 2 - 2 and then we'll keep the others the same. So this is 8 - 2 = 6, 2, 2, 2. Another solution.
Okay. So let's see if I can do this in my head. So this is going to be 36 + 4 + 4 + 4. So that sounds like 48 to me on that side.
Yeah. And then I should get 6 * 2 * 2 * 2. 6 * 8 is 48. Okay. So I didn't lie. All right. My method worked. All right. So we've got So there are there are two ways to crack this nut.
Okay. Yeah. I I mean I I have a way to go from one solution to another. Then I check that the solution worked. Okay. I don't need to check that, but I just wanted to convince you. So we we can keep going. We can keep going with this. Okay. So now that I've found this new solution, I can now transform this coordinate by the same rule. So I can change this to six times. So it's always the rule is the product of the other three minus this guy. So it's 6 * 2 * 2 - 2, 2, 2. And then let's see if I can do this in my head. This looks like 6 * 4 = 24 - 2 = 22, 2, 2, 2. And we should notice several things that are happening. The first thing is we keep getting integer solutions. We always have whole numbers.
Okay. And that's going to keep working. If y, z, and w and x are whole numbers, then this will also be a whole number. No fractions or any bothersome things like that.
Yeah. The other thing is that because we had a factor of two everywhere at the start. We always have a factor of two everywhere here. Now, presumably you can just keep doing this forever. There must be an infinite number of solutions. So, it turns out there are an infinite number of solutions. Just to get an idea of what they look like, let me just give you one more. So if I uh if I now modify this coordinate, I'll go to 6, 2, 2, 2; 6, 2, 2, 2. So what we notice is that the new numbers that we're getting, they're growing pretty quickly, right? So you changed the x, then you changed the y, then you change the z, then you'll change the w. Well, now I could now I could even change this. I can keep this two forever if I want, but I I can I can change I can change them all. Now, what you should notice, though, is if you you shouldn't change the same guy twice in a row because you just go back, right? because I'm flipping the root. So if I flip the root and then I flip the root back, then I go backwards, right? So if I if I do my transformation here to the 262, I go back to here. You want to always do the move at a at a position that you didn't just do it in if you want to get something bigger. Never change that same variable twice.
Okay. Right. Cool. But but other than that, we can just keep fiddling forever and come up with all sorts of
Right. Right. That's exactly right. Does this process have a name? Uh yeah. So the process of um flipping the root of a quadratic equation is called a Vieta involution or Vieta jumping. Where might they have seen that a number far? I believe it's related to a number video about an IMO problem from 1986. I hope I convince you that there are infinite number of solutions to this equation in in whole numbers. Now what I want to do is in in some sense count them. So what I'll do is I'll put a upper bound on x, y, z, and w and I'll ask how many solutions are there with each of x, y, z, and w less than or equal to R. Let me write down this quantity. So this is the number of fours x, y, z, and w whole positive with the equation holding and x, y, z, and w all less than or equal to R. R is your upper bound. And so what we're going to be interested in is we know there are infinitely many. So we know as R gets larger and larger, this quantity gets larger and larger. The number of sets that work, right? Yeah. But how how fast does it get larger and larger? And so it turns out this quantity, if I call this V(R). So here's the first weird thing about this equation. V(R) grows like a power of log R. So the reason a mathematician might be slightly surprised at this log R is if you have a Diophantine equation that has many variables compared to its degree, like for example here we have degree four and you expect the growth rate to be a polynomial of R. So this log is a little bit surprising and weird.
Okay, it's not it's not what we expect. Okay, but that's not the only weird thing about this equation. So not only is the growth rate like going to be a power of log R, but it's a power of log R that we don't understand. So let's call it β. We don't know an exact value for β. We can calculate to whatever precision we want. And we know that β is roughly between 2.43 and 2.48. And you can know that because you can just calculate solutions and and count them and see, right? So one way to calculate this number would be just to use the algorithm I gave you for generating solutions from 2, 2, 2, 2. You could put this in your computer, generate all the solutions you can find up to a certain limit like say 1000, then count how many there are. Take log of this, divide by log of 1000, and you'll get some number. And if you make that number 1000 larger and larger, the number that you get from this will converge to this β and it it'll converge pretty quickly.
Okay. So this is this β is sort of a a mystery number. We don't know it exactly. Yeah. It's describing the how the number of solutions below a given bound grows. So it's this it's this strange mystery number. So we know that the growth rate is the growth is like log R to the β and we actually know by a theorem that this quantity is asymptotic to a constant, a positive constant times log R^β as R tends to infinity. Okay. So the remaining question which we don't know how to answer is that it certainly seems as though β is is probably not going to be a rational number. We don't know how to prove this. So it's a question of Silverman from 1995. Is β a rational number? That's something that is of interest to you. Is is that something you're trying to crack or
Yeah. I I think what is what it what it's asking is that we started from a very simple question, which is when are the sum of the squares of four whole numbers equal to the product, and we ended up with this strange growth rate, but we don't know how strange it really is. Is it possible that really β is 245/100? In which case we should probably try to answer where did 245 come from, or is it even stranger than this that there is it is not a rational number, it's not a fraction of whole numbers, and it's such a fundamental question that I think this is important.
Nice. Michael's research receives support from the Leverhulme Trust via one of its prestigious Philip Leverhulme Prizes. One of the trust's main purposes is supporting Blue Skies Research, seeking breakthroughs which will hopefully benefit society, but maybe in less obvious or unexpected ways. It's an independent charity that's been operating for over a hundred years. Their website is filled with great information about how it works, who it supports, even how to apply for funding. I'll include all those links below. Our thanks to the Leverhulme Trust for everything it does to push research forward and for making this video possible. I've actually covered up part of what what you need to understand. Okay, let me just write it out for you. Okay. So, so what this says is that first of all, a and b uh can be elements. They can be 0, 1, 2, 3, blah blah blah. They can be any whole number including zero. Let's stick them into this equation. We can see that we could reduce the larger number down. So, can't we now reduce 30 down? Let's try. And I know the formula for reducing the a's.