Have you memorized the three numbers?

I have it.

You don't know that.

I don't know.

I'm sorry that I didn't even notice I I learned from somebody else's posting that one of the numbers of prime, so I hadn't noticed that we found three injures Who's Cube?

Some 2 33 This number was immortalized in a paper by Bjorn Putin.

He pointed out that if you wanna solve that problem for the number 29 it's really easy.

So 29 27 plus one plus one.

If you move a bit up the number line and you ask about 30 we know a solution for that.

But the numbers are in the billions on that was only discovered in 1999 and Andrew's understand there are some numbers that you will never find one for.

That's right.

So if your number, if you divide it by nine and you get a remainder of four or five that we know you can't do it.

So you made a brilliant video with Tim Browning on this att the time it was an unsolved problem.

There have been some attempts toe prove that this number isn't on the list, but those attempts failed and, well, I've just solved it.

I first saw the video, You know, around the time that it came out and I thought a bit about it, but didn't get very far.

It was actually the follow up video on 74.

I think it was made a couple of years ago, but I only saw it last month.

And that, um, spurred me to think more about it.

And I got hooked that this last number here 74 is actually representatives of some of three cues.

Maybe I'll show you something.

So about these numbers, I've got sand paper here.

So if you look at the end of the paper, he's put all of the numbers below 1000 for which we don't know the answer.

Now, here's a test.

Brady, Do you notice anything about these numbers?

Peculiar.

Now, if you stare at it a while, um, you might notice that their old divisible by three.

Okay.

And if you take this a step further and divide them by nine.

Actually, this is a good plug for your casting out nine video.

Okay, so this one has remained or six when you divided by nine.

This is another six.

Another six starting to see a pattern here breaks with this one.

This is remainder three.

Okay, um, this is remainder three.

This is remainder three on so on and so forth.

So it turns out they all have remained or three or six when you divide them by nine.

That's not an accident.

So, uh, it's Tim explained in one of the of the follow up videos.

It matters what the remainder is.

When you divide by nine, we know that if you get remainder four or five, then actually there's no solutions at all.

Um so it's the same kind of analysis shows that if you get remainder three or six, it's possible.

But only just it means the solutions for these numbers are likely to be more sports.

Okay, And that's what ends up, making them hard to find the modern conjectures.

Every one of these numbers they'll be infinitely many solutions.

Um, it's just they're extremely sparse on dso.

You know, the 1st 1 could be enormous.

As we've seen, we want to solve an equation like X cubed plus y cubed plus Z cubed equals, say 33 on.

We're looking for solutions where all of the numbers are below some bounce.

Okay, so say absolute value of X.

Absolute value of why and absolutely Uzi are all below some bounds.

Say be okay.

So imagine B is ah, 1,000,000 or something.

Now, on the face of it, this is a three dimensional problem.

Okay, we've got three variables to solve for X, y and Z.

And if you just went through all possibilities up to be that somewhere on the order of be cubed numbers to check, I said you can imagine B is a 1,000,000.

To put things in perspective, your smartphone could do a 1,000,000 things in the blink of an eye.

A 1,000,000 calculations like this.

Okay, but if I say a 1,000,000 cubed all right, that's in the realm of possible but very expensive, um, to do on modern computers.

So this is actually very wasteful.

So it's not hard to see that in the search range.

There's loads of places where there's no point in looking Okay, So if I imagine that X, Y and Z are old, positive and large, and of course, there's no way that, uh, some of their cubes is gonna be a small number.

Okay, So already there's there's big parts of the search range that you that you can cut out.

If you take this idea to its logical conclusion, it's It's not hard to see how to get from be cubes down to be squared.

This was realized pretty early on even the very first computer investigations that people did in the fifties.

On this.

They had algorithms that were basically quadratic would work in this time.

Now that wasn't improved on for about 40 years.

And then a guy called No Mel Keys and Harvard.

Um, actually, in a newsgroup posting, um, he proposed another our rhythm, which is pretty amazing.

Actually, it manages to get down from B Square to basically be to the one power.

So be to one plus something very small.

And it's basically what enables us to go to the range is that we are today, you know, tend to the 15 10 to the 16.

It turns out that l Keys algorithm is most efficient when you're looking for a solution for lots of numbers, um, a CZ we were for a while But you can see we're getting down to a pretty short list at this point.

So I started to think, Well, maybe there are other approaches that we could look at.

My approach uses more algebra.

So what I'll do is I'll take this equation and then we'll move Z cubed over to the other side.

Okay, So that gives us X cubed, plus y cubed equals 33 minus Z cubed.

This is now X plus Y X squared minus x y.

That's why squared equals 33.

Minus said Cute.

Okay, now, why have I done that?

Well, um, we have this factor that that we know is there.

Okay, I'm gonna give us a name.

I call it d say for divisor.

All right, that one more step.

And I promise we're getting to the end of elder breath.

Let's divide both sides by D.

Okay, so I'm gonna have on the left side just expert minus X y plus y squared right side 33 minus Z cubed over D.

Now it probably looks like I've made things a lot more complicated, right.

We started with three variables.

Now we before, but I want you to imagine for a moment that we know the values of Z and of D.

Or maybe we have some guests for Well, that means that everything in the right hand side is something I know or I can compute.

On the left hand side, I have a quadratic equation with two unknowns.

As everybody knows, if you got to variable to Seoul for you need two equations.

I've got one here, but there's another one that I used implicitly but didn't write down.

And that's this one.

So X plus Y is equal to D.

All right, so now I've got +22 equations to unknowns.

We know how to solve this.

Or more to the point, if I guess values for Z and D.

I can tell pretty easily whether there are corresponding integer values of X and y okay, and this is the idea.

We just go through all possible values frizzy and D, and see if there's any corresponding X and y.

And now that again sounds like it should be something like B squared, right, because there's two variables, but the key is, once I've picked a value for Z, the values for D actually quite restricted because they have to be devise er's of this number 33 month.

Is he cute?

And there aren't many of those.

So once you've worked out, see, you can work out all the possible values for D pretty easily, and then just try all of them.

I've lost over a few things here, So in actual practice, you do it the other way around.

You go through values of D, and then you work out Z from there.

But yeah, that's the idea.

In a nutshell.

I use lots of computers, actually, a few separate clusters on, and the computation that found the solution was the big computer at the University of Bristol Blue Crystal phase three.

This happened way faster than I was expecting.

So I started this.

I thought I was in for, you know, six months of computation.

My initial fear was that nothing would turn up and then I'd have to justify why I was using so many CPU cycles.

Yeah.

Soon as I started using computer within a few days, it found the solution.

The balance I used was 10 to the 16.

So that's one followed by 16 zeros.

I had actually planned to do 10 to 17.

Okay, but but one pops out before he expected.

Uh, you'll see here at 905 in the morning on February 27th.

So this was running on 512 CBU course, and at the same time, one of them went through one of these values of D.

Actually, this is the number D in the middle here.

I don't know what that is.

A 87 trillion.

Okay, it turns out to be a prime number from that.

It did all of his algebra, and it found the corresponding values of, uh, X, y and Z.

And here they are.

The computer found the solution at 905 which is just after I drop my kids off at school.

And as I was walking to the university and I noticed it around 9 30 when I got to my office.

No, I didn't rig it to send me an email, so I have to check it.

Makes it more fun because there's an element of, uh, you know, uncertainty in the thing.

How often were you checking?

We'd already check in, like, every hour.

Or do you check once a day one today or something like that.

A soon as I saw the file was there, I knew that it had a solution.

Of course.

Then I checked it.

You never 100% sure that, you know, you haven't made a mistake somewhere.

So I put these numbers into the computer cube them and added them up in about 33.

What did that feel?

Like?

I said, a jump for joy.

Yeah, literally D s side at this.

You made an excellent point in the first video on this, which is that we can't hope to solve this, but just computing, because there's always another number for us.

The next one is 42 42.

The new The new white whale is 42.

Is the new 33?

Yeah, that's right.

Um and so if we just look at the numbers up to 100 so let me say K less than or equal to 100.

There's 20 do numbers in that range that give a remainder of four or five when you divide by nine.

So they're so they're not possible.

That's right.

We know the answer for those on there.

Not possible that leaves 78 numbers to account for the very first computer investigation of this was done way back in 1954.

That's pretty early as these things go, and they found 69 solutions for 69 out of the 78.

Well, of course, that's not the end of the story.

So the computer technology advanced rapidly from there, and within a decade another group of mathematicians rented on a bigger computer.

But they were disappointed because they found only one more, despite going about 20 times further in terms of the variables.

But remember, they were using a quadratic time algorithm, so it's about 400 times as much work.

So it's impressive from an engineering standpoint.

Yes, they found only one more solution.

So they took this as evidence that there probably aren't any more the hedge, their bets a little bit.

They didn't say no more solutions.

But they said that not all of the remaining eight candidates would have solutions.

That was the prevailing wisdom for awhile for over 20 years.

Now we think that conjecture is wrong.

Okay, um, things have kind of turned around, but it wasn't until the early nineties that Heath Brown said, Well, hang on eso.

If there are solutions, they're likely to be extremely sparse.

In fact, he does some calculations and he shows that for some of these small numbers, you would expect the numbers from one solution to the next to increase by a factor of millions.

Okay, so that means, you know, each time you find one, you're gonna have to work a 1,000,000 times harder, um, to find the next one.

So the fact that you that you've expanded the search range by 20 and not found anything is not by itself very good evidence that there's nothing more to be found.

And so people looked again.

From 1991 to this year, 19 we found solutions for seven Out of the eight, there's only one number left.

43 is the only one left below 100.

So we have not completely disproved their conjecture from 63.

But it's looking pretty good for Heath Brown's counter conjecture.

Andrew, you got the computer looking for 42 now.

I already looked for 42 in the same range and and nothing turned up.

But yeah, maybe we shouldn't stop there, so to find 42.

You gotta go bigger.

That's right.

Yeah.

There's a definite purpose here because this is a problem where we're not even sure or it's not even obvious what the right conjecture should be.

Okay.

And that's evidenced by the fact that the conjectures changed in our lifetimes.

Does that mean, you know, a mathematician like you instead of just picking them off one by one with the computer?

Yeah.

Should you be sitting down and trying to come up with a rigorous proof?

I mean, what's the point of just picking these off one at a time?