So today I’m gonna tell you about what’s called Catalan’s Conjecture.
So even though it’s called the conjecture, it’s actually known.
So this was proved by a mathematician about 15 years ago,
and it’s a question about or a statement about perfect powers,
So, what do I mean by perfect powers first of all?
也就是任取一个整数 如1 2 3等
I mean take some whole number, 1 2 3 so on,
and raise it to a power which is larger than 1, okay?
So squares cubes and so on.
So, if you start to write these things down,
1 is a perfect power,
in fact a power of itself to any exponent.
2和3都不是完全幂 而4是的 它是2的平方
2 is not, 3 is not, but then we get 4 which is 2 squared.
Then we have to go up to, I guess, 8,
which is 2 cubed.
And 9 is 3 squared,
and 16, 25,
继续这样下去 对 还有27
I gonna run into any cube soon, let’s see, oh, 27,
and 49 and so on.
所有这些数都由两个整数组成 一个底数 一个指数
So these are numbers that can be built by taking two integers and raising one to the power.
That’s exactly right, yeah.
So of course all we’ve seen so far mostly squares and cubes,
I guess 16 is also a fourth power.
But you can take the exponent to be as large as you want,
raise it as high of a power as you want,
and you can take the number to be as large as you want to.
I’m just trying to put them in order.
So do these numbers become more common or less common
the further we go down the number line.
So they become less common
渐渐地 你会发现它们更少 对不？
and generally you think they’re spread out, right?
Like just think about squares, for example.
If you hand me a big number like a million,
and you want to know about how many squares are less than a million,
well, it’s about square root of a million.
Right? Because that’s how small the number has to be,
to be less than a million after you square it.
They’re cool numbers,
they’re kind of spread out as you go along and get bigger and bigger,
but notice that in the beginning here,
some of them are close together.
So we’ve got actually a pair of consecutive powers,
so separated only by one,
I guess here we have separated by two
and three and four and so on,
so there’s some separation here.
But what Catalan conjectured
and this is a question people have been interested in for hundreds of years actually,
is this particular pair of eight and nine.
So the fact that three squared minus two cubed is equal to one,
or it really that you have any two powers which differ by one,
no other example of that was known.
And so what Catalan conjectured is that
this is the only time that this happens
that you have two powers whose difference is exactly one.
But it was unknown if like a gajillion and four, a gajillion and five.
Yeah, that’s exactly right,
that’s what we didn’t know for a long time,
whether or not this conjecture was true.
So back a few decades ago we did know that
if it wasn’t true,
there would be only finitely many exceptions.
So this tends to be easier to show in general, right? That… okay.
I know that there’s a really large number of things happening
or only finitely many times it happens.
But to show that there’s only one is what we didn’t know until recently.
这个例子很有名 对 是由数学家Mihailescu证实的
So it’s known, yeah, so this is what was proved by this mathematician Mihailescu.
In fact, these are the only consecutive perfect powers, okay?
So this notion that,
okay, they probably kind of spread out as time goes on,
is it least true if spreading out means bigger than one?
I mean the proof of this is really advanced and so,
okay, we don’t have time for.
I don’t know, the next couple of years to go through it.
But I want to tell you at least about sort of a special case of this,
and something that’s been known for a while
but it gives you an idea of kind of the type of manipulation you can do
to understand this kind of problem.
All right, so solving this equation showing that there’s only one solution,
and that’s it, it’s pretty hard.
But let me show you a special case.
let’s look at the case of
x squared minus y cubed equals one.
So for example, that has some solution we know,
because we have these consecutive powers.
So we’re really just asking about a square differing from a cube by one.
Okay, so why doesn’t this have any other solutions?
Well, here’s what you want to do,
general rule, adding and understanding factors at the same time is hard,
that’s why Fermat’s Last Theorem is hard,
that’s why this kind of question is hard.
So let’s change it to a question about multiplication,
So how do we do that?
Let’s move things to the other side,
so let’s say instead, we’re gonna solve this equation.
Now why have I just made my life better?
The reason why is because this thing breaks up into two factors,
anytime you see a difference of squares like this,
you can factorize it in this way.
And so this just changes from a very hard problem about
addition and multiplication interacting together to just multiplication.
Okay, so now why is this better?
Why do I find out whether or not this has solutions?
Here’s the idea,
the factors of y,
any number that divides Y has gotta divide one of these two things.
And so let’s ignore two for the moment,
let’s pretend that Y is odd,
so two doesn’t divide Y.
So if Y is odd that means then
that any factor of Y has to divide one of these two things,
but it can’t divide both of them at once.
And the reason why is that whatever these two numbers are,
they differ only by two.
So for the same reason, they can’t have a common factor,
other than possibly two.
So if Y is odd, all the factors of Y either divide X minus one,
or a factor of Y divides X plus one,
but not both at the same time,
which means that both of these numbers have to be cubes.
So these are both cubes,
because any number which divides Y,
we know there’s at least three copies of,
but all of those copies have to go to the same one of these two numbers.
And so we’ve got two cubes,
which are only two apart, Right?
So the difference between these two cubes is only two,
but that never happens with cubes, right?
Because we know what the cubes are,
1 8 27等等
1 8 27 and so on.
And that for sure spreads out as you go along,
and so you can’t have any cubes which are separated by two,
and so you can’t have any solution to this equation, at least if Y is odd.
What If it was even?
哦 那就有点难了 但一切皆有可能嘛
Well, then it’s a little bit harder, but not that much.
You haven’t thought that proof.
不 我有证据 不过通过这个方案找不到证据
No, so in general, of course the proof is not via this plan.
Although changing from addition to multiplication is the important part of the proof actually,
and important part of the proof.
But this is just sort of
the smallest tiniest piece the step in the right direction
towards proving this kind of thing.
Hi, everyone, thanks for watching this video.
I’m just working on brilliants problem of the week,
which triangle has the greatest area.
I can’t decide if this is a trick question
or do they have the same area?
Let me give C.
啊 它说我回答错了 不过这里有句话说得很对
Ah, it says that I got it wrong but it says here and this is true,
getting stumped is part of learning,
and now we get to continue and discuss the solutions.
Yeah, see I probably should have done better working out like that person did,
but you know we’re recording here in a hurry.
Now, brilliant.org is a fantastic problem-solving website,
it’s also the sponsor of today’s video in case you hadn’t noticed.
And the thing I like about brilliant is
小测验 课程 自创课程
it’s full of all sorts of quizzes and courses and curated lessons
that doesn’t just tell you stuff so that you’ll know it,
it’ll help you understand that it helps you go deeper,
and I think that’s a really important thing.
You can watch loads of videos and learn lots of new things,
but to really understand
that sometimes you actually have to do the problems
have someone hold your hand as you go through it,
and that’s what’s great about brilliant.
They obviously cover loads of mathematics,
they also cover science,
they also cover computer science,
there’s physics, there’s all sorts of things here.
You really should have a look.
By the way, if you enjoyed today’s video with Holly,
我会为你推荐Brilliant网站的Open Problems Group版块
I’ll tell the Open Problems Group here on brilliant
could also be a great place for you to be hanging out.
So what are you gonna have a look at that,
I’ll include a link down in the video description.
Now while I get back into these triangles and figure things out,
what you should do is go to brilliant. org/numberphile,
you can sign up to brilliant for free,
but if you use the slash numberphile I just mentioned
and there’ll be a link in the description,
you’ll also get 20% off a premium membership.
Go and have a look, and don’t just get told stuff.
There’s a difference.