数学中最古老的未解之谜:奇完全数之问 veritasium 2024-03-08

数学中最古老的未解之谜

这是一个关于数学中最古老的未解之谜的视频,它可追溯到两千年前。

View/Hide Original English

This is a video about the oldest unsolved problem in math that dates back 2000 years.

历史上一些最杰出的数学家曾试图破解它,但都失败了。
View/Hide Original English

Some of the brightest mathematicians of all time have tried to crack it, but all of them failed.

在2000年,意大利数学家皮耶尔乔治·奥迪弗雷迪(Piergiorgio Odifreddi)将其列为当时四个最紧迫的开放问题之一。
View/Hide Original English

In the year 2000 the Italian mathematician, Piergiorgio Odifreddi, listed it among four of the most pressing open problems at the time.

解决这个问题可能就像找到一个数字一样简单。
View/Hide Original English

Solving this problem could be as simple as finding a single number.

因此,数学家们使用计算机检查了高达10的2200次方的数字,但到目前为止一无所获。
View/Hide Original English

So mathematicians have used computers and checked numbers up to 10 to the power of 2,200, but so far they've come up empty handed.

你认为为什么这个问题能吸引如此多数学家的想象力?
View/Hide Original English

Why do you think this problem has captured the imaginations of so many mathematicians?

它古老、简单、优美。
View/Hide Original English

It's old, it's simple, it's beautiful.

你还能要求什么呢?
View/Hide Original English

What else could you want?

所以问题是:是否存在任何**奇完全数**(Odd Perfect Numbers: 指除了自身以外的所有约数之和等于它本身的奇数)?
View/Hide Original English

So the problem is this. Do any odd perfect numbers exist?

什么是完全数?

那么,什么是完全数(Perfect Number: 指其所有真因子(即除了自身以外的约数)之和等于它本身的自然数)呢?

View/Hide Original English

So what is a perfect number?

以数字6为例。
View/Hide Original English

Well take the number six for example.

你可以用1、2、3和6来除它,但我们忽略6,因为那是数字本身。
View/Hide Original English

You can divide it by 1, 2, 3, and 6, but let's ignore 6 because that's the number itself,

现在我们只剩下**真因子**(Proper Divisors: 指除了自身以外的约数)。
View/Hide Original English

and now we're left with just the proper divisors.

如果你把它们都加起来,你会发现它们的和是6,也就是数字本身。
View/Hide Original English

If you add them all up, you find that they add to six, which is the number itself.

所以像这样的数字被称为完全数。
View/Hide Original English

So numbers like this are called perfect.

你也可以用其他数字来尝试,比如10。
View/Hide Original English

You can also try this with other numbers like 10.

10的真因子是1、2和5。
View/Hide Original English

10 has the proper divisors one, two, and five.

如果你把它们加起来,你只会得到8。
View/Hide Original English

If you add those up, you only get eight.

所以10不是一个完全数。
View/Hide Original English

So 10 is not a perfect number.

现在你可以对所有其他数字重复这个过程,你会发现大多数数字在1到100之间要么过高要么过低,只有6和28是完全数。
View/Hide Original English

Now you can repeat this for all other numbers, and what you find is that most numbers either overshoot or undershoot between 1 and a 100, only 6 and 28 are perfect numbers.

如果你检查到10,000,你会发现接下来的两个完全数是496和8,128。
View/Hide Original English

Go up to 10,000 and you find the next two perfect numbers 496 and 8,128.

这些是古希腊人所知的唯一完全数,并且在一千多年里它们都是唯一已知的。
View/Hide Original English

These were the only perfect numbers known by the ancient Greeks, and they would be the only known ones for over a thousand years.

如果我们能找到一个生成这些数字的模式,那么我们就可以用它来预测更多的完全数。
View/Hide Original English

If only we could find a pattern that makes these numbers, then we could use that to predict more of them.

那么这些数字有什么共同之处呢?
View/Hide Original English

So what do these numbers have in common?

一个值得注意的特点是,每一个下一个完全数都比前一个多一位数字。
View/Hide Original English

Well, one thing to notice is that each next perfect number is one digit longer than the number that came before it.

它们共享的另一个特点是,末位数字在6和8之间交替,这也意味着它们都是偶数。
View/Hide Original English

Another thing they share is that the ending digit alternates between 6 and 8, which also means they are all even.

但真正奇怪的地方在这里。
View/Hide Original English

But here's where things get really weird.

你可以将6写成1加2加3的和,将28写成1加2加3加4加5加6加7的和,其他数字也一样,它们都是连续数字的和。
View/Hide Original English

You can write 6 as the sum of 1 plus 2 plus 3 and 28 as the sum of one, plus 2, plus 3, plus 4 plus 5 plus 6 plus 7, and so on for the others as well, they're all just the sum of consecutive numbers

你可以把每个增加的数字看作是增加了一个新层。
View/Hide Original English

and you can think of each additional number as adding a new layer.

因此,这些数字形成了一个三角形,这就是为什么它们被称为**三角形数**(Triangular Numbers: 可以排成等边三角形的数字,是连续自然数的和)。
View/Hide Original English

And so these create a triangle, which is why these numbers are called triangular numbers.

此外,除了6以外的每个数字都是连续奇数立方和。
View/Hide Original English

Also, every number except for six is the sum of consecutive odd cubes.

所以28是1的立方加3的立方。
View/Hide Original English

So 28 is 1 cubed plus 3 cubed.

496等于1的立方加3的立方加5的立方加7的立方。
View/Hide Original English

496 is equal to 1 cubed plus 3 cubed plus 5 cubed plus 7 cubed.

而8,128等于1的立方加3的立方加5的立方加7的立方加9的立方一直到15的立方。
View/Hide Original English

And 8,128 is equal to 1 cubed plus 3 cubed plus 5 cubed plus 7 cubed plus 9 cubed all the way up to 15 cubed.

但真正让我震惊的是这个。
View/Hide Original English

But here's the one that really blows my mind.

如果你把这些数字写成二进制,6变成110。
View/Hide Original English

If you write these numbers in binary, six becomes 110,

28变成11100。
View/Hide Original English

and 28 becomes 11100.

496变成111110000。
View/Hide Original English

496 becomes 111110000.

而8,128,你猜对了。
View/Hide Original English

And 8,128, you guessed it.

它也是一串1后面跟着一串0。
View/Hide Original English

It is also a string of ones followed by a series of zeros.

所以如果你把它们写出来,它们都只是连续的2的幂。
View/Hide Original English

So if you write them out, they are all just consecutive powers of two.

欧几里得的偶完全数公式

大约在公元前300年,欧几里得(Euclid)在发现这些完全数生成模式时,实际上也在思考类似的问题。

View/Hide Original English

What now around 300 BC Euclid was actually thinking along similar lines when he discovered the pattern that makes these perfect numbers.

取数字1并将其加倍,你得到2,现在继续加倍。
View/Hide Original English

Take the number one and double it, you get two now, keep doubling it.

你得到4、8、16、32、64等等。
View/Hide Original English

You get 4, 8, 16, 32, 64, and so on.

现在从1开始,将下一个数字加到它上面。
View/Hide Original English

Now starting from one, add the next number to it.

所以1加2等于3。
View/Hide Original English

So 1 plus 2 equals 3.

如果这个和是一个**素数**(Prime: 只能被1和它本身整除的正整数),那么你把它乘以序列中的最后一个数字,就得到一个完全数。
View/Hide Original English

If that adds up to a prime, then you multiply it by the last number in the sequence to get a perfect number.

所以2乘以3等于6,第一个完全数。
View/Hide Original English

So two times three equals six, the first perfect number.

现在我们继续这样做。
View/Hide Original English

Now let's keep doing this.

加1加2加4,你得到7,这又是一个素数。
View/Hide Original English

Add 1 plus 2 plus 4, and you get 7, which is again prime.

所以把它乘以最后一个数字4,你得到28。
View/Hide Original English

So multiply it by the last number four, and you get 28.

下一个完全数。
View/Hide Original English

The next perfect number.

接下来,加1加2加4加8等于15,但15不是素数,所以我们继续加16得到31。
View/Hide Original English

Next, add 1 plus 2 plus 4 plus 8 equals 15, but 15 isn't prime, so we continue add 16 to get 31,

这是素数。
View/Hide Original English

this is prime.

所以你把它乘以16,得到496。
View/Hide Original English

So you multiply it by 16 and you get 496.

第三个完全数。
View/Hide Original English

The third perfect number.

现在你可以继续这样做,找到越来越大的完全数,利用这个方法我们可以重写前三个。
View/Hide Original English

Now you can keep doing this to find bigger and bigger perfect numbers, and using this we can rewrite the first three.

所以6等于(1加2)乘以2的1次方。
View/Hide Original English

So 6 equals 1 plus 2 times 2 to the power of 1

28等于(1加2加4)乘以2的2次方。
View/Hide Original English

and 28 equals 1 plus 2 plus 4 times 2 squared

496等于(1加2加4加8加16)乘以2的4次方,其中第一项是素数。
View/Hide Original English

and 496 equals 1 plus 2 plus 4 plus 8 plus 16 times 2 to the power of 4 where the first term is prime.

但还有一种更方便的写法。
View/Hide Original English

But there's a more convenient way to write this still.

取任何连续的2的幂的和。
View/Hide Original English

Take any sum of consecutive powers of 2.

所以2的0次方(即1)加2的1次方加2的2次方,一直到2的n减1次方。
View/Hide Original English

So 2 to the power of zero which is 1 plus 2 to the 1 plus 2 to the 2, all the way up to 2 to the n minus 1.

现在因为你不知道n,你不知道它等于什么,但它会等于某个值。
View/Hide Original English

And now because you don't know n, you don't know what that is equal to, but it will be equal to something.

所以我们称之为T。
View/Hide Original English

So let's call that T.

现在将整个方程乘以2。
View/Hide Original English

Now multiply this whole equation by two.

所以你得到2的1次方加2的2次方,一直到2的n次方,这等于2T。
View/Hide Original English

So you get 2 to the 1 plus 2 to the 2, all the way up to 2 to the n, and this is equal to 2T.

如果你现在从第二个方程中减去第一个方程,几乎所有的项都会抵消,你剩下T等于2的n次方减1。
View/Hide Original English

If you now subtract the first equation from the second, almost all the terms will cancel out and you're left with T equals 2 to the n minus 1.

所以你可以用下一个2的幂减1来替换整个数列。
View/Hide Original English

So you can replace this whole series with one less than the next power of 2.

所以6变成(2的平方减1)乘以2的1次方。
View/Hide Original English

So six becomes 2 squared minus 1 times 2 to the 1.

28变成(2的立方减1)乘以2的平方。
View/Hide Original English

28 becomes 2 cubed minus 1 times 2 squared,

496变成(2的5次方减1)乘以2的4次方。
View/Hide Original English

and 496 becomes 2 to the 5 minus 1 times 2 to the 4.

你看到规律了吗?
View/Hide Original English

Do you see the pattern?

这个数字总是比那个数字大1。
View/Hide Original English

This number is always one more than this.

所以如果我们称之为P,那么欧几里得给出完全数的公式是(2的P次方减1)乘以2的P减1次方,只要前者是素数。
View/Hide Original English

So if we call this P, then Euclid formula that gives a perfect number is 2 to the P minus 1 times 2 to the P minus 1 whenever this is prime.

现在,因为你把它乘以2的P减1次方,这是一个偶数,所以这总是会得到一个偶数。
View/Hide Original English

Now, because you're multiplying it by 2 to the P minus 1, which is even, this will always give an even number.

欧几里得找到了一种生成偶完全数的方法,但他没有证明这是唯一的方法。
View/Hide Original English

Euclid had found a way to generate even perfect numbers, but he didn't prove that this was the only way.

所以可能还有其他方法可以得到完全数,包括潜在的奇完全数。
View/Hide Original English

So there could be other ways to get perfect numbers, including potentially ones that are odd.

尼科马库斯的猜想与反驳

400年后,希腊哲学家尼科马库斯(Nicomachus)出版了《算术入门》(Introdutio Arithmetica),这是接下来一千年里的标准算术教材。

View/Hide Original English

400 years later, the Greek philosopher nicomchaus published Introdutio Arithmetica, the standard arithmetic text for the next thousand years.

在书中,他提出了五个猜想,他认为这些陈述是正确的,但没有费心去证明。
View/Hide Original English

In it, he stated five conjectures statements he believed to be true, but did not bother actually trying to prove.

他的猜想是:一、第n个完全数有n位数字。
View/Hide Original English

His conjectures were one, the nth perfect number has n digits.

二、所有完全数都是偶数。
View/Hide Original English

Two, all perfect numbers are even.

三、所有完全数都交替以6和8结尾。
View/Hide Original English

Three, all perfect numbers end in 6 and 8 alternately.

四、欧几里得的算法产生每一个偶完全数。
View/Hide Original English

Four, Euclid algorithm produces every even perfect number.

五、存在无限多个完全数。
View/Hide Original English

And five, there are infinitely many perfect numbers.

在接下来的一千年里,没有人能证明或反驳这些猜想,它们被认为是事实。
View/Hide Original English

For the next thousand years no one could prove or disprove any of these conjectures, and they were considered facts.

但在13世纪,埃及数学家伊本·法卢斯(Ibn Fallus)发表了一份包含10个完全数及其P值的列表。
View/Hide Original English

But in the 13th century, Egyptian mathematician Ibn Fallus published a list with 10 perfect numbers and their values of P.

其中三个完全数结果根本不是完全数。
View/Hide Original English

Three of these perfect numbers turned out not to be perfect at all.

但其余的是。
View/Hide Original English

But the remaining ones are.

第五个完全数有八位数字长,这反驳了尼科马库斯的第一个猜想。
View/Hide Original English

The fifth perfect number is eight digits long, which disproves Nicomachus's first conjecture.

接下来要注意的是,第五个和第六个完全数都以6结尾。
View/Hide Original English

And the next thing to notice is that both the fifth and sixth perfect number end in a 6.

这反驳了尼科马库斯的第三个猜想,即所有完全数都交替以6或8结尾。
View/Hide Original English

So that disproves Nicomachus's third conjecture that all perfect numbers end in a 6 or 8 alternately.

两个猜想被证明是错误的。
View/Hide Original English

Two conjectures were proven false.

但其他三个呢?
View/Hide Original English

But what about the other three?

梅森素数与欧拉的突破

两个世纪后,这个问题传到了文艺复兴时期的欧洲,在那里人们重新发现了第五、第六和第七个完全数。

View/Hide Original English

Two centuries later, the problem reached Renaissance Europe where they rediscovered the fifth, sixth, and seventh perfect numbers.

到目前为止,每一个完全数都具有欧几里得的形式。
View/Hide Original English

So far every perfect number had Euclid's form.

寻找新完全数的最佳方法是找到使2的P次方减1为素数的P值。
View/Hide Original English

And the best way to find new ones was by finding the values of P that make 2 to the P minus 1 prime.

法国博学家马林·梅森(Marin Mersenne)广泛研究了这种形式的数字。
View/Hide Original English

So French polymath Marin Mersenne extensively studied numbers of this form.

1644年,他在一本书中发表了他的研究,其中包括一个他声称对应于素数的11个P值列表。
View/Hide Original English

In 1644, he published his in a book including a list of 11 values of P for which he claimed they corresponded to primes.

满足这个条件的数字现在被称为**梅森素数**(Mersenne Primes: 形如2^p - 1的素数,其中p也是素数)。
View/Hide Original English

Numbers for which this is true are now called Mersenne Primes.

在他的列表中,前七个P的指数确实产生了素数,它们对应于前七个完全数。
View/Hide Original English

Of his list the first seven exponents of P do result in primes and they correspond to the first seven perfect numbers.

但对于一些更大的数字,比如2的67次方减1,梅森承认他甚至没有检查它们是否是素数。
View/Hide Original English

But for some of the larger numbers like 2 to the 67 minus 1, Mersenne admitted to not even checking whether they were prime.

“要判断一个15到20位的给定数字是否是素数,所有时间都不足以进行测试。”
View/Hide Original English

"To tell if a given number of 15 to 20 digits is prime or not all time would not suffice for the test."

梅森与当时的其他杰出人物讨论了完全数问题,包括皮埃尔·德·费马(Pierre de Fermat)和勒内·笛卡尔(Rene Descartes)。
View/Hide Original English

Mersenne discussed the problem of perfect numbers with other luminaries of the time, including Pierre de Fermat and Rene Descartes.

1638年,笛卡尔写信给梅森:“我想我能证明除了欧几里得的形式之外,没有其他的偶完全数。”
View/Hide Original English

In 1638, Descartes wrote to Mersenne, I think I can show that there are no even perfect numbers except those of Euclid.

他还认为,如果存在一个奇完全数,它必须具有特殊的形式。
View/Hide Original English

He also believed that if an odd perfect number does exist, it must have a special form.

它必须是一个素数与另一个不同数字的平方的乘积。
View/Hide Original English

It must be the product of a prime and the square of a different number.

如果他是对的,这将是自2000年前欧几里得以来,这个问题上最大的突破。
View/Hide Original English

If he was right, these would easily have been the biggest breakthroughs on the problem since Euclid 2000 years earlier.

但笛卡尔无法证明这两个陈述。
View/Hide Original English

But Descartes couldn't prove either of those statements.

相反,他写道:“至于我,我认为可以找到真正的奇完全数。但无论你使用什么方法,寻找这些数字都需要很长时间。”
View/Hide Original English

Instead, he wrote "As for me, I judge that one can find real odd perfect numbers. But whatever method you use, it takes a long time to look for these."

大约一百年后,在圣彼得堡科学院,普鲁士数学家克里斯蒂安·哥德巴赫(Christian Goldbach)遇到了一位20岁的数学神童。
View/Hide Original English

Around a hundred years later at the St. petersburg Academy, the Prussian mathematician Christian Goldbach met a 20-year-old math prodigy.

两人通过书信保持联系,1729年,哥德巴赫向这位年轻人介绍了费马的工作。
View/Hide Original English

The two stayed in touch corresponding by mail, and in 1729, Goldbach introduced this young man to the work of Fermat.

起初,他似乎无动于衷,但在哥德巴赫的进一步鼓励下,他对数论产生了热情。
View/Hide Original English

At first, he seemed indifferent, but after a little more prodding by Goldbach he became passionate about number theory

他花了接下来的40年时间研究该领域的不同问题,其中包括完全数问题。
View/Hide Original English

and he spent the next 40 years working on different problems in the field among them was the problem of perfect numbers.

这位神童的名字是莱昂哈德·欧拉(Leonhard Euler)。
View/Hide Original English

This Prodigy's name was Leonhard Euler.

欧拉从笛卡尔停下的地方继续,但取得了更大的成功。
View/Hide Original English

Euler picked up where Descartes had left off, but with more success.

在此过程中,他在这个问题上取得了三项突破。
View/Hide Original English

In doing so, he made three breakthroughs on this problem.

首先在1732年,他发现了第八个完全数,他是通过验证2的31次方减1是素数来完成的。
View/Hide Original English

First in 1732, he discovered the eighth perfect number, which he had done by verifying that 2 to the 31 minus 1 is prime.

正如梅森所预测的那样。
View/Hide Original English

Just as Mersenne had predicted.

为了他的另外两项突破,他发明了一种新武器:**Sigma函数**(Sigma Function: 一个数的所有约数(包括自身)之和)。
View/Hide Original English

For his other two breakthroughs, he invented a new weapon, the sigma function.

这个函数所做的就是它取一个数字的所有约数,包括数字本身,并将它们加起来。
View/Hide Original English

All this function does is it takes all the divisors of a number, including the number itself and adds them up.

所以取任何一个数字,比如6,将其所有约数加起来,你得到12,这是我们开始的数字的两倍。
View/Hide Original English

So take any number, say six, sum up all its divisors and you get 12, which is twice the number we started with.

这对于所有完全数都成立。
View/Hide Original English

And this will be true for all perfect numbers.

一个完全数的Sigma函数总是给出数字本身的两倍,因为Sigma函数将数字本身包含在其约数中。
View/Hide Original English

The Sigma function of a perfect number will always give twice the number itself because the sigma function includes the number as one of its divisors.

现在这可能看起来是一个微小的变化,但它最终变得极其强大。
View/Hide Original English

Now this may seem like a small change, but it ends up being extremely powerful.

所以让我们看几个例子。
View/Hide Original English

So let's look at a few examples.

取一个素数,比如7。
View/Hide Original English

Take a prime number like seven.

现在,因为它是素数,你不能把它重新排列成一个矩形,因此唯一的约数是1和素数本身。
View/Hide Original English

Now, because it's prime, you can't rearrange it into a rectangle, therefore the only divisors are one and the prime itself.

所以Sigma 7是1加7,等于8。
View/Hide Original English

So Sigma seven is 1 plus 7, which is equal to 8.

现在,为了更容易理解,我们只关注数字。
View/Hide Original English

Now, to keep things easier to follow, we'll just stick to the numbers.

但如果不是7,而是7的立方呢?
View/Hide Original English

But what if instead of seven, you had seven cubed?

同样,约数之和非常简单。
View/Hide Original English

Well, again, the sum of the divisors is really simple.

它只是1加7加7的平方加7的立方。
View/Hide Original English

It's just 1 plus 7 plus 7 squared plus 7 cubed.

现在让我们在一个不同的数字上使用它,比如20。
View/Hide Original English

Now let's use it on a different number, say 20.

它的约数之和是1加2加4加5加10加20,等于42。
View/Hide Original English

The sum of its divisors is 1 plus 2 plus 4 plus 5 plus 10 plus 20, which equals 42.

但你也可以把它写成(1加2加4)乘以(1加5)。
View/Hide Original English

But you can also write this as 1 plus 2 plus 4 times 1 plus 5.

这就是Sigma函数如此强大的原因。
View/Hide Original English

And this is what really makes the sigma function so powerful.

如果你有一个由不共享因子的其他数字组成的数字,那么你可以将Sigma函数分解为组成它的素数幂的Sigma函数。
View/Hide Original English

If you have a number that is made up of other numbers that don't share factors with each other, then you can split up the sigma function into the sigma functions of the prime powers that make it up.

所以Sigma(2的平方)乘以Sigma 5等于Sigma 20。
View/Hide Original English

So sigma of 2 squared times sigma 5 is equal to sigma 20.

既然任何数字都可以写成素数幂的乘积,你就可以将任何**合数**(Composite Number: 除了1和它本身以外还能被其他正整数整除的自然数)的Sigma函数分解为其素数幂的Sigma函数。
View/Hide Original English

And since any number can be written as the product of prime powers, you can split up the sigma function of any composite number into the sigma functions of its prime powers.

欧拉手握他的新函数,取得了他的第二项突破,并完成了笛卡尔未能完成的任务。
View/Hide Original English

With his new function in hand, Euler achieved his second breakthrough and did what Descartes couldn't.

他证明了每一个偶完全数都具有欧几里得的形式。
View/Hide Original English

He proved that every even perfect number has Euclid's form.

这个**欧几里得-欧拉定理**(Euclid-Euler Theorem: 证明了所有偶完全数都具有欧几里得发现的特定形式)解决了一个1600年的问题,并证明了尼科马库斯的第四个猜想。
View/Hide Original English

This Euclid-Euler theorem solved a 1600-year-old problem and proved Nicomachus's fourth conjecture.

数学史学家威廉·邓纳姆(William Dunham)称之为历史上最伟大的数学合作。
View/Hide Original English

Math historian William Dunham called it the greatest mathematical collaboration in history.

但欧拉还没有结束。
View/Hide Original English

But Euler wasn't finished yet.

他还想解决奇完全数的问题。
View/Hide Original English

He also wanted to solve the problem of odd perfect numbers.

所以为了他的第三项突破,他着手证明笛卡尔的另一个陈述,即每一个奇完全数都必须具有特定的形式。
View/Hide Original English

So for his third breakthrough, he set out to prove Descartes other statement that every odd perfect number must have a specific form.

因为如果一个奇完全数确实存在,你知道两件事:首先,n是奇数。
View/Hide Original English

Because if an odd perfect number does exist, you know two things first n is odd.

其次,Sigma n等于2n。
View/Hide Original English

And second sigma of n equals 2n.

现在任何数字n,你都可以写成不同素数的乘积,每个素数都可以是某个幂次。
View/Hide Original English

Now any number n, you can write as a product of different prime numbers and each prime can be to some power.

所以我们把它代入欧拉的Sigma函数。
View/Hide Original English

So let's take that and put it into Euler sigma function.

所以你得到Sigma n等于所有这些素数幂的Sigma,它等于2n。
View/Hide Original English

So you get sigma of n equals sigma of all of those primes to their powers, which equals 2n.

但由于所有这些因子都是素数,你实际上可以将Sigma函数分解为各个素数幂的Sigma。
View/Hide Original English

But since all of these factors are primes, you can actually split up the sigma function into the sigmas of the individual prime powers.

一个值得注意的特点是,如果一个素数被提高到奇数次幂,例如7的1次方,那么Sigma函数将是偶数。
View/Hide Original English

Now one thing to notice is that if you have a prime number raised to an odd power, for example seven to the power of 1, then the sigma function will be even

因为1加7等于8,你总是会得到一个偶数,因为奇数加奇数是偶数。
View/Hide Original English

because 1 plus 7 equals 8, you'll always get an even number because odd plus odd is even

如果素数被提高到偶数次幂,比如7的平方,那么Sigma函数返回一个奇数。
View/Hide Original English

if the prime number is instead raised to an even power like seven squared, then the sigma function returns an odd number.

Sigma(7的平方)等于1加7加7的平方,等于57。
View/Hide Original English

Sigma of 7 squared equals 1 plus 7 plus 7 squared, which equals 57.

因为奇数加奇数加奇数等于奇数。
View/Hide Original English

Because odd plus odd plus odd equals odd.

所以如果一个奇素数被提高到奇数次幂的Sigma函数,它将给出一个偶数。
View/Hide Original English

So if you have the sigma function of an odd prime raised to an odd power, it will give an even number.

如果它被提高到偶数次幂,你得到一个奇数。
View/Hide Original English

If instead it's raised to an even power, you get an odd number.

这就是欧拉天才洞察力发挥作用的地方,因为在右侧你有2乘以n,其中n是一个奇完全数,而2是偶数。
View/Hide Original English

And this is where Euler's genius insight comes in because here on the right side you've got 2 times n where n is an odd perfect number, and 2 is even.

这意味着在左侧必须只有一个偶数,因为如果有两个偶数,你可以分解出4。
View/Hide Original English

Well, what that means is that on the left side there must only be one even number because if there were two even numbers, you could factor out four.

但这表示你也应该能够在右侧分解出4,而你不能,因为n是奇数,这里只有一个2。
View/Hide Original English

But that means you should also be able to factor out four on the right side, which you can't because n is odd and there's only a single 2 here.

所以这里这些Sigma中只有一个能给出一个偶数,这意味着只有一个素数是奇数次幂,而所有其他的都必须是偶数次幂,正如笛卡尔所预测的那样。
View/Hide Original English

So only one of these sigmas here can give an even number, which means that there is exactly one prime that is to an odd power and all the others must be to an even power just as Descartes had predicted.

现在,欧拉进一步完善了形式,并表明一个奇完全数必须满足这个条件,但即使是欧拉也无法证明它们是否存在。
View/Hide Original English

Now, Euler refined the form a bit more and showed that an odd perfect number must satisfy this condition, but even Euler couldn't prove whether they existed or not.

他写道:“是否存在任何奇完全数是一个最困难的问题。”
View/Hide Original English

He wrote "Whether there are any odd perfect numbers is a most difficult question."

现代计算搜索与奇完全数的持续谜团

在接下来的150年里,进展甚微,没有发现新的完全数。

View/Hide Original English

For the next 150 years very little progress was made and no new perfect numbers were discovered.

英国数学家彼得·巴洛(Peter Barlow)写道,欧拉的第八个完全数“是迄今为止发现的最伟大的,因为它们仅仅是好奇的,而没有用处,所以不太可能有人会试图去寻找超越它的数字。”
View/Hide Original English

English mathematician Peter Barlow wrote that Euler eighth perfect number "Is the greatest that ever will be discovered for as they are merely curious without being useful, it is not likely that any person will ever attempt to find one beyond it."

但巴洛错了。
View/Hide Original English

But Barlow was wrong.

数学家们继续追寻这些难以捉摸的完全数,大多数人从梅森提出的素数列表开始。
View/Hide Original English

Mathematicians kept pursuing these elusive perfect numbers and most started with Mersenne's list of proposed primes.

他列表上的下一个是2的67次方减1。
View/Hide Original English

The next on his list was 2 to the 67 minus 1.

到目前为止,梅森做得非常出色。
View/Hide Original English

So far, Mersenne had done an excellent job.

他包含了欧拉的第八个完全数,同时避免了像29这样最终没有导致完全数的数字。
View/Hide Original English

He had included Euler's eighth perfect number while avoiding others like 29 that turned out not to lead to a perfect number,

但在梅森发表他的列表230年后,爱德华·卢卡斯(Edouard Lucas)证明了2的67次方减1不是素数,尽管他无法找到它的因子。
View/Hide Original English

but 230 years after Mersenne published his list, Edouard Lucas proved that 2 to the 67 minus 1 was not prime, although he was unable to find its factors.

27年后,弗兰克·尼尔森·科尔(Frank Nelson Cole)在美国数学学会(American Mathematical Society)发表了一场演讲,一言不发。
View/Hide Original English

27 years later, Frank Nelson Cole gave a talk to the American mathematical society without saying a word,

他走到黑板的一侧,写下2的67次方减1等于147,573,952,589,676,412,927。
View/Hide Original English

he walked to one side of the blackboard and wrote down 2 to the 67 minus 1 equals 147,573,952,589,676,412,927.

然后他走到黑板的另一侧,将193,707,721乘以761,838,257,287,得到了相同的答案。
View/Hide Original English

He then walked to the other side of the blackboard and multiplied 193,707,721 times 761,838,257,287 giving the same answer.

他一言不发地坐下,观众爆发出了掌声。
View/Hide Original English

He sat down without saying a word and the audience erupted in applause.

他后来承认,他花了三年时间,利用周日工作才解决了这个问题。
View/Hide Original English

He later admitted it took him three years working on Sundays to solve this.

一台现代计算机可以在不到一秒钟内解决这个问题。
View/Hide Original English

A modern computer could solve this in less than a second.

从公元前500年到1952年,人们只发现了12个梅森素数,因此也只有12个完全数。
View/Hide Original English

From 500 BC until 1952 people had discovered just 12 Mersenne primes and therefore only 12 perfect numbers.

主要困难在于检查大的梅森数是否真的是素数。
View/Hide Original English

The main difficulty was checking whether large Mersenne numbers were actually prime.

但在1952年,美国数学家拉斐尔·罗宾逊(Raphael Robinson)编写了一个计算机程序来执行这项任务。
View/Hide Original English

But in 1952, American mathematician Raphael Robinson wrote a computer program to perform this task

他在当时最快的计算机SWAC上运行了它。
View/Hide Original English

and he ran it on the fastest computer at the time, the SWAC.

在10个月内,他发现了接下来的五个梅森素数以及相应的完全数。
View/Hide Original English

Within 10 months, he found the next five Mersenne primes and so corresponding perfect numbers.

在接下来的50年里,新的梅森素数被迅速发现,所有这些都使用了计算机。
View/Hide Original English

And over the next 50 years, new Mersenne primes were discovered in rapid succession, all using computers.

1952年底最大的梅森素数是2的2281次方减1,它有687位数字长。
View/Hide Original English

The largest Mersenne prime at the end of 1952 was 2 to the power of 2,281 minus 1, which is 687 digits long.

到1994年底,最大的梅森素数是2的859,433次方减1,它有258,716位数字长。
View/Hide Original English

By the end of 1994, the largest Mersenne prime was 2 to the power of 859,433 minus 1, which is 258,716 digits long.

由于这些数字变得如此天文数字般巨大,寻找大量素数的任务变得越来越困难,即使对于超级计算机也是如此。
View/Hide Original English

Since these numbers were getting so astronomically large, the task of finding numerous end primes became more and more difficult even for supercomputers.

所以在1996年,计算机科学家乔治·沃尔特曼(George Woltman)启动了**大互联网梅森素数搜索**(GIMPS: Great Internet Mersenne Prime Search)项目。
View/Hide Original English

So in 1996, computer scientist George Woltman launched the Great Internet Mersenne Prime Search or GIMPS.

GIMPS将工作分配给许多计算机,允许任何人志愿贡献他们的计算机算力来帮助搜索梅森素数。
View/Hide Original English

GIMPS distributes the work over many computers allowing anyone to volunteer their computer power to help search for Mersenne primes.

该项目迄今为止非常成功,已经发现了17个新的梅森素数,其中15个是当时已知的最大素数。
View/Hide Original English

The project has been highly successful so far, having discovered 17 new Mersenne primes, 15 of which were the largest known primes at that time.

最棒的是,如果你的计算机发现了一个新的梅森素数,你将被列为它的发现者,将自己加入到包括一些有史以来最优秀的数学家在内的名单中。
View/Hide Original English

And the best part, if your computer discovers a new Mersenne prime, you'll be listed as its discoverer, adding yourself to a list that includes some of the best mathematicians of all time.

甚至还有一个25万美元的奖金,奖励给第一个发现十亿位数字的素数的人。
View/Hide Original English

There's even a $250,000 prize for the first billion-digit prime.

2017年,教会执事约翰·佩斯(John Pace)通过使用GIMPS发现了第50个梅森素数。
View/Hide Original English

In 2017 Church Deacon John Pace discovered the 50th Mersenne Prime by using GIMPS.

数字2的77,232,917次方减1有超过2300万位数字长,它也是当时已知的最大素数。
View/Hide Original English

The number 2 to the 77,232,917 minus 1 is more than 23 million digits long, and it was also the largest known prime at the time.

为了庆祝这一成就,日本出版商Nanairosha出版了这本书,《2017年最大素数》。
View/Hide Original English

To celebrate this achievement the Japanese publishing house, Nanairosha published this book, "The Largest Prime number of 2017."

它就是那个数字,铺满了719页辉煌的页面。
View/Hide Original English

And all it is is that number spread over 719 glorious pages.

太疯狂了。
View/Hide Original English

It's wild.

这个字体大小非常小。
View/Hide Original English

The size of this font is so tiny.

这本书迅速登上亚马逊畅销榜第一名,并在四天内售罄。
View/Hide Original English

The book quickly rose to the number one spot on Amazon and sold out in four days.

一年后,第51个梅森素数被发现。
View/Hide Original English

A year later, the 51st Mersenne Prime was discovered.

它是2的82,589,933次方减1,这个数字有24,000,860位数字。
View/Hide Original English

It's 2 to the 82,589,933 minus 1, and this number has 24,000,860 2048 digits.

但这种荒谬之处让我感到享受,就像这里面有知识,但它不是那种任何人会从书中读出来的知识。
View/Hide Original English

But there's something I enjoy about the absurdity, like there is knowledge in here, but it's not the kind of knowledge that anyone's ever gonna read out of a book.

但在某种程度上,有这样一个实物,上面有这个数字,如果有一天我们失去了所有的素数,那也很好。
View/Hide Original English

But in some way it's nice that there's this physical artifact that like has the number, if ever we lost all the prime numbers.

你知道,有人可能会找到这本书,然后说:“这是最大的一个。”
View/Hide Original English

You know, someone could find this book be like, here's the big one.

截至今天,这仍然是已知最大的素数。
View/Hide Original English

As of today, this is still the largest known prime.

而且由于这种形式的数字增长如此迅速,最大的梅森素数几乎总是已知最大的素数。
View/Hide Original English

And since numbers of this form grow so rapidly, the largest Mersenne Prime is almost always the largest known prime.

计算机在寻找新的梅森素数及其相应的完全数方面取得了令人难以置信的成功,但到目前为止我们只找到了51个。
View/Hide Original English

Computers have been incredibly successful at finding new Mersenne primes and their corresponding perfect numbers, but we've still only found 51 so far.

所以你可能会怀疑它们只有有限的数量,这意味着尼科马库斯的第五个猜想是错误的,即不存在无限多个完全数。
View/Hide Original English

So you might suspect that there are only a finite number of them, which would mean that Nicomachus's fifth conjecture would be false, that there aren't infinitely many perfect numbers,

但这可能并非如此。
View/Hide Original English

but that might not be the case.

伦斯特拉(Lenstra)、波梅兰斯(Pomerance)和瓦格斯塔夫(Wagstaff)的猜想预测了应该出现多少个梅森素数,基于P的大小。
View/Hide Original English

The Lenstra and Pomerance Wagstaff conjecture predicts how many Mersenne primes should appear based on how large P is.

现在这是实际数据,这个猜想表现得非常出色。
View/Hide Original English

Now this is the actual data the conjecture performs remarkably well.

但更重要的是,它预测存在无限多个梅森素数,因此也存在无限多个偶完全数。
View/Hide Original English

But more importantly, it predicts that there are infinitely many Mersenne primes and so infinitely many even perfect numbers.

梅森素数只是太大太稀有,以至于需要大量时间和计算机资源才能找到。
View/Hide Original English

The Mersenne primes are just so large and rare that they take a lot of time and computer resources to find.

但猜想不是证明。
View/Hide Original English

But a conjecture is not a proof.

直到今天,这个问题仍然与另一个开放问题共享“数学中最古老的未解之谜”的称号:是否存在任何奇完全数?
View/Hide Original English

And up until this day, this problem shares the title of oldest unsolved problem in math with the other open problem. Do any odd perfect numbers exist?

解决这个问题的最简单方法是找到一个例子。
View/Hide Original English

The easiest way to solve this problem is by finding an example.

所以也许我们可以检查不同的奇数,看看其中是否有完全数。
View/Hide Original English

So maybe we could just check different odd numbers and see if one of them is perfect.

这正是研究人员在1991年尝试的。
View/Hide Original English

That's exactly what researchers tried in 1991.

通过使用一种称为因子链的智能算法,他们能够证明如果奇完全数确实存在,它必须大于10的300次方。
View/Hide Original English

By using a smart algorithm called a factor chain, they were able to show that if an odd perfect number does exist, it must be larger than 10 to the power of 300.

21年后,帕斯卡尔·奥切姆(Pascal Ochem)和迈克尔·拉奥(Michael Rao)将这个下限提高到10的1500次方。
View/Hide Original English

21 years later, Pascal Ochem and Michael Rao raised that lower bound to 10 to the 1,500

最近的进展将这个数字推高到10的2200次方。
View/Hide Original English

with recent progress pushing that number up to 10 to the 2,200.

对于如此大的数字,计算机不太可能很快找到一个。
View/Hide Original English

With numbers that large, it's unlikely that a computer will find one anytime soon.

所以我们需要变得更聪明。
View/Hide Original English

So we'll need to get smart.

一个证明会是什么样子?
View/Hide Original English

What would a proof look like?

我们如何才能真正证明这一点?
View/Hide Original English

Like how could we actually prove this?

我认为人们一直在尝试解决这个问题的主要思路是提出越来越多的奇完全数必须满足的条件,这被称为“条件之网”,我们现在知道它必须有10个素因子,也许还有数千个非独特素因子,并且必须大于10的3000次方。
View/Hide Original English

I think the main idea that people have been trying to approach this problem with is coming up with more and more conditions odd perfect numbers have to satisfy, it's called this web of conditions where it has to have 10 prime factors now that we know and maybe thousands of non distinct prime factors and has to be bigger than 10 to the 3000.

它必须做所有这些不同的事情,我们希望最终条件如此之多,以至于它们不能存在。
View/Hide Original English

And it has to do all these different things and we hope that eventually there's just so many conditions that can strain the numbers so much that they can't exist.

自欧拉以来,数学家们一直在为这个网络添加新的条件。
View/Hide Original English

Since Euler, mathematicians have kept adding new conditions to this web.

但到目前为止,它还没有奏效。
View/Hide Original English

But so far it hasn't worked.

但可能还有另一条路。
View/Hide Original English

But there might be another path.

当笛卡尔寻找奇完全数时,他发现了198,585,576,189,你可以将其分解为3的平方乘以7的平方乘以11的平方乘以13的平方乘以22021。
View/Hide Original English

When Descartes was looking for odd perfect numbers, he came across 198,585,576,189, which you can factor as 3 squared times 7 squared times 11 squared times 13 squared times 22021.

将此代入欧拉Sigma函数,你会发现它等于原始数字的两倍。
View/Hide Original English

Put this into Euler sigma function and you find it is equal to two times the original number.

换句话说,它是完全数。
View/Hide Original English

In other words, it is perfect.

前提是22021是素数,但它不是,因为它等于19的平方乘以61。
View/Hide Original English

That is if 22021 were prime, but it's not because it is equal to 19 squared times 61.

填入这些因子表明它不是完全数。
View/Hide Original English

And filling that in shows that it is not perfect.

像这样非常接近奇完全数的数字被称为**伪完全数**(Spoofs: 那些非常接近奇完全数,满足奇完全数某些性质但本身不是奇完全数的数字)。
View/Hide Original English

Numbers like this that are very close to being odd perfect numbers are called spoofs.

伪完全数是一个更大的数字群。
View/Hide Original English

Spoofs are a larger group of numbers.

所以奇完全数共享伪完全数的所有属性,然后还有一些额外的属性。
View/Hide Original English

So odd perfect numbers share all properties of spoofs and then a few extra ones.

目标是找到伪完全数的属性,这些属性最终阻止它们成为奇完全数。
View/Hide Original English

And the goal is to find properties of spoofs that ultimately prevent them from being odd perfect numbers.

例如,奇完全数的一个条件是它们不能被105整除。
View/Hide Original English

For example, one condition of odd perfect numbers is that they can't be divided by 105.

所以如果你发现伪完全数必须能被105整除,那么这将证明奇完全数不能存在。
View/Hide Original English

So if you find that spoofs must be divisible by 105, then this would prove that odd perfect numbers can't exist.

2022年,佩斯·尼尔森(Pace Nielsen)和杨百翰大学(BYU)的一个团队发现了21个伪完全数,包括笛卡尔的数字。
View/Hide Original English

In 2022 Pace Nielsen and a team at BYU found 21 spoof numbers including Descartes number,

虽然他们发现了一些伪完全数的新属性,但他们没有找到任何能排除奇完全数的属性。
View/Hide Original English

and while they discovered some new properties of spoofs, they didn't find any that rule out odd, perfect numbers.

那么一个奇完全数需要有多大呢?
View/Hide Original English

So how large would an odd perfect number have to be?

它们不存在。
View/Hide Original English

They don't exist.

你不认为奇完全数存在?
View/Hide Original English

You don't think odd perfect numbers exist?

不,它们不存在。
View/Hide Original English

No, they don't exist.

我希望它们存在。
View/Hide Original English

I wish they did.

如果宇宙中真的存在这样一个巨大的奇完全数,那会非常酷。
View/Hide Original English

That'd be really cool if if there was just this one gigantic odd, perfect number out in the universe.

它们不存在。
View/Hide Original English

They don't exist.

不。
View/Hide Original English

No.

你为什么确信它们不存在?
View/Hide Original English

How are you convinced that they don't exist?

有一种**启发式论证**(Heuristic Argument: 一种基于经验或直觉的论证方法,而非严格的数学证明),它不是一个证明。
View/Hide Original English

There is something called a heuristic argument where it's not a proof.

所以如果我们有一个证明,我们就完成了。
View/Hide Original English

So if we had a proof, we'd be done.

这只是一个基于“我们认为这种类型的素数出现的频率”的论证。
View/Hide Original English

It's just an argument from, okay, we think primes occur this often of this type.

你把这些信息放在一起,然后思考:“平均而言,应该有多少个数字是完全数?”
View/Hide Original English

And you put that those pieces of information together and you think, okay, on average how many numbers should be perfect.

卡尔·波梅兰斯(Carl Pomerance)提出的这个论证预测,在10的2200次方到无穷大之间,形式为N等于pm平方的完全数不会超过10的负540次方。
View/Hide Original English

This argument, which was made by Carl Pomerance predicts that between 10 to the 2,200 and infinity, there are no more than 10 to the negative 540 perfect numbers of the form N equals pm squared

对于奇完全数,启发式论证说我们不应该期望有任何一个。
View/Hide Original English

With odd perfect numbers the heuristic says we shouldn't expect any.

我们已经搜索得足够高,现在我们认为有足够的证据表明它们不应该再存在了。
View/Hide Original English

We've searched high enough now that we think we have enough evidence they shouldn't exist anymore.

我的理解是这个启发式论证也预测不存在大的完全数,无论是偶数还是奇数。
View/Hide Original English

My understanding is this heuristic argument. It also predicts that there are no large perfect numbers even or odd.

所以……
View/Hide Original English

So...

那是真的。
View/Hide Original English

That's true.

所以有一个缺点。
View/Hide Original English

So there's a downside.

是的,有一个缺点,因为它说不应该有大的偶完全数,而我们实际上期望有无限多个。
View/Hide Original English

Yeah, there's a downside because it says there shouldn't be large, even perfect numbers and we actually expect there to be infinitely many.

所以,好吧,那么……
View/Hide Original English

And so, okay, so...

为什么我在这两种情况下都相信启发式论证,而不是只相信其中一种?
View/Hide Original English

why do I believe the heuristic in this case and not this case?

你说得对。
View/Hide Original English

You're right.

我在这方面是不是自相矛盾?
View/Hide Original English

Am I being hypocritical about that?

你可以为启发式论证添加其他方面,使其更强大。
View/Hide Original English

There are other aspects you can add on to the heuristic and make it stronger.

让我这样说吧。
View/Hide Original English

Let me put it that way.

但你说得对,它不是一个证明。
View/Hide Original English

But you're right, it's not a proof.

目前这仍然是数学中最古老的未解之谜。
View/Hide Original English

For now this is still the oldest unsolved problem in math.

欧拉说“是否存在任何奇完全数是一个最困难的问题”时是对的。
View/Hide Original English

Euler was right when he said whether there are any odd perfect numbers is a most difficult question.

纯数学研究的价值

那么这个问题有什么应用吗?

View/Hide Original English

So are there any applications of this problem?

我可以说没有。
View/Hide Original English

I can say no.

现在,许多人可能认为,如果对现实世界没有应用,那么研究它就没有意义。
View/Hide Original English

Now, many people may think that if there are no applications to the real world, then there's no point studying it.

为什么有人会关心一个古老的未解之谜呢?
View/Hide Original English

Why should anyone care about some old unsolved problem?

但我认为那是错误的方法。
View/Hide Original English

But I think that's the wrong approach.

两千多年来,数论没有真正的实际应用。
View/Hide Original English

For more than 2000 years, number theory had no real world applications.

数学家们只是追随他们的好奇心,解决他们觉得有趣的问题,一个接一个地证明结果,并建立了一个无用数学的基础。
View/Hide Original English

It was just mathematicians following their curiosity and solving problems they found interesting proving one result after another and building a foundation of useless mathematics.

但到了20世纪,我们意识到我们可以利用这个基础,并将其作为我们**密码学**(Cryptography: 研究信息加密、解密和安全传输的科学)的基础。
View/Hide Original English

But then in the 20th century, we realized that we could take this foundation and base our cryptography on it.

这就是保护从短信到政府机密的一切信息的基础。
View/Hide Original English

This is what protects everything from text messages to government secrets.

每当一群人把他们的思想投入到一个问题上时,总会产生一些好的结果。
View/Hide Original English

Whenever you have a group of people put their minds towards a problem, something good's gonna come out of it.

如果一开始它不起作用,那也没关系。
View/Hide Original English

If it's only, if it's only at the beginning, this doesn't work.

好吧,正如爱迪生所说,我学会了999种不制造灯泡的方法。
View/Hide Original English

Okay, well, as Edison said, I learned 999 ways of not making a light bulb.

最终我找到了一种好方法。
View/Hide Original English

Eventually I got a good way to do it.

数学也是如此。
View/Hide Original English

It's the same with math.

你有一个问题,你投入你的思想,其他人也一样。
View/Hide Original English

You have a problem and you throw your mind at it and others do too.

你提出新的想法,最终这个过程会产生好的结果。
View/Hide Original English

And you come up with new ideas and eventually something good comes from that process.

爱因斯坦的广义相对论建立在**非欧几何**(Non-Euclidean Geometries: 区别于欧几里得几何的几何学,其平行公设与欧氏几何不同)之上,这些几何学是作为智力好奇心而发展起来的,当时并没有预见到它们有一天会改变我们理解宇宙的方式。
View/Hide Original English

Einstein's general relativity was built on non-Euclidean geometries, geometries that were developed as intellectual curiosities without foresight of how they would one day change the way we understand the universe.

你认为现在有多少人正在研究完全数问题?
View/Hide Original English

How many people do you think are working on the problem of perfect numbers right now?

我猜目前大约有10到15人在这个领域发表论文。
View/Hide Original English

I'd guess around 10 people currently have papers in the area, 10 to 15.

如果你是一名高中生,你热爱数学,并且你认为“我想要一个可以思考的问题”,那么这是一个很好的思考问题。
View/Hide Original English

If you're a high schooler and you just love mathematics and you think, I want a problem to think about, this one's a great problem to think about.

你可以取得进展。
View/Hide Original English

And you can make progress.

你可以发现新事物。
View/Hide Original English

You can figure out new things.

是的,不要害怕。
View/Hide Original English

Yeah, don't be scared.

数百年来,成千上万的人思考过这个问题。
View/Hide Original English

Hundreds of people have thought about this problem for thousands of years.

我能做什么?
View/Hide Original English

What can I do?

你能做一些事情。
View/Hide Original English

You can do something.

如果你不知道它会走向何方,为什么还要学习数学呢?
View/Hide Original English

Why should you do math if you don't know that it will lead anywhere?

嗯,因为做数学是唯一确定的方法。
View/Hide Original English

Well, because doing the math is the only way to know for sure.

你无法提前知道结果会是什么。
View/Hide Original English

You can't tell in advance what the outcome will be.

就像这个问题可能最终一无是处。
View/Hide Original English

Like this problem might turn out to be a dud.

我们可能解决了它,但它对任何人都没有意义,或者它可能变得非常有帮助。
View/Hide Original English

We might solve it and it might not mean anything to anyone, or it could turn out to be remarkably helpful.

唯一确定的方法就是尝试。
View/Hide Original English

The only way to know for sure is to try

在当今世界,人们常常觉得你必须在追随好奇心和培养你可以应用的实际技能之间做出选择。
View/Hide Original English

In today's world, it often feels like you've gotta choose between following your curiosity and building real skills you can apply.

但事实是两者都至关重要。
View/Hide Original English

But the truth is it's essential to do both.

幸运的是,有一个学习平台可以让你做到这一点。
View/Hide Original English

Fortunately, there's one learning platform which allows you to do just that.

那就是本视频的赞助商Brilliant。
View/Hide Original English

And it's this video sponsor, Brilliant.

Brilliant将通过帮助你培养从数学和数据科学到编程技术等各个领域的实际技能,让你成为一个更好的思考者和问题解决者。
View/Hide Original English

Brilliant will make you a better thinker and problem solver by helping you build real skills and everything from math and data science to programming technology.

你可以说出任何领域,Brilliant都有数千节课程供你探索,这样你就可以追随你的好奇心,无论它把你带到哪里。
View/Hide Original English

You name it, Brilliant has thousands of lessons for you to explore so you can follow your curiosity wherever it leads you,

你不仅会学习关键概念。
View/Hide Original English

and you won't just learn key concepts.

你还会将这些知识应用到现实世界中,给你真正的实践直觉。
View/Hide Original English

You'll actually apply that knowledge to real world situations, giving you real hands-on intuition.

通过Brilliant,你可以通过亲自动手尝试来学习。
View/Hide Original English

With Brilliant you can learn by trying things out for yourself.

这就是Brilliant学习如此强大的原因。
View/Hide Original English

That's what makes learning with brilliant just so powerful.

在此过程中,你可以设定目标,跟踪你的进度,并通过引导式学习路径提升自己,让你深入研究特定主题。
View/Hide Original English

And along the way, you can set goals, track your progress, and level up with guided learning paths that let you go deep into specific topics.

我认为每天学习新事物是你能给自己最好的礼物之一。
View/Hide Original English

I think that learning something new every day is one of the best gifts you can give yourself.

Brilliant是做到这一点的完美方式,它提供你可以直接在手机上完成的短小课程。
View/Hide Original English

And Brilliant is the perfect way to do it with bite-sized lessons that you can do right from your phone.

所以即使你只有几分钟时间,你也可以满足你的好奇心,磨砺你的思维,并培养新技能。
View/Hide Original English

So even if you only have a few minutes, you can feed your curiosity, sharpen your mind, and build new skills.

要免费试用Brilliant提供的所有内容,为期30天,请访问brilliant.org/veritasium,扫描此二维码,或点击描述中的链接。
View/Hide Original English

To try everything Brilliant has to offer for free, for a full 30 days, visit brilliant.org/veritasium, scan this QR code or click the link in the description

前200名用户将获得Brilliant年度高级订阅的八折优惠。
View/Hide Original English

and the first 200 of you will get 20% off Brilliant's annual premium subscription.

所以我要感谢Brilliant赞助本视频,也要感谢你的观看。
View/Hide Original English

So I wanna thank Brilliant for sponsoring this video and I want to thank you for watching.

📌 文中提及的人物和组织

关键字: canada perfect-number problem science theory