And in the case of Einstein, he didn't develop Riemannian geometry, he just used them.

至于爱因坦，他并未开创黎几何，只是加以运用。

Some of you may know that this is what the famous Riemann hypothesis is all about.

你们中有些人可能知道，这正是着名的黎假设所涉及的内容。

And so I don't... I'm more interested in knowing about the Riemann hypothesis or something.

所以我对了解黎猜想之类的东西更感兴趣。

One is Riemannian, one is Erismannian.

一个是黎的，一个是埃的。

The techniques in his case get a lot more sophisticated, after  all, Riemann was a pioneer in complex analysis.

在他的情况下， 技术变得复杂得多，毕竟， 黎是复分析领域的先驱。

There actually is something called the Riemann sphere that involves dividing by zero by a different method, but that's a story for another day.

事实上，有个叫黎球面的东西，它涉及到通过不同的方法来除以 0，但今天我们且不谈这个。

Mr.Jonathan promised to secure the release of those still held. From Abuja the head of the BBC's Hausa Service Mansur Liman has this assessment.

乔纳森承诺解救目前仍被绑架的学生，BBC驻豪萨分支机构记者苏尔·黎报道。

And that's why so many people gravitate towards these like Riemann hypotheses or Fermat's last theorem or that simple paper, which is like just one page.

这就是为什么许多人会被诸如黎猜想、费马大定理或那篇简短的论文所吸引，那篇论文仅有一页， 极为简洁。

Of course our puzzle, which I promise  I'll share in just a moment, is a lot more innocent than the Riemann hypothesis; it's a toy problem.

当然， 我们的谜题， 我保证马上会分享，比黎假设要简单得多；它只是一个练习题。

So for example, if you want to find the complex roots of a polynomial, or some of those million dollar zeros of the Riemann zeta function, this algorithm will do it for you.

所以例如， 如果你想找到多项式的复数根，或者黎zeta函数中那些价值百万美元的零点，这个算法就能帮你找到。

OpenAI's Kevin Wheel, who used to be their Chief Product Officer but is now their VP of Science, about a week ago tweeted, Now he clarified, I'm not claiming GPT-5 is ready to prove the Riemann hypothesis.

OpenAI的凯文·惠尔，他曾是该公司的首席产品官，如今担任科学副总裁，大约一周前发推称：“我现在澄清一下，我并不是说GPT-5已经准备好去证明黎猜想。”

Just like our subsets puzzle, the way that  Riemann studied primes involved a discrete sequence we want to understand, something  carrying information about prime numbers, and then considering a function whose  coefficients are the terms in that sequence.

就像我们的子谜题一样， 黎研究素数的方法涉及一个我们想要理解的离散序列，这个序列携带着关于素数的信息，然后考虑一个其系数为该序列项的函数。