What sort of maths are LLMs good at?
What sort of maths are LLMs good at?
大语言模型擅长什么样的数学？
For the sake of anyone who might read this blog post in the distant future (a month from now, say), let me mention that I am writing it a few days after OpenAI announced that it had solved ten major problems in mathematics and theoretical computer science, including the first construction of a non-sofic group, and a proof that the multicolour Ramsey number (where there are 3’s) grows superexponentially in $n$. 为了方便未来（比如一个月后）阅读这篇博文的读者，我先说明一下：我写下这些文字时，距离 OpenAI 宣布其解决了数学和理论计算机科学领域的十大难题仅过去几天。这些成就包括首次构造出非索菲克群（non-sofic group），以及证明了多色拉姆齐数（含 3 个颜色时）随 $n$ 超指数增长。
The first was, to judge from various talks I have been to, one of the most important unsolved problems in group theory, and the second was a major open problem in Ramsey theory that I didn’t necessarily expect to see solved in my lifetime, though of course such expectations now have to be revised. 根据我参加过的各种讲座来看，前者是群论中最悬而未决的重要问题之一；而后者是拉姆齐理论中的一个重大公开问题，我原本没指望在有生之年能看到它被解决，当然，现在这种预期必须得修正了。
The reason I want to be clear about the timing is that I shall be discussing the current capabilities of LLMs in the full expectation that those will continue to change rapidly. So it is likely that in not too long from now, if there is anything interesting in what I write, it will be interesting mainly as a record of what the situation looked like in early August 2026. 我之所以要明确时间点，是因为我在讨论大语言模型（LLM）当前能力时，完全预见到这些能力会持续快速演变。因此，如果我写的内容还有什么趣味性，那很可能仅仅是作为 2026 年 8 月初当时情况的一份记录。
These results, and the other eight on the list, are extraordinarily impressive, but it still doesn’t seem to be the case that LLMs are better than all humans at all aspects of mathematics. If they were, then their big speed advantage over us would mean that there would be much more of a flood of results. So it is natural to wonder about what kinds of problems LLMs are good at, and about where there is still room for improvement. 这些成果以及名单上的其他八项成就令人叹为观止，但大语言模型似乎并未在数学的所有方面都超越全人类。如果它们真的做到了，那么凭借其相对于人类巨大的速度优势，数学界理应涌现出更多的成果。因此，探讨大语言模型擅长解决哪类问题，以及哪些方面仍有改进空间，是顺理成章的事。
I don’t pretend to have a good answer to this question, where a good answer would be a crisp classification that would fit the current examples well, but it is an interesting exercise to try to rule out some bad answers, and to try to identify potential answers that aren’t obviously contradicted by the evidence. 我不敢说自己对这个问题有完美的答案——一个完美的答案应该是一套能很好地涵盖现有案例的清晰分类法——但尝试排除一些错误的答案，并找出那些不被现有证据明显反驳的潜在答案，是一个有趣的练习。
Are LLMs particularly good at finding counterexamples? A first remark here is that LLMs are not just good at finding counterexamples: they can find proofs of difficult statements as well. However, it is notable that the most famous problems they have solved have almost all been with counterexamples rather than proofs. That is true of the two problems mentioned above, and also of the Jacobian conjecture and the unit distance conjecture. 大语言模型是否特别擅长寻找反例？首先要指出的是，大语言模型不仅擅长寻找反例，它们也能找到困难命题的证明。然而值得注意的是，它们所解决的最著名的问题几乎全是通过寻找反例而非证明完成的。上述提到的两个问题，以及雅可比猜想（Jacobian conjecture）和单位距离猜想（unit distance conjecture）都是如此。
If one wants to theorize that LLMs are particularly good at finding counterexamples, then there are two things it would be good to do to make the theory more convincing. The first may sound unproblematic: it is to decide when solving a problem counts as finding a counterexample. Once that is sorted out, the second is to come up with a potential explanation of why LLMs would be particularly well suited to solving problems of that particular kind. 如果有人想提出“大语言模型特别擅长寻找反例”这一理论，那么为了让该理论更具说服力，有两件事需要做。第一件事听起来很简单：确定在什么情况下解决一个问题才算作“寻找反例”。一旦明确了这一点，第二件事就是给出一个合理的解释，说明为什么大语言模型特别适合解决这类问题。
What does it mean to find a counterexample? Why am I suggesting that it is not completely obvious what it means to find a counterexample? Surely, one might suggest, all it means is that you have a statement of the form “Every object of such and such a type has such and such a property,” and you exhibit an object of the given type that does not have the given property. 什么是寻找反例？为什么我说“寻找反例”的定义并不完全显而易见？有人可能会说，这不就是指当你面对一个形式为“每一类某某对象都具有某某性质”的命题时，你展示出一个不具备该性质的特定对象吗？
However, this doesn’t always work. Consider a famous result of Vinogradov, which states that every sufficiently large positive integer is a sum of three primes. The negation of this statement is (or is equivalent to) the statement that for every positive integer $N$ there exists an integer $n > N$ such that $n$ is not a sum of three primes. In other words, it states that every positive integer has a certain property. Seen in this light, Vinogradov found an example of a positive integer that does not have the given property. 然而，这种定义并不总是奏效。考虑维诺格拉多夫（Vinogradov）的一个著名结论：每个足够大的正整数都是三个素数之和。该命题的否定形式（或等价形式）是：对于每一个正整数 $N$，都存在一个整数 $n > N$，使得 $n$ 不能表示为三个素数之和。换句话说，它陈述了每一个正整数都具有某种性质。从这个角度看，维诺格拉多夫找到的其实是一个“不具备该性质的正整数”的例子。
Do we want to say that Vinogradov found a counterexample? Clearly not — the result should obviously be classified as a theorem and not a counterexample. Thus, we cannot just naively say that LLMs are particularly good at negating universally quantified statements: there has to be something about the nature of the universal quantification. 我们能说维诺格拉多夫找到了一个反例吗？显然不能——这个结果显然应该被归类为定理，而不是反例。因此，我们不能天真地认为大语言模型只是擅长否定全称量化命题：全称量化的本质必然包含某种更深层的含义。
With the three-primes example, it is clear that Vinogradov did not think, “How am I going to find $n$ with this property?” Rather, what he thought would have been more like, “I’ve got an integer that is very large. How am I going to show that it is a sum of three primes?” In other words, all his focus would have been on the universally quantified $n$, with the existentially quantified being a sort of afterthought once the details of the proof have been worked out. 在三个素数的例子中，很明显维诺格拉多夫思考的不是“我该如何找到具有这种性质的 $n$？”，而是“我有一个非常大的整数，我该如何证明它是三个素数之和？”换句话说，他的全部注意力都集中在全称量化的 $n$ 上，而存在量化的部分，是在证明细节梳理清楚后的一种“事后补充”。
In general, many interesting results, when they are stated formally, begin with an alternation of two or three (or more) quantifiers. The question then becomes to determine which is the first “interesting” quantified variable in some sense. Here’s another example to illustrate the point, from the theory of finite-dimensional normed spaces. I’ll give a few mathematical details for those curious, but if you don’t care about those, then you can skip the next three paragraphs and should get the gist of what I am saying about this example. 通常情况下，许多有趣的数学结论在形式化表述时，往往以两三个（或更多）量词的交替出现为开端。问题随之转化为：从某种意义上讲，哪一个是第一个“有趣的”量化变量？下面我用有限维赋范空间理论中的另一个例子来说明这一点。我会为感兴趣的读者提供一些数学细节，但如果你不关心这些，可以跳过接下来的三段，依然能领会我举这个例子的核心意图。
Let $X$ and $Y$ be two $n$-dimensional normed spaces and let $T$ be a linear map from $X$ to $Y$. We say that $T$ is a $\lambda$-isomorphism if there exists $c > 0$ such that for every $x \in X$, $c|x| \le |Tx| \le \lambda c|x|$. By rescaling we can always take $c$ to be 1, in which case we have that for every $x \in X$, $|x| \le |Tx| \le \lambda |x|$. If $\lambda = 1$, then this tells us that $T$ is an isometry. 设 $X$ 和 $Y$ 是两个 $n$ 维赋范空间，$T$ 是从 $X$ 到 $Y$ 的线性映射。如果存在 $c > 0$，使得对于每个 $x \in X$，都有 $c|x| \le |Tx| \le \lambda c|x|$，我们就称 $T$ 为 $\lambda$-同构。通过重缩放，我们总可以令 $c=1$，此时对于每个 $x \in X$，都有 $|x| \le |Tx| \le \lambda |x|$。如果 $\lambda = 1$，则说明 $T$ 是等距映射。
In general, the Banach-Mazur distance between $X$ and $Y$ is defined to be the smallest $\lambda$ such that there exists a $\lambda$-isomorphism from $X$ to $Y$. It is easy to see that the logarithm of the Banach-Mazur distance is a metric on the set of isometry classes of $n$-dimensional normed spaces. A less easy fact, but still not too hard, is that the resulting metric space is compact: in fact, it is known as the Banach-Mazur compactum. 通常，Banach-Mazur 距离定义为使得从 $X$ 到 $Y$ 存在 $\lambda$-同构的最小 $\lambda$ 值。容易看出，Banach-Mazur 距离的对数是 $n$ 维赋范空间等距类集合上的一个度量。一个稍难但并不太难的事实是，由此产生的度量空间是紧致的：事实上，它被称为 Banach-Mazur 紧致空间。
It is natural to wonder what the diameter of the Banach-Mazur compactum is, and here things get interesting. A result of Fritz John states that every $n$-dimensional space has distance at most $\sqrt{n}$ from $\ell_2^n$. (The idea of the proof is as follows: pick $E$ inside $t$ 人们自然会好奇 Banach-Mazur 紧致空间的直径是多少，而这正是问题变得有趣的地方。Fritz John 的一个结论指出，每个 $n$ 维空间与 $\ell_2^n$ 的距离最多为 $\sqrt{n}$。（证明思路如下：在 $t$ 中选取 $E$……）