I've "solved" many math problems with LLMs, with LLMs giving full confidence in subtly or significantly incorrect solutions.

I'm very curious here. The Open AI memory orders and claims about capacity limits restricting access to better models are interesting too.

"Very nice! ... actually the thing that impresses me more than the proof method is the avoidance of errors, such as making mistakes with interchanges of limits or quantifiers (which is the main pitfall to avoid here). Previous generations of LLMs would almost certainly have fumbled these delicate issues.

...

I am going ahead and placing this result on the wiki as a Section 1 result (perhaps the most unambiguous instance of such, to date)"

The pace of change in math is going to be something to watch closely. Many minor theorems will fall. Next major milestone: Can LLMs generate useful abstractions?

The answer is yes. Assume, for the sake of contradiction, that there exists an \(\epsilon > 0\) such that for every \(k\), there exists a choice of congruence classes \(a_1^{(k)}, \dots, a_k^{(k)}\) for which the set of integers not covered by the first \(k\) congruences has density at least \(\epsilon\).

For each \(k\), let \(F_k\) be the set of all infinite sequences of residues \((a_i)_{i=1}^\infty\) such that the uncovered set from the first \(k\) congruences has density at least \(\epsilon\). Each \(F_k\) is nonempty (by assumption) and closed in the product topology (since it depends only on the first \(k\) coordinates). Moreover, \(F_{k+1} \subseteq F_k\) because adding a congruence can only reduce the uncovered set. By the compactness of the product of finite sets, \(\bigcap_{k \ge 1} F_k\) is nonempty.

Choose an infinite sequence \((a_i) \in \bigcap_{k \ge 1} F_k\). For this sequence, let \(U_k\) be the set of integers not covered by the first \(k\) congruences, and let \(d_k\) be the density of \(U_k\). Then \(d_k \ge \epsilon\) for all \(k\). Since \(U_{k+1} \subseteq U_k\), the sets \(U_k\) are decreasing and periodic, and their intersection \(U = \bigcap_{k \ge 1} U_k\) has density \(d = \lim_{k \to \infty} d_k \ge \epsilon\). However, by hypothesis, for any choice of residues, the uncovered set has density \(0\), a contradiction.

Therefore, for every \(\epsilon > 0\), there exists a \(k\) such that for every choice of congruence classes \(a_i\), the density of integers not covered by the first \(k\) congruences is less than \(\epsilon\).

\boxed{\text{Yes}}

EDIT: After reading a link someone else posted to Terrance Tao's wiki page, he has a paragraph that somewhat answers this question:

> Erdős problems vary widely in difficulty (by several orders of magnitude), with a core of very interesting, but extremely difficult problems at one end of the spectrum, and a "long tail" of under-explored problems at the other, many of which are "low hanging fruit" that are very suitable for being attacked by current AI tools. Unfortunately, it is hard to tell in advance which category a given problem falls into, short of an expert literature review. (However, if an Erdős problem is only stated once in the literature, and there is scant record of any followup work on the problem, this suggests that the problem may be of the second category.)

from here: https://github.com/teorth/erdosproblems/wiki/AI-contribution...

This is no longer true, a prior solution has just been found[1], so the LLM proof has been moved to the Section 2 of Terence Tao's wiki[2].

[1] - https://www.erdosproblems.com/forum/thread/281#post-3325

[2] - https://github.com/teorth/erdosproblems/wiki/AI-contribution...