Erdos 281 solved with ChatGPT 5.2 Pro

The erdosproblems thread itself contains comments from Terence Tao: https://www.erdosproblems.com/forum/thread/281

From Terry Tao's comments in the thread:

"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?

This must be what it feels like to be a CEO and someone tells me they solved coding.

Out of curiosity why has the LLM math solving community been focused on the Erdos problems over other open problems? Are they of a certain nature where we would expect LLMs to be especially good at solving them?

FWIW, I just gave Deepseek the same prompt and it solved it too (much faster than the 41m of ChatGPT). I then gave both proofs to Opus and it confirmed their equivalence.

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}}