Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. The answer to Problem 871
is no: there is an additive basis of order with
that cannot be partitioned into two disjoint
additive bases of order . In the Lean statement not_erdos_871, is a
set of natural numbers such that every sufficiently large is a sum
with , such that for every every sufficiently large has at
least pairs in with , and such that no disjoint
with both represent every sufficiently large integer as a
sum of two of their elements. Counting pairs with differs from
by at most a factor of two, so the divergence is the one the
problem asks for.
Submission note. Posted to the site's forum by Daniel Larsen on 5 January 2026:
This seems to be an LLM-generated formal proof. (Statement on line 4930.)
It follows the Erdős-Nathanson method extremely closely, but lets the size of go to infinity very slowly. The same method should resolve the first half of [868] (courtesy of Claude Opus 4.5). In particular, the instead of being an enumeration of -element subsets of over all should be an enumeration of -element subsets of . For , being a basis is equivalent to having fewer than elements in for all but finitely many . This property is stable under removal of finite subsets.
Argument. Erdős and Nathanson had proved (erdos_1989_additive_bases_many_representations) that for every fixed there is a basis of order with for all large that cannot be split into two disjoint bases, and had asked (erdos_1988_partitions_bases_into_disjoint_unions_bases) whether divergence suffices, having proved that it does once the number of representations with in is at least for all large , for some constant . The disproof keeps their construction and lets grow slowly. A base set , a union of long and widely spaced intervals with a few elements removed, has a representation function that vanishes on a rapidly growing sequence and tends to infinity elsewhere. To one adds a few elements so that each gains exactly representations, through a -element set with , where slowly and the sets run through all -element subsets of successive blocks of the set. Whenever is split into two parts, the pigeonhole principle puts some -subset of each block wholly inside one part, and the other part then contains no element of that ; so for infinitely many one fixed part contains no element of and cannot represent , and that part is not a basis. The curator's remark that only a small modification of the 1989 argument is needed, and a sketch of this shape posted on the thread by Tao on 2026-01-05, are the sources of this outline; it is a reading aid, not proof coverage.
The postings. The claimant posted the Lean proof to the problem's thread
on 2026-01-05 as a Lean playground link, with the statement on line 4930 of
the file (the first formalization link is that post; the playground address
encodes the whole file and is too long to reproduce), and the system's
write-up the same day (the first preprint link, a GitHub upload of
2026-01-05 that Larsen described as stylistically flawed); the system's original
solution, a LaTeX write-up produced from the problem statement and the 1989
paper, was uploaded on 2026-01-06 (the second preprint link). On the thread
Larsen described the system as Claude and Gemini agents with assigned roles and
Larsen's own part as reading the Erdős–Nathanson paper, directing the system to
formalize the relevant parts and to extend the construction, and intervening
when the formalization went off track. Larsen's later preprint
(larsen_2026_three_questions_erdos_nathanson_asymptotic_bases,
arXiv:2603.03472, 2026-03-03, 7 pages) records the result as the author's,
obtained with the multi-agent system, and proves more by a different
construction: a divergent representation function, decomposability into two
disjoint bases, and containing a minimal basis are mutually independent
properties of asymptotic bases of order , every combination being
realized (its Theorem 1). That preprint is a second, unrefereed proof of the
same answer and is linked here rather than given its own page.
Acceptance. The reviewed evidence is the site's acceptance: its
curator, Thomas Bloom, credits the disproof to Larsen using Claude Opus 4.5
in the problem's remarks, and the page carries the label DISPROVED (LEAN)
(last edited 2026-01-06). The formal-conjectures statement
file (the record link, pinned at its revision of 2026-10-06) tags
erdos_871 research solved with the answer False and records a formal proof
at the Lean file below; the statement itself is left as sorry there, so
the record is a catalog entry and not a formalization. On the thread, Tao
noted on 2026-01-05 that the result is a disproof rather than a proof and
sketched the argument; Tao did not report checking the Lean file, which
exceeded the heartbeat limit when Tao ran it. Alexeev reported that every
native_decide call in the first Lean file could be replaced by decide or
norm_num. No named
mathematician has reviewed the write-ups and there is no refereed publication.
Formalization. The second formalization link is the file
src/v4.29.1/ErdosProblems/Erdos871.lean of Boris Alexeev's repository
https://github.com/plby/lean-proofs, a later revision of the playground file,
shorter and without native_decide, added to the repository on 2026-05-12
(the link's date) and pinned at the commit of 2026-06-24 that gave the folder
its name (Lean 4.29.1). Its header declares it a Lean
formalization of a solution to the problem and names Erdős, Nathanson, Larsen
and Claude Opus 4.5 as informal authors and Claude Opus 4.5, Gemini 3 Pro and
Larsen as formal authors, so it is a formalization of the claimant's result
and is linked here rather than given its own page. At the pinned revision
the file imports only Mathlib, its 2,415 lines contain no sorry and no
native_decide, and its closing
#print axioms comment records propext, Classical.choice and
Quot.sound for not_erdos_871, whose conclusion is the negation of the
universal clause of the catalog's erdos_871. This corpus has not built the
file, printed its axioms or audited its definitions, so the claim carries no
formalized evidence and the formalization is a link, not a warrant.
Depends on. Nothing in this wiki; the construction is self-contained apart from the 1989 paper it modifies.