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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The answer to Problem 431 is no: there are no two infinite sets AA and BB of nonnegative integers whose sumset A+BA+B agrees with the set of primes outside a finite set. This is Theorem 2.3 of The additive indecomposability of the primes, a manuscript of the OpenAI mathematics release dated 24 September 2026 with the author line "OpenAI"; the release's README says its manuscripts were produced by an internal OpenAI model and come at different stages of verification, not all with Lean formalizations. The manuscript's main statement, Theorem 1.1, is Ostmann's inverse Goldbach conjecture in full: for any A,B⊆N0A,B\subseteq\mathbb N_0 with at least two elements each, the symmetric difference of A+BA+B and the primes is infinite. Lemma 2.2 reduces it to the two-infinite-summands case by a sieve count, so the problem's question is the case the rest of the 80-page manuscript proves. The problem leaves the ambient set unstated; sets of positive integers are the special case 0∉A∪B0\notin A\cup B, so the theorem covers either reading. The manuscript's library card compiles the two statements on the result pages Theorem 1.1 and Theorem 2.3. The proof argues by contradiction from a decomposition: for each prime pp the residues of the large elements of AA and of −B-B are disjoint, which partitions Fp\mathbb F_p; a collision estimate makes the two parts nearly equal in size; quadratic and higher-order character sums are shown to decorrelate from the partition; prime coverage supplies additive transforms that are not small; and a finite-field comparison of binary trees built by repeated Cauchy--Schwarz transfers, averaged over permutations of the variables, contradicts a positive statistic. The proof has not been reviewed in this corpus.

If the claim stands, it supersedes as partial progress the square-root counting bounds of Elsholtz (2001) and of Elsholtz and Harper (2015) on a hypothetical decomposition. The site's commentary records the Elsholtz–Harper bounds as the best result toward the expected negative answer, and cites Elsholtz (2001) for the impossibility of three summands: no sets A,B,CA,B,C with at least two elements each have A+B+CA+B+C agreeing with the primes up to finitely many exceptions.

Acceptance. Formalized. The release pairs the manuscript with a Lean development under lean/OAI/NumberTheory/Ostmann/ whose comparator statements, OAI.Ostmann.inverseGoldbach and OAI.Ostmann.twoInfiniteSummandsImpossible in lean/ComparatorChallenges/OstmannComplete.lean and OAI.Ostmann.main in lean/ComparatorChallenges/OstmannPrimes.lean, state Theorem 1.1 and Theorem 2.3 for sets of natural numbers with Mathlib's primes; the second of them is the problem's question with the answer no, and the other two are the stronger form for sets with at least two elements each. This corpus's verification built the three declarations at the pinned revision of 6 October 2026 with the toolchain leanprover/lean4:v4.34.1 and checked their axioms, which are exactly propext, Classical.choice and Quot.sound, with no sorry; the two challenges pin them, and each fingerprint was found identical to its challenge. The statement audit found twoInfiniteSummandsImpossible exactly the question with the answer no: for all infinite A,B⊆NA,B\subseteq\mathbb N, 00 allowed, there is no NN such that every n≥Nn\ge N lies in A+BA+B exactly when nn is prime, and over N\mathbb N agreement from some point on is agreement up to finitely many exceptions, since a finite set of naturals is bounded; the sumset is Mathlib's pointwise addition, the hypotheses are only that the two sets are infinite, and nothing is vacuous. It found inverseGoldbach and main the same proposition, written with and without the symmetric-difference notation: for all A,B⊆NA,B\subseteq\mathbb N with at least two elements each, the symmetric difference of A+BA+B and the primes is infinite, which implies the answer no and is strictly stronger. All three work in N\mathbb N, so sets of positive integers are covered as the case 0∉A∪B0\notin A\cup B; a reading over the integers with negative elements is not literally covered and does not reduce to N\mathbb N by a translation, but the problem's standard reading, the formal-conjectures statement and this page take nonnegative integers. Not reviewed or refereed: the manuscript has no refereed publication, no arXiv version and no reviewer independent of the release, its proof has not been reviewed in this corpus, and the site's commentary (last edited 8 April 2026) did not record it.

Depends on. Nothing on this wiki; the result is the manuscript's own.