Wiki
Wiki

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

Updated


Statement

Let P\mathcal P be the set of positive primes and N0={0,1,2,… }\mathbb N_0=\{0,1,2,\dots\}; for A,B⊆N0A,B\subseteq\mathbb N_0 write A+B={a+b:a∈A, b∈B}A+B=\{a+b:a\in A,\ b\in B\}. Theorem 1.1 (named "Ostmann's conjecture" in the source). If A,B⊆N0A,B\subseteq\mathbb N_0 have ∣A∣≥2|A|\ge2 and ∣B∣≥2|B|\ge2, then the symmetric difference

(A+B)△P(A+B)\mathbin{\triangle}\mathcal P

is infinite. The source states the equivalent form: no set obtained from P\mathcal P by changing finitely many elements is A+BA+B with both summands of size at least two. The summands may be finite or infinite and may contain 00; the source's introduction adds that the conclusion is not a density statement: a sumset of two sets with at least two elements each that contains every large prime contains infinitely many composites.

Source. OpenAI, The additive indecomposability of the primes, release folder the-additive-indecomposability-of-the-primes-September-24-2026; TeX sections/01-introduction.tex lines 15--20 (label cor:ostmann), PDF p. 1; the reduction to Theorem 2.3 is sections/02-preliminaries.tex lines 91--121 (Lemma 2.2, label lem:finite-factor, PDF pp. 4--5). Read 2026-10-07.

Read depth. Claims checked: the statement, its equivalent form and the statements of Lemmas 2.1 and 2.2 were read clause by clause in the TeX source. The proof was read for its structure (below) and no step was checked. Nothing here is independently reviewed, and the problem page's status is not changed by this record.

Proof pointer

Section 2 (PDF pp. 3--5) reduces the theorem to Theorem 2.3, and the manuscript's Sections 2--9 prove that theorem. The reduction is Lemma 2.2: suppose A+BA+B agrees with P\mathcal P outside a finite set and AA is finite. Two distinct elements of AA are two shifts of BB that are prime for all large b∈Bb\in B, so Lemma 2.1 (a large-sieve bound for sets with jj fixed prime shifts, C(Y)≪Y/(log⁡Y)jC(Y)\ll Y/(\log Y)^j) gives B(Y)≪Y/(log⁡Y)2B(Y)\ll Y/(\log Y)^2; coverage of every large prime up to YY gives Y/log⁡Y≪∣A∣ B(Y)Y/\log Y\ll|A|\,B(Y), a contradiction, and the case of finite BB is symmetric. Hence a counterexample to Theorem 1.1 has two infinite summands and contradicts Theorem 2.3. The source notes that Laffer and Mann (1964, Theorem 12) already proved the two-infinite-summands reduction and that Lemma 2.2 is a short sieve proof of it. Lemma 2.1 is proved from the additive large sieve in the Montgomery--Vaughan form: a mean-zero local test on the allowed classes modulo each prime in (p0,Y](p_0,\sqrt Y], a primitive-mode energy lower bound tensored over squarefree moduli, and Mertens' estimate for the total weight.

Dependencies

The additive large sieve (Montgomery and Vaughan 1973, Theorem 1; Green and Harper, Proposition 3.1), Mertens' estimates and the prime number theorem for the reduction; everything the proof of Theorem 2.3 depends on for the main step. External premises are taken at statement level; none was checked here.

Bears on

  • Problem 431: a claimed stronger form of the negative answer. The problem asks for two infinite sets whose sumset agrees with the primes up to finitely many exceptions; this theorem claims that no two sets with at least two elements each, finite or infinite, have that property. The claim is unverified here, and the page's status rests on acceptance evidence.