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Problem 334

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claims/: The 1 claim page of Problem 334, one per claimant's result; the problem's standing derives from them.


Statement. Find the best function f(n)f(n) such that every nn can be written as n=a+bn=a+b where both a,ba,b are f(n)f(n)-smooth (that is, are not divisible by any prime p>f(n)p>f(n).)

Formulation. Erdős posed the problem as a yes-or-no question. In [Er76e], p. 272, he asks whether for every α\alpha there is n0(α)n_0(\alpha) such that every n>n0(α)n>n_0(\alpha) is a sum a+ba+b of two positive integers with P(a⋅b)<n1/αP(a\cdot b)<n^{1/\alpha}, where P(m)P(m) is the largest prime factor of mm, and writes that he has been unable to prove this for α>2\alpha>2. [ErGr80], p. 70, asks the same with nϵn^\epsilon for every ϵ>0\epsilon>0, and [Er82d], p. 55, item 4, asks it in the equivalent form that the least integer not a sum a+ba+b with no prime factor of abab above nn exceeds nkn^k for every kk and n>n0(k)n>n_0(k). In the notation of the Statement, the question is whether f(n)≤no(1)f(n)\le n^{o(1)}, which the site expects and no source settles. The site records Erdős's original question as whether even f(n)≤n1/3f(n)\le n^{1/3}, the case α=3\alpha=3; Balog's bound f(n)≪ϵn4/(9e)+ϵf(n)\ll_\epsilon n^{4/(9\sqrt e)+\epsilon}, with 4/(9e)=0.2695…<1/34/(9\sqrt e)=0.2695\ldots<1/3, answers that case yes. The site asks instead for the best function ff, and that question sets the standing.

Status. Open, the site's label (OPEN; page last edited 2026-04-03). The case the site records as Erdős's original question, whether f(n)≤n1/3f(n)\le n^{1/3}, is answered yes by Balog [Ba89], recorded as the accepted partial claim [[problems/arithmetic_functions/E0334/claims/1989_09_01_balog|Balog's bound with exponent 4/(9e)4/(9\sqrt e)]]; the best function ff is not determined, so the problem stays open.

Source. erdosproblems.com/334, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #334, https://www.erdosproblems.com/334.

References.

Formalization. None recorded.

Current assessment

The site's formulation above asks for the best function ff such that every nn is a sum of two f(n)f(n)-smooth integers. The one credited result is Balog's theorem that f(n)≪ϵn4/(9e)+ϵf(n)\ll_\epsilon n^{4/(9\sqrt e)+\epsilon} for every ϵ>0\epsilon>0, where 4/(9e)=0.2695…4/(9\sqrt e)=0.2695\ldots, recorded at [[problems/arithmetic_functions/E0334/claims/1989_09_01_balog|Balog's bound with exponent 4/(9e)4/(9\sqrt e)]] as an accepted partial claim on its refereed publication; it answers yes the question whether f(n)≤n1/3f(n)\le n^{1/3}, which the site records as Erdős's original form of the problem, and the site expects f(n)≤no(1)f(n)\le n^{o(1)}. The site also points to Problem 59 of Green's open problems list. No result determines ff, so the problem stays open.

The thread's two comments settle nothing and are not claims. A comment of 2025-11-19 by the user Woett reports that a literature search made with Gemini's Deep Research tool found nothing beyond Balog's bound; it notes Sárközy's bound f(n)<exp⁡(clog⁡nlog⁡log⁡n)f(n)<\exp(c\sqrt{\log n\log\log n}) for sums of three smooth summands, with Erdős's conjecture, as Sárközy reports it, that the same bound holds for two summands, and it notes that a bound f(n)≤n0.1f(n)\le n^{0.1} would give, for large primes p≡3(mod4)p\equiv3\pmod4, a quadratic non-residue of size at most p0.1p^{0.1}, below the known exponent 1/(4e)+ϵ1/(4\sqrt e)+\epsilon. A comment of 2026-03-03 by the user my99n links the OEIS entry A062241 for the problem and a note, written with AI assistance, bounding that sequence by A045535, which is easier to compute.

Search scope (2026-10-07). The account above rests on the site's problem page and forum thread, on Erdős's statements of the question in [Er76e], [Er82d] and [ErGr80], and on the publisher's record of Balog's paper. Balog's proof is not checked here, and no literature search beyond these sources is recorded.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.