Skip to content

Erdős problem 283

Start work

Let p ⁣:Z→Zp\colon \mathbb{Z} \rightarrow \mathbb{Z} be a polynomial whose leading coefficient is positive and such that there exists no d≥2d≥2 with d∣p(n)d ∣ p(n) for all n≥1n≥1. Is it true that, for all sufficiently large mm, there exist integers 1≤n1<⋯<nk1≤n_1<\dots < n_k such that 1=1n1+⋯+1nk1=\frac{1}{n_1}+\cdots+\frac{1}{n_k} and m=p(n1)+⋯+p(nk)m=p(n_1)+\cdots+p(n_k)?

Sources

The source's own material, retained exactly.

8 retained statements36c6d0fef71a

Open selected source

FormalConjectures/ErdosProblems/

283.lean

Retained formal statement6 of 8

Graham also conjectures that this remains true with 11 replaced by an arbitrary rational α>0\alpha>0 (provided mm is taken sufficiently large depending on α\alpha).

FormalConjectures/ErdosProblems/283.leanErdos283.erdos_283.variants.graham_alpha5 linesExact file
∀ (p : Polynomial ℤ) (α : ℚ),  0 < p.leadingCoeff →    (¬∃ d ≥ 2, ∀ n ≥ 1, d ∣ Polynomial.eval n p) →      α > 0 →        ∀ᶠ (m : ℕ) in Filter.atTop, ∃ S, (∀ n ∈ S, 1 ≤ n) ∧ ∑ n ∈ S, 1 / ↑n = α ∧ ∑ n ∈ S, Polynomial.eval (↑n) p = ↑m
SolvedStatement only, no proof