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Erdős problem 283

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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)?

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8 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

283.lean

Retained formal statement4 of 8

Cassels [Ca60] has proved that these conditions on the polynomial imply every sufficiently large integer is the sum of p(ni)p(n_i) with distinct nin_i.

FormalConjectures/ErdosProblems/283.leanErdos283.erdos_283.variants.cassels4 linesExact file
∀ (p : Polynomial ℤ),  0 < p.leadingCoeff →    (¬∃ d ≥ 2, ∀ n ≥ 1, d ∣ Polynomial.eval n p) →      ∀ᶠ (m : ℕ) in Filter.atTop, ∃ S, (∀ n ∈ S, 1 ≤ n) ∧ ∑ n ∈ S, Polynomial.eval (↑n) p = ↑m
SolvedStatement only, no proof