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Erdős problem 283

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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The source's own material, retained exactly.

8 retained statements36c6d0fef71a

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FormalConjectures/ErdosProblems/

283.lean

Retained formal statement5 of 8

Graham [Gr63] has proved this when p(x)=xp(x)=x.

FormalConjectures/ErdosProblems/283.leanErdos283.erdos_283.variants.graham1 lineExact file
Erdos283.Condition Polynomial.X
SolvedStatement only, no proof