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Erdős problem 283

Let p ⁣:ZZp\colon \mathbb{Z} \rightarrow \mathbb{Z} be a polynomial whose leading coefficient is positive and such that there exists no d2d≥2 with dp(n)d ∣ p(n) for all n1n≥1. Is it true that, for all sufficiently large mm, there exist integers 1n1<<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 statements2415f78e850a

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FormalConjectures/ErdosProblems/

283.lean

Retained formal statement7 of 8

van Doorn [vD25] has investigated the question of what 'sufficiently large' means for p(x)=xp(x)=x. van Doorn has also proved the original conjecture for many linear and quadratic polynomials. For example, if p(x)=x+bp(x) = x + b with 1b50001 \leq b \leq 5000, then the conjecture is true.

FormalConjectures/ErdosProblems/283.leanErdos283.erdos_283.variants.van_doorn_linear1 lineExact file
∀ (b : ℤ), 1 ≤ bb ≤ 5000 → Erdos283.Condition (Polynomial.X + Polynomial.C b)
SolvedStatement only, no proof

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