Skip to content

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)?

Sources

Browse retained paths and inspect the exact material available for this Problem.

8 retained statements2415f78e850a

Open selected source

FormalConjectures/ErdosProblems/

283.lean

Retained formal statement8 of 8

van Doorn [vD25] has proved the original conjecture for many linear and quadratic polynomials. For example, if p(x)=x2+bp(x) = x^2 + b with 1b8001 \leq b \leq 800, then the conjecture is true.

FormalConjectures/ErdosProblems/283.leanErdos283.erdos_283.variants.van_doorn_quadratic1 lineExact file
∀ (b : ℤ), 1 ≤ bb ≤ 800 → Erdos283.Condition (Polynomial.X ^ 2 + Polynomial.C b)
SolvedStatement only, no proof

Search problems.science

Find a Problem, Result, source, or page