Erdős problem 283 · problems.science

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

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 statement6 of 8

Graham also conjectures that this remains true with $1$ replaced by an arbitrary rational $\alpha>0$ (provided $m$ is taken sufficiently large depending on $\alpha$ ).

FormalConjectures/ErdosProblems/283.leanErdos283.erdos_283.variants.graham_alpha5 linesExact file

1∀ (p : Polynomial ℤ) (α : ℚ),2  0 < p.leadingCoeff →3    (¬∃ d ≥ 2, ∀ n ≥ 1, d ∣ Polynomial.eval n p) →4      α > 0 →5        ∀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

Sources

Search problems.science