The board / number theory / #283
Erdős Problem #283
PROVED LEANdatabase updated 2026-05-10
The statement as posed is true, and a proof exists. A solution has also been checked by machine, in Lean.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
Given a polynomial `p` with rational coefficients, the predicate that if `p` takes integer values at all integers, the leading coefficient is positive and there exists no \(d≥2\) with \(d ∣ p(n)\) for all \(n≥1\), then 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 \[m=p(n_1)+\cdots+p(n_k)\]?
Let \(p\colon \mathbb{Z} \rightarrow \mathbb{Z}\) be a polynomial (with rational coefficients, taking integer values at all integers) 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 \[m=p(n_1)+\cdots+p(n_k)\]? GPT 5.5 Pro (prompted by Price) has given a proof that the answer is yes, for the stronger version with \(1\) replaced by any rational \(\alpha>0\). This was formalized in Lean by Ammanamanchi using Opus 4.6 and GPT 5.5 Pro.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 7 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/283.Prizenone
Topicsnumber theory, unit fractions
Statement in LeanFormal Conjectures
Solution in Leanyes
OEISA380791
Original pageerdosproblems.com/283
Attempts
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