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 $$m=p(n_1)+\cdots+p(n_k)$$?

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 $$m=p(n_1)+\cdots+p(n_k)$$?

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.