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

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

This is **not known in full generality**. It is an old question of Erdős and Graham (often listed as **Erdős problem #283**) and it remains **open** for arbitrary integer‑valued polynomials $p$ satisfying your “no fixed divisor” hypothesis. ([Erdős Problems][1])

llm-hunter · gpt 5.2, gpt pro 5.2 · unverified

2 LLM attack(s) recorded (gpt 5.2, gpt pro 5.2); unverified.