Erdős Problem #283 — Let p:Z→Zp:\mathbb{Z}\to \mathbb{Z} be a polynomial whose leading coefficient is positive and such that there exists no d≥2d\geq 2 with d∣p(n)d\mid p(n) for all n≥1n\geq 1.

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Let p:Z→Zp:\mathbb{Z}\to \mathbb{Z} be a polynomial whose leading coefficient is positive and such that there exists no d≥2d\geq 2 with d∣p(n)d\mid p(n) for all n≥1n\geq 1. Is it true that, for all sufficiently large mm, there exist integers 1≤n1<⋯<nk1\leq n_1<\cdots <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)?

References

Progress summary

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Special cases are known, but the full conjecture remains unverified despite a reported proof by GPT-5.5 Pro.

The conjecture asks whether every sufficiently large integer can be represented by polynomial values at distinct positive integers whose reciprocals sum to 11. It is recorded as Erdős Problem #283\#283; the retrieved sources do not give a posing date.

Known results

  • Graham proved the linear case p(x)=xp(x)=x, in a stronger form for every positive rational reciprocal sum and lower bound.
  • Cassels proved that the polynomial hypotheses suffice without the reciprocal-sum constraint.
  • Burr proved a power-polynomial variant when repeated indices are allowed.
  • Alekseyev proved p(x)=x2p(x)=x^2 for every m>8542m>8542; van Doorn proved many linear and quadratic cases, including p(x)=x+5p(x)=x+5 and p(x)=x2+100p(x)=x^2+100.

Undated GPT-5.5 Pro proof claim

A discussion reports that GPT-5.5 Pro, prompted by Price, produced a proof of the stronger statement with 11 replaced by any rational α>0\alpha>0. This remains an AI-generated claim: no retrieved preprint or independently verified proof corroborates it.

Current status (as of March 2026): Special cases are established, but the full conjecture has no verified proof; the reported GPT-5.5 Pro proof remains unconfirmed.

Sources

Solutions 0

No solutions have been posted yet.