Erdős Problem #543 — Define f(N)f(N) be the minimal kk such that the following holds: if GG is an abelian group of size NN and A⊆GA\subseteq G is a random set of size kk then, with probability ≥1/2\geq 1/2, all elements of…

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Define f(N)f(N) be the minimal kk such that the following holds: if GG is an abelian group of size NN and A⊆GA\subseteq G is a random set of size kk then, with probability ≥1/2\geq 1/2, all elements of GG can be written as ∑x∈Sx\sum_{x\in S}x for some S⊆AS\subseteq A. Is f(N)≤log⁡2N+o(log⁡log⁡N)?f(N) \leq \log_2 N+o(\log\log N)?

References

Progress summary

Refreshed
Claimed solved

A 2026 paper shows that the proposed bound is false in general: groups of prime order require an extra term growing like the logarithm of the logarithm.

Erdős and Rényi established the original upper bound in 1965; Erdős conjectured in 1973 that its double-logarithmic error term could not be reduced to a smaller-order term. The problem is now answered negatively using additive groups of prime order.

Known results

  • Erdős and Rényi, 1965: f(N)≤log⁡2N+1log⁡2log⁡log⁡N+O(1)f(N)\leq\log_2N+\frac{1}{\log 2}\log\log N+O(1) for finite abelian groups.
  • Erdős and Hall: the stronger bound f(N)≤log⁡2N+o(log⁡log⁡log⁡N)f(N)\leq\log_2N+o(\log\log\log N) is false.

February 2026 negative resolution

The paper proves that for every fixed 0<c<12log⁡20<c<\frac{1}{2\log 2}, random subsets of Fp\mathbb{F}_p of size ⌊log⁡2p+clog⁡log⁡p⌋\lfloor\log_2p+c\log\log p\rfloor fail to cover all subset sums with probability tending to 11. Hence f(p)≥log⁡2p+(12log⁡2+o(1))log⁡log⁡pf(p)\geq\log_2p+\left(\frac{1}{2\log 2}+o(1)\right)\log\log p, disproving the proposed universal bound. ChatGPT-5.2 Pro supplied the initial qualitative proof idea; the authors repaired it and developed the quantitative refinement.

Current status (as of February 2026): The proposed universal bound is false, with a quantitative obstruction proved for prime-order groups; the exact threshold and behavior for other group families remain open.

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