znams_problem

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Proved status carries no information about that problem. Do not import this node or use it as a dependency.

Znám's Problem: For every $k \geq 2$, does there exist a set of $k$ positive integers $\{x_1, \ldots, x_k\}$ such that each $x_i$ divides $1 + \prod_{j \neq i} x_j$?

Štefan Znám posed this in 1975. Solutions are known for $k = 2, 3, \ldots, 9$. Example for $k=2$: $\{2, 3\}$ since $2 \mid 1+3=4$ and $3 \mid 1+2=3$. The question is whether solutions exist for all $k$. Closely related to Egyptian fraction representations and Sylvester sequences.

Source: Znám, Š. (1975). J. Number Theory 7, 303–310. Also: Sun, Z.W. (2009). On Znám's problem and generalized Znám's problem. Integers 9, A44.

Why this node was retired

The posted statement is

theorem znams_problem (k : ℕ) (hk : 2 ≤ k) :
    ∃ s : Fin k → ℕ,
      (∀ i, 0 < s i) ∧
      ∀ i, s i ∣ (1 + ∏ j ∈ Finset.univ.erase i, s j) := by
  sorry

The site states the improper divisor version of Znám's problem. The accepted Sylvester-sequence construction uses its largest term being exactly one plus the product of the others, which the proper-divisor condition would exclude.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Distinguish the improper problem actually proved from the proper version: determine those k for which k distinct integers xᵢ>1 satisfy xᵢ | 1+∏ⱼ≠ᵢxⱼ and xᵢ<1+∏ⱼ≠ᵢxⱼ for every i. Do not assert the proper version for every k≥2 without checking its small-case failures.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.