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RESEARCH / Focused target

Sun's Conjecture 3.8: circular permutations with primitive adjacent products

Literature status: not recheckedRepository record: refuted

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Problem

The question

Zhi-Wei Sun, “Some new problems in additive combinatorics”, arXiv:1309.1679, Conjecture 3.8:

Let F_q be a finite field with q > 7 elements, and let a_0 be any element of F_q. Then there is a circular permutation (a_1, …, a_{q−1}) of all the nonzero elements of F_q such that all the q − 1 elements a_0 + a_1a_2, a_0 + a_2a_3, …, a_0 + a_{q−2}a_{q−1}, a_0 + a_{q−1}a_1 are primitive elements of the field F_q.

The Lean definition claim : Prop quantifies over every finite field and every offset. An equivalence from Fin (q - 1) to the nonzero subtype expresses that every nonzero element appears exactly once. finRotate advances cyclically, including the last-to-first edge. IsPrimitiveRoot z (q - 1) means exact multiplicative order q - 1, hence precisely that z generates the multiplicative group, whose order is q - 1.

Motivation

Our foothold

The assertion seeks one circular arrangement satisfying all adjacent primitive-element constraints simultaneously. Its universal field and offset quantifiers permit refutation by one admissible pair.

Gap

Missing bridges

Sun stated the assertion in 2013 together with a numerical remark for the offset a₀ = −1 only. Luo and She, “A conjecture on circular permutations over finite fields”, arXiv:2610.06289v1 (2026), restate it as their Conjecture 1.1 with the same wording, prove it for every prime power q > 18 888 871, and prove the offset a₀ = 0 for all q > 4; they give no table of small fields. The OEIS entry A229232 records only the offsets ∓1. Within these sources the fields of order at most 18 888 871 with nonzero offset are unsettled.

Route

Proposed approach

Take F = ZMod 11 and a₀ = 2. Every primitive element has order ten. The residues outside {2,6,7,8} are zero or have fifth or second power one, so none has order ten. A cyclic neighbour is distinct from the vertex itself. Direct finite arithmetic with this condition then implies that each neighbour of vertex 2 or vertex 9 must belong to {3,8}. Multiplication is commutative, so this holds for both cyclic neighbours.

For any cycle of length greater than four, the two neighbours of each vertex are distinct. If two vertices have both neighbours in a common set of at most two elements, their neighbour pairs coincide. Matching successors or predecessors identifies the vertices. Interchanging the pairs forces a four-step return, impossible for a cycle of length greater than four. The required ten-cycle therefore cannot exist.

Falsifier

What would falsify this route

The counterexample would fail if eleven were not a prime field order greater than seven, the offset two were inadmissible, the primitive-element exclusions or forced-neighbour arithmetic were incorrect, the predecessor restriction did not follow from the cyclic conditions, or the twin-neighbour obstruction failed at length ten. The Lean proof discharges each obligation without enumerating all permutations or adding assumptions to result.

Evidence

Evidence to collect

D5/S3/Arith/SunCircularPrimitivePermutationRefutation.result : ¬ claim specializes the universal assertion to ZMod 11 and offset two. The reusable theorem twin_neighbour_obstruction applies to arbitrary vertex types and all cycle lengths greater than four. Finite arithmetic uses kernel-checked decide; the cyclic obstruction uses finite-set cardinality and exact integer reasoning. No native_decide, sorry, or new axiom is used. The axiom closure of result is propext, Classical.choice, Quot.sound.

Triage

Scope assessment

  • [proved: D5/S3/Arith/SunCircularPrimitivePermutationRefutation.result] The conjecture is false at field order eleven and offset two.
  • [proved: D5/S3/Arith/SunCircularPrimitivePermutationRefutation.twin_neighbour_obstruction] Two distinct vertices of a cycle of length greater than four cannot both have all their cyclic neighbours in one set of at most two vertices.
  • [computed] Exhaustive Hamilton-cycle checks over every offset for the fields of order 8, 9, 11, 13, 17 and 19 find exactly one failing pair, (q, a₀) = (11, 2); at q = 11, a₀ = 2 no Hamilton path even starts at 1.
  • [open] Classification of all exceptional field orders and offsets is not supplied by this refutation. No statement about other pairs follows from this single counterexample.

ASSUMED-UNVERIFIED

Unverified assumptions

Forward citations of arXiv:1309.1679 were screened through web search and the OEIS only; no citation index was exhaustively read. The journal version (Nanjing Univ. J. Math. Biquarterly 36 (2019)) was not compared line by line with the arXiv text. Neither condition is a hypothesis of the refutation.

Dossier & anchor history

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