slug: fiebig-mbirika-spilker-vertical-slice bibkey: fiebigmbirikaspilker2025lucas doi: null url: https://arxiv.org/abs/2408.14632v2 triage: theorem motivation_gids:

• D5/S1/Recurrence/LucasVerticalSlice

Necessary and sufficient conditions for the vertical slice result

Problem

Write V_n = V_n(p, q) for the companion Lucas sequence and pi_V(m) for its least period modulo m. Drawing one fundamental period around a circle, Fiebig, Mbirika and Spilker call it the vertical slice result when a line through the circle meets two terms that are additive inverses of one another. Their Remark 4.12 records that it holds in many cases where the entry point e_V(m) does not exist at all, and their Theorem 4.11 proves it for q = 1 and order statistic omega_V(m) = 4, where the axis passes through the two zeros. Question 5.5 of arXiv:2408.14632v2 asks:

Can we find necessary and sufficient conditions for the “vertical slice result” to hold for (V_n(p,q)) modulo m with q = 1 and omega_V != 4?

The sentence quoted from Remark 4.12 is looser than the property the paper actually tracks. Modulo 28 with p = 7 and q = 1 the period is 2, 7, 19, 14, 23, 7, 26, 7, 23, 14, 19, 7, and the chord joining positions 0 and 6 does join two additive inverses, 2 + 26 = 28; the paper nonetheless lists m = 28 as a case where the result fails. So the property is that every pair reflected in one axis cancels, not that some chord joins a cancelling pair.

Motivation

The entry point is the paper’s organising statistic and every criterion it proves is phrased through it. The question asks for a criterion in the regime where that statistic is undefined, so an answer has to be stated without it.

Gap

The paper’s own examples rule out the naive guesses. For (V_n(7,1)) the result holds at m = 41 and fails at m = 28, and the residue of pi_V(m) modulo four does not separate them: m = 33 has pi_V = 10 = 2 (mod 4) like m = 41, and fails. No criterion in the paper reaches the case where e_V(m) is undefined.

Route

Pinning the predicate down comes first. The paper’s m = 41 numbers are the pairs V_j + V_{pi_V/2 - j}, and V_{-k} = V_k turns that into V_n + V_{n + pi_V/2} = 0 for all n, which refers to no entry point. Dropping the period gives the predicate exists s, forall n, V_n + V_{n+s} = 0.

Two boundary readings come with that predicate. It is the global cancellation, so the single cancelling chord at m = 28 does not satisfy it, which is what makes it agree with the paper there. And the degenerate shift s = 0 satisfies it exactly when 4 = 0 and 2p = 0 in the coefficient ring, since s = 0 says 2 V_n = 0 for every n and the first two values are 2 and p. Modulo 4 with p = 4 both hold: the period is 2, 0, every term is its own additive inverse, and no pair of distinct positions cancels. Modulo 4 with p = 1 the second fails and s = 0 does not satisfy the predicate. The half-period criterion’s hypothesis (4 : R) != 0 excludes the first of the two conditions and so excludes the degeneracy.

Two structural facts then carry the answer. Cayley-Hamilton makes {1, M} span the algebra generated by the companion matrix M, so the infinitely many congruences are decided by two: V_s = -2 and V_{s+1} = -p. That equivalence needs no hypothesis. Lifting it from “orthogonal to the algebra” to “equal to minus the identity” needs the trace form on that algebra to be nondegenerate, and the determinant of its Gram matrix in the basis {1, M} is exactly p^2 - 4q. When that is a unit, the vertical slice result holds exactly when some integer power of the companion matrix is minus the identity.

Falsifier

A commutative ring, a parameter p, and a unit q with p^2 - 4q a unit, for which a shift cancels every companion trace while no integer companion power is minus the identity, or the converse, would contradict the criterion. For the hypothesis-free criterion, any p, q and shift s with V_s = -2 and V_{s+1} = -p but some n with V_n + V_{n+s} != 0 would do.

Evidence

• Module: D5/S1/Recurrence/LucasVerticalSlice.lean.
• Answer to the question, carrying no hypothesis: verticalSlice_iff_two_traces. This is the declaration the Blueprint records as resolving this entry.
• Structural refinement, assuming IsUnit (p^2 - 4q): verticalSlice_iff_neg_one_mem_zpowers.
• Half-period form, assuming (4 : R) != 0: verticalSlice_iff_half_companionPeriod.
• Non-removability of the unit hypothesis: result refutes claim, witness p = 4, q = 1, modulus 6.
• Orchestrator scans over ZMod m, periods computed from the state pair and never assumed: over p <= 30, 3 <= m <= 40, with q ranging over every unit modulo m, the hypothesis-free reduction agrees on 15128 cases (6722 holding, 8406 failing, 0 disagreements) and the unit-discriminant criterion agrees on 10308 cases (4936 holding, 5372 failing, 0 disagreements); over p <= 60, 3 <= m <= 80 at q = 1, dropping the unit hypothesis produces 309 disagreements out of 4758 cases.
• Neither criterion carries an omega_V hypothesis, so neither is confined to the omega_V != 4 regime the question restricts to, and both also cover the regime Theorem 4.11 already handled. Coverage of the parameters themselves differs: the hypothesis-free reduction reaches every commutative ring, parameter and unit, while the unit-discriminant criterion reaches only the parameters with p^2 - 4q a unit and so misses part of the regime the question mentions, the witness p = 4, q = 1, m = 6 included.

Triage

theorem. The two criteria have different scopes and the distinction matters for what is claimed settled.

verticalSlice_iff_two_traces carries no hypothesis: for every commutative ring, every parameter p and every unit q, a vertical slice exists exactly when two consecutive companion traces take the values -2 and -p. That is a necessary and sufficient condition with no restriction on q, on the order statistic, or on the discriminant, so it answers the question as asked and beyond it. This is the declaration the Blueprint records as resolving this entry.

verticalSlice_iff_neg_one_mem_zpowers refines that to a group-theoretic condition — minus one lies in the cyclic group generated by the companion matrix — but only when p^2 - 4q is a unit, and result proves that hypothesis is not removable. The refinement therefore does not cover the whole regime the question mentions: at p = 4, q = 1, m = 6 the companion sequence is 2, 4, 2, 4, ... with no entry point at all, which is the missing-entry-point regime, while its discriminant 12 is not a unit modulo 6.

What is not settled here: a structural criterion of the second kind valid when p^2 - 4q is not a unit, and the vertical slice question for the companion sequence’s own zero set rather than its trace.

ASSUMED-UNVERIFIED

The identification of the paper’s informal “vertical slice result” with the predicate exists s, forall n, V_n + V_{n+s} = 0 rests on an orchestrator reading of Remark 4.12 checked against the paper’s three printed examples at m = 41, 28 and 33; it is not a machine proof of that identification. The non-removability of the (4 : R) != 0 hypothesis in the half-period criterion is measured, not proved: over p <= 40, 3 <= m <= 60 at q = 1, restricted to the paper’s omega_V != 4 regime, 21 counterexamples were found and every one is at modulus 4. That count is specific to that window and that filter; widening either changes it. The module carries no machine certificate for the claim because the utility header admits one claim and one result. No MathOverflow, Math.SE or Zulip search was performed; the citation graph of the source paper was checked and contains one work, which does not address this question. First-publication priority is not established.