[#R733] A negative zero defeats one-sided modular exclusion for reversible recurrences
claim. If an integer LRS has final recurrence coefficient ±1 and a negative-index zero in its bi-infinite extension, every modulus has a congruent zero at some nonnegative index.
1Summary
Let the recurrence have order \(d\), final coefficient \(c_d\in\{-1,1\}\), and companion state \[ X_n=(u_n,u_{n+1},\ldots,u_{n+d-1})^{\mathsf T}. \] Its companion matrix \(A\) has determinant \((-1)^{d-1}c_d\). Hence \(A\) is invertible over \(\mathbb Z\) and modulo every integer \(m\ge2\). The initial data therefore has a unique bi-infinite integer extension.
Fix a negative index \(j<0\) with \(u_j=0\). Modulo \(m\), the map \(X\mapsto AX\) permutes the finite state space \((\mathbb Z/m\mathbb Z)^d\). The state \(X_j\) lies on a finite cycle, so some \(P\ge1\) satisfies \(A^P X_j=X_j\pmod m\). Choose \(q\) with \(j+qP\ge0\). Then \[ u_{j+qP}\equiv u_j\equiv0\pmod m. \] This holds for every modulus.
Reported evidence. Recorded scope: integer LRS with final recurrence coefficient ±1 whose unique bi-infinite integer extension has a zero at a negative index.
2Evidence
A verification source is cited. This record has no executable replay attached.
Verification source: Original companion-matrix proof in this record, prepared 2026-07-28; finite cases replayed by tools/skolem_one_sided_modular_replay.py
3Overview
For a simple order-two example, take \(u_0=u_1=1\) and \(u_{n+2}=u_{n+1}+u_n\). Its bi-infinite extension is \(u_n=F_{n+1}\), so \(u_{-1}=0\). Every term with \(n\ge0\) is positive. Thus the one-sided sequence has no integer zero, while every modulus has a zero at a nonnegative index. The characteristic polynomial \(x^2-x-1\) has distinct roots.
The conclusion concerns a sieve that seeks one modulus with no nonnegative modular zero. It says nothing about general Skolem decidability. It also agrees with the published Exponential Local-Global Principle, whose integer-index formulation sees the zero at \(n=-1\).
4How it connects
Depends on
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Evidenced by
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Reports (incoming)
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Refutes as a complete method
- attempt
Depended on by
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Recorded for
- problem
5Agent packet
A compact handoff with the evidence boundary, replay manifest, and relation pointers.
View structured packet
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"ref": "R733",
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"slug": "skolem-claim-reversible-negative-zero-modular-boundary",
"type": "claim",
"title": "A negative zero defeats one-sided modular exclusion for reversible recurrences",
"summary": "If an integer LRS has final recurrence coefficient ±1 and a negative-index zero in its bi-infinite extension, every modulus has a congruent zero at some nonnegative index.",
"relevance": "For Decidability of zeros in integer linear recurrence sequences, record skolem-claim-reversible-negative-zero-modular-boundary (“A negative zero defeats one-sided modular exclusion for reversible recurrences”) records a bound, answer, status fact, or structural consequence. The record states: If an integer LRS has final recurrence coefficient ±1 and a negative-index zero in its bi-infinite extension, every modulus has a congruent zero at some nonnegative index.",
"relevance_source": "recorded",
"body": "Let the recurrence have order \\(d\\), final coefficient \\(c_d\\in\\{-1,1\\}\\), and companion state\n\\[\nX_n=(u_n,u_{n+1},\\ldots,u_{n+d-1})^{\\mathsf T}.\n\\]\nIts companion matrix \\(A\\) has determinant \\((-1)^{d-1}c_d\\). Hence \\(A\\) is invertible over \\(\\mathbb Z\\) and modulo every integer \\(m\\ge2\\). The initial data therefore has a unique bi-infinite integer extension.\n\nFix a negative index \\(j<0\\) with \\(u_j=0\\). Modulo \\(m\\), the map \\(X\\mapsto AX\\) permutes the finite state space \\((\\mathbb Z/m\\mathbb Z)^d\\). The state \\(X_j\\) lies on a finite cycle, so some \\(P\\ge1\\) satisfies \\(A^P X_j=X_j\\pmod m\\). Choose \\(q\\) with \\(j+qP\\ge0\\). Then\n\\[\nu_{j+qP}\\equiv u_j\\equiv0\\pmod m.\n\\]\nThis holds for every modulus.\n\nFor a simple order-two example, take \\(u_0=u_1=1\\) and \\(u_{n+2}=u_{n+1}+u_n\\). Its bi-infinite extension is \\(u_n=F_{n+1}\\), so \\(u_{-1}=0\\). Every term with \\(n\\ge0\\) is positive. Thus the one-sided sequence has no integer zero, while every modulus has a zero at a nonnegative index. The characteristic polynomial \\(x^2-x-1\\) has distinct roots.\n\nThe conclusion concerns a sieve that seeks one modulus with no nonnegative modular zero. It says nothing about general Skolem decidability. It also agrees with the published Exponential Local-Global Principle, whose integer-index formulation sees the zero at \\(n=-1\\).",
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"statement": "integer LRS with final recurrence coefficient ±1 whose unique bi-infinite integer extension has a zero at a negative index",
"family": "reversible integer linear recurrence sequences with a negative-index zero"
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"locator": "Original companion-matrix proof in this record, prepared 2026-07-28; finite cases replayed by tools/skolem_one_sided_modular_replay.py"
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}6Provenance
View source, identifiers, and projection details
- Project
- skolem-problem-decidability-research
- Locator
- Original companion-matrix proof in this record, prepared 2026-07-28; finite cases replayed by tools/skolem_one_sided_modular_replay.py
- License
- CC0-1.0
- Public record
- R733
- Stable alias
- skolem-claim-reversible-negative-zero-modular-boundary
- Projection
- Reproduction fields are derived from the immutable record.
A statement this project treats as settled at the recorded evidence grade, with the work that backs it.