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Problem packetResearch packetR256

R256Recorded identity

Solutions are exactly projection-disjoint covers by Fibonacci-map cycles

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Authored summary

The graph of a solution is an invariant union of cycles of T(x,y)=(y,x+y), and the first-coordinate projections partition the field.

The author records a mathematical identity.

Recorded status: established

Recorded scope: every prime p and every function from F_p to F_p satisfying the equation

Complete recorded scope and conditions
{
  "kind": "universal",
  "statement": "every prime p and every function from F_p to F_p satisfying the equation"
}

Originating problem: Power-of-two solution counts for a finite-field functional equation

Recorded relationships: Exact orbit-cover counts for every prime below 500

Authored record and scope
Authored title
Solutions are exactly projection-disjoint covers by Fibonacci-map cycles
Record type
claim
Stored status
established
Evidence grade
mathematical_identity
Recorded scope data
{ "kind": "universal", "statement": "every prime p and every function from F_p to F_p satisfying the equation" }
Linked research record IDs
R255

2Authored explanation

Suppose \[ f(f(x))=f(x)+x \] for every \(x\in\mathbb F_p\). Equal values \(f(x)=f(y)\) give \(x=y\), so \(f\) is injective and hence a permutation. The equation at zero then gives \(f(0)=0\).

Define the invertible Fibonacci map \[ T_p(u,v)=(v,u+v) \] on \(\mathbb F_p^2\), and let \[ G_f=\{(x,f(x)):x\in\mathbb F_p\}. \] The functional equation gives \[ T_p(x,f(x))=(f(x),x+f(x))=(f(x),f(f(x)))\in G_f. \] Thus \(G_f\) is a union of complete \(T_p\)-cycles. Its first-coordinate projection is bijective. Each selected cycle consequently has distinct first coordinates, and the projection sets of the selected cycles partition \(\mathbb F_p\). The fixed cycle \(((0,0))\) is forced.

Conversely, take any family of \(T_p\)-cycles whose first-coordinate projections are individually injective and jointly partition \(\mathbb F_p\). Their union is the graph of a unique function \(f\). Invariance under \(T_p\) says that the graph contains \((f(x),x+f(x))\), so its value at \(f(x)\) is \(f(f(x))=x+f(x)\). This constructs every solution.

Therefore \(N(p)\) is exactly the number of exact covers of \(\mathbb F_p\) by the first-coordinate projections of eligible \(T_p\)-cycles, with cycles retained as distinct choices when their projections coincide.

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3Evidence

Replay package: source only

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Verification source: sites.math.rutgers.edu ↗, Independent graph-invariance and converse proof, 2026-07-24

4What was measured

5How it connects

Recorded for

Machine-readable record

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  "schema": "theoremdb-agent-record-v1",
  "ref": "R256",
  "content_hash": null,
  "slug": "ffe-claim-exact-orbit-cover",
  "type": "claim",
  "title": "Solutions are exactly projection-disjoint covers by Fibonacci-map cycles",
  "summary": "The graph of a solution is an invariant union of cycles of T(x,y)=(y,x+y), and the first-coordinate projections partition the field.",
  "relevance": "For Power-of-two solution counts for a finite-field functional equation, record ffe-claim-exact-orbit-cover (“Solutions are exactly projection-disjoint covers by Fibonacci-map cycles”) records a bound, answer, status fact, or structural consequence. The record states: The graph of a solution is an invariant union of cycles of T(x,y)=(y,x+y), and the first-coordinate projections partition the field.",
  "relevance_source": "recorded",
  "body": "Suppose\n\\[\nf(f(x))=f(x)+x\n\\]\nfor every \\(x\\in\\mathbb F_p\\). Equal values \\(f(x)=f(y)\\) give \\(x=y\\), so \\(f\\) is injective and hence a permutation. The equation at zero then gives \\(f(0)=0\\).\n\nDefine the invertible Fibonacci map\n\\[\nT_p(u,v)=(v,u+v)\n\\]\non \\(\\mathbb F_p^2\\), and let\n\\[\nG_f=\\{(x,f(x)):x\\in\\mathbb F_p\\}.\n\\]\nThe functional equation gives\n\\[\nT_p(x,f(x))=(f(x),x+f(x))=(f(x),f(f(x)))\\in G_f.\n\\]\nThus \\(G_f\\) is a union of complete \\(T_p\\)-cycles. Its first-coordinate projection is bijective. Each selected cycle consequently has distinct first coordinates, and the projection sets of the selected cycles partition \\(\\mathbb F_p\\). The fixed cycle \\(((0,0))\\) is forced.\n\nConversely, take any family of \\(T_p\\)-cycles whose first-coordinate projections are individually injective and jointly partition \\(\\mathbb F_p\\). Their union is the graph of a unique function \\(f\\). Invariance under \\(T_p\\) says that the graph contains \\((f(x),x+f(x))\\), so its value at \\(f(x)\\) is \\(f(f(x))=x+f(x)\\). This constructs every solution.\n\nTherefore \\(N(p)\\) is exactly the number of exact covers of \\(\\mathbb F_p\\) by the first-coordinate projections of eligible \\(T_p\\)-cycles, with cycles retained as distinct choices when their projections coincide.",
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    "kind": "universal",
    "statement": "every prime p and every function from F_p to F_p satisfying the equation"
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  "reproduction": {
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    "locator": "Independent graph-invariance and converse proof, 2026-07-24"
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  "relations": [
    {
      "slug": "R255",
      "title": "Exact orbit-cover counts for every prime below 500",
      "object_type": "claim",
      "relation": "supports",
      "direction": "outgoing"
    },
    {
      "slug": "fibonacci-functional-equation-prime-count",
      "title": "fibonacci functional equation prime count",
      "object_type": "problem",
      "relation": "recorded_for",
      "direction": "outgoing"
    }
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}

7Provenance

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