Problem packetResearch packetR256
Solutions are exactly projection-disjoint covers by Fibonacci-map cycles
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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
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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
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"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.",
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"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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"slug": "R255",
"title": "Exact orbit-cover counts for every prime below 500",
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}7Provenance
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