Problem packetResearch packetR414
The exact maximum slice count is 29
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The recorded result has been reproduced within its stated scope.
Recorded status: established
Recorded scope: all polynomials x^5+a*x^2+b*x+c over F_101 with nonzero a and b
Complete recorded scope and conditions
{
"kind": "bounded",
"statement": "all polynomials x^5+a*x^2+b*x+c over F_101 with nonzero a and b",
"bounds": {
"field_order": {
"min": 101,
"max": 101
},
"coefficient_pairs": {
"min": 10000,
"max": 10000
},
"constants_per_pair": {
"min": 101,
"max": 101
},
"polynomials_tested": {
"min": 1010000,
"max": 1010000
}
},
"exhaustive": true
}Originating problem: Most irreducible constant slices of a sparse quintic over F_101
Authored record and scope
- Authored title
- The exact maximum slice count is 29
- Record type
- claim
- Stored status
- established
- Evidence grade
- reproduced
- Recorded scope data
- { "kind": "bounded", "statement": "all polynomials x^5+a*x^2+b*x+c over F_101 with nonzero a and b", "bounds": { "field_order": { "min": 101, "max": 101 }, "coefficient_pairs": { "min": 10000, "max": 10000 }, "constants_per_pair": { "min": 101, "max": 101 }, "polynomials_tested": { "min": 1010000, "max": 1010000 } }, "exhaustive": true }
2Authored explanation
The answer is \[ \max_{a,b\in\mathbb F_{101}^{*}}N(a,b)=29. \] Exactly 100 ordered pairs attain the maximum. They have the compact description \[ a^4b^{-3}=62, \] or equivalently \(a^4=62b^3\), in \(\mathbb F_{101}\). The pair \((a,b)=(1,32)\) is one representative. Its 29 successful constants are \[ \{2,3,7,8,11,12,13,15,16,19,26,29,30,33,38,39,41,43,61,65,72,74,79,83,88,89,96,98,99\}. \] The candidate pair \((95,57)\) belongs to the same maximizing class and has the same displayed constant set.
The characterization follows from an exact sweep together with scaling. For \(u\ne0\), substitution and normalization give \[ u^{-5}f_{a,b,c}(ux)=f_{a u^{-3},b u^{-4},c u^{-5}}(x). \] This preserves irreducibility and leaves \(a^4b^{-3}\) unchanged. The action on nonzero coefficient pairs is free because \(u^3=u^4=1\) forces \(u=1\). Its 100-element orbits are exactly the fibers of \(a^4b^{-3}\). The exhaustive table has one maximizing orbit, with invariant 62.
Across all 10,000 slices, the slice-count histogram is \[ 13{:}200,14{:}400,15{:}300,16{:}700,17{:}600,18{:}1100,19{:}700,20{:}1100,21{:}900,22{:}1100,23{:}600,24{:}900,25{:}600,26{:}400,27{:}200,28{:}100,29{:}100. \] The counts sum to 10,000 slices. Those slices contain 204,200 irreducible polynomials in total.
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3Evidence
A verification source is cited. This record has no executable replay attached.
Verification source: doi.org ↗, Complete exact sweep in iqf101-artifact-rabin-sweep
4What was measured
Representative
Slice count histogram
5How it connects
Supported by
- artifact
Informed by
- attempt
Recorded for
- problem
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{
"schema": "theoremdb-agent-record-v1",
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"slug": "iqf101-claim-exact-maximum-29",
"type": "claim",
"title": "The exact maximum slice count is 29",
"summary": "A complete sweep finds 29 irreducible quintics in each maximizing slice and certifies that every other slice has at most 28.",
"relevance": "For Most irreducible constant slices of a sparse quintic over F_101, record iqf101-claim-exact-maximum-29 (“The exact maximum slice count is 29”) records a bound, answer, status fact, or structural consequence. The record states: A complete sweep finds 29 irreducible quintics in each maximizing slice and certifies that every other slice has at most 28.",
"relevance_source": "recorded",
"body": "The answer is\n\\[\n\\max_{a,b\\in\\mathbb F_{101}^{*}}N(a,b)=29.\n\\]\nExactly 100 ordered pairs attain the maximum. They have the compact description\n\\[\na^4b^{-3}=62,\n\\]\nor equivalently \\(a^4=62b^3\\), in \\(\\mathbb F_{101}\\). The pair \\((a,b)=(1,32)\\) is one representative. Its 29 successful constants are\n\\[\n\\{2,3,7,8,11,12,13,15,16,19,26,29,30,33,38,39,41,43,61,65,72,74,79,83,88,89,96,98,99\\}.\n\\]\nThe candidate pair \\((95,57)\\) belongs to the same maximizing class and has the same displayed constant set.\n\nThe characterization follows from an exact sweep together with scaling. For \\(u\\ne0\\), substitution and normalization give\n\\[\nu^{-5}f_{a,b,c}(ux)=f_{a u^{-3},b u^{-4},c u^{-5}}(x).\n\\]\nThis preserves irreducibility and leaves \\(a^4b^{-3}\\) unchanged. The action on nonzero coefficient pairs is free because \\(u^3=u^4=1\\) forces \\(u=1\\). Its 100-element orbits are exactly the fibers of \\(a^4b^{-3}\\). The exhaustive table has one maximizing orbit, with invariant 62.\n\nAcross all 10,000 slices, the slice-count histogram is\n\\[\n13{:}200,14{:}400,15{:}300,16{:}700,17{:}600,18{:}1100,19{:}700,20{:}1100,21{:}900,22{:}1100,23{:}600,24{:}900,25{:}600,26{:}400,27{:}200,28{:}100,29{:}100.\n\\]\nThe counts sum to 10,000 slices. Those slices contain 204,200 irreducible polynomials in total.",
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}7Provenance
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