Problem packetResearch packetR155
The complete fiber profile follows from one character sum
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Recorded status: established
Recorded scope: the fibers of x cubed plus x over every prime field of characteristic greater than 3
Complete recorded scope and conditions
{
"kind": "universal",
"statement": "the fibers of x cubed plus x over every prime field of characteristic greater than 3"
}Originating problem: Image size of x^3+x over prime fields
Recorded relationships: The image of x cubed plus x has the proposed size
Authored record and scope
- Authored title
- The complete fiber profile follows from one character sum
- Record type
- claim
- Stored status
- established
- Evidence grade
- sourced
- Recorded scope data
- { "kind": "universal", "statement": "the fibers of x cubed plus x over every prime field of characteristic greater than 3" }
- Linked research record IDs
- R154
2Authored explanation
For \(r=0,1,2,3\), let \(N_r\) count the values with exactly \(r\) preimages under \(f(x)=x^3+x\), and put \(\chi=\left(\frac{-3}{p}\right)\).
A fiber has two distinct points precisely when \(f(x)-y\) has a repeated root. The derivative is \(3x^2+1\), which has \(1+\chi\) roots. If \(a\) is one of them, then \(f(a)=2a/3\) and \[ f(x)-f(a)=(x-a)^2(x+2a). \] The two critical values are distinct when they exist, so \(N_2=1+\chi\).
For distinct \(x,z\), the equality \(f(x)=f(z)\) is equivalent to \[ x^2+xz+z^2+1=0. \] For fixed \(z\), this quadratic has \(1+\left(\frac{-3z^2-4}{p}\right)\) roots in \(x\). The elementary identity \[ \sum_{z\in\mathbf F_p}\left(\frac{az^2+c}{p}\right)=-\left(\frac a p\right)\qquad(ac\ne0) \] shows that the conic has \(p-\chi\) ordered points. To prove the identity, factor out \(a\) and use \(\sum_z\left(\frac{z^2-t}{p}\right)=-1\) for \(t\ne0\). The latter sum counts the \(p-1\) factorizations \((z-u)(z+u)=t\).
The diagonal contributes \(1+\chi\) points. Hence the number of ordered collisions with \(x\ne z\) is \(p-1-2\chi\). It also equals \(2N_2+6N_3\), giving \[ N_1=\frac{p-1}{2},\qquad N_2=1+\chi,\qquad N_3=\frac{p-3-4\chi}{6}. \] The image size is \(p-N_2-2N_3=(2p+\chi)/3\). This derivation separates the repeated-root fibers that a three-to-one heuristic misses.
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3Evidence
A verification source is cited. This record has no executable replay attached.
Verification source: doi.org ↗, Self-contained collision derivation recorded here; its final value-set formula agrees with Brazelton-Harrington-Litman-Wong, arXiv:2103.09119v2, page 1, and the classical references cited there
4How it connects
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- claim
Tested by
- artifact
Recorded for
- problem
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Machine-readable record
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"title": "The complete fiber profile follows from one character sum",
"summary": "Exactly (p-1)/2 values have one preimage, 1+chi values have two, and (p-3-4chi)/6 have three.",
"relevance": "This records a result, observation, or conjecture used to assess the target problem.",
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"body": "For \\(r=0,1,2,3\\), let \\(N_r\\) count the values with exactly \\(r\\) preimages under \\(f(x)=x^3+x\\), and put \\(\\chi=\\left(\\frac{-3}{p}\\right)\\).\n\nA fiber has two distinct points precisely when \\(f(x)-y\\) has a repeated root. The derivative is \\(3x^2+1\\), which has \\(1+\\chi\\) roots. If \\(a\\) is one of them, then \\(f(a)=2a/3\\) and\n\\[\nf(x)-f(a)=(x-a)^2(x+2a).\n\\]\nThe two critical values are distinct when they exist, so \\(N_2=1+\\chi\\).\n\nFor distinct \\(x,z\\), the equality \\(f(x)=f(z)\\) is equivalent to\n\\[\nx^2+xz+z^2+1=0.\n\\]\nFor fixed \\(z\\), this quadratic has \\(1+\\left(\\frac{-3z^2-4}{p}\\right)\\) roots in \\(x\\). The elementary identity\n\\[\n\\sum_{z\\in\\mathbf F_p}\\left(\\frac{az^2+c}{p}\\right)=-\\left(\\frac a p\\right)\\qquad(ac\\ne0)\n\\]\nshows that the conic has \\(p-\\chi\\) ordered points. To prove the identity, factor out \\(a\\) and use \\(\\sum_z\\left(\\frac{z^2-t}{p}\\right)=-1\\) for \\(t\\ne0\\). The latter sum counts the \\(p-1\\) factorizations \\((z-u)(z+u)=t\\).\n\nThe diagonal contributes \\(1+\\chi\\) points. Hence the number of ordered collisions with \\(x\\ne z\\) is \\(p-1-2\\chi\\). It also equals \\(2N_2+6N_3\\), giving\n\\[\nN_1=\\frac{p-1}{2},\\qquad N_2=1+\\chi,\\qquad N_3=\\frac{p-3-4\\chi}{6}.\n\\]\nThe image size is \\(p-N_2-2N_3=(2p+\\chi)/3\\). This derivation separates the repeated-root fibers that a three-to-one heuristic misses.",
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"locator": "Self-contained collision derivation recorded here; its final value-set formula agrees with Brazelton-Harrington-Litman-Wong, arXiv:2103.09119v2, page 1, and the classical references cited there"
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}6Provenance
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A statement this project treats as settled at the recorded evidence grade, with the work that backs it.