Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

Will Cook

Long paper

Paths in Polynomial Lemniscates: A Degree-Seven Counterexample and Radial Connections

Erdős #1041 By Will Cook 29 September 2026 Authorship, AI use and citation

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Summary. The record gives the rational estimates and geometric proof for ani's degree-seven polynomial: every connected set in its strict unit lemniscate joining two roots has Hausdorff measure greater than two. It develops contained paths for trinomials, collinear roots and two sparse families. Ordinary analytic arguments for critical-value criteria retain their unformalised inputs beside the statements. Critical-value moments, counterexamples to prescribed paths and questions about flow-strip geometry complete the account. The one-polynomial formal statement and correspondence with the historical wording remain distinct.

This paper covers the exact geometry of ani's one-polynomial counterexample, proved positive families and critical-value arguments with their unformalised inputs identified; formal-statement status and historical correspondence remain owned by the claims registry.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or the current status of Erdős #1041, including the later one-polynomial Formal Conjectures path-image Hausdorff-length check and its unreviewed historical curve-length correspondence; those distinctions belong to docs/claims.json in the mathematics repository.

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In this paper

Two failures of straight segments.

Proposition 10.7 (two counterexamples to straight-path assertions). There is a monic quintic with all roots in the open unit disc and a non-root critical point $c$ whose unique nearest root has a point on the straight segment to $c$ outside ${| f | < 1}$ . There is also a monic cubic with all roots in the open unit disc such that the midpoint of every pair of distinct roots lies outside ${| f | < 1}$ .

Proof. For the quintic, put $p = 999 / 1000$ and $a = (901 / 902) p$ , and take

Q (z) = (z - a) (z^{2} + p^{2}) (z^{2} + \frac{902}{901} p z + p^{2}) .

Its roots are

a, i p, - i p, p \frac{- 451 + 780 i}{901}, p \frac{- 451 - 780 i}{901} .

Since $451^{2} + 780^{2} = 901^{2}$ , the last four roots have modulus $p < 1$ , while $0 < a < p$ . Also $Q^{'} (0) = 0$ and $Q (0) = - a p^{4} \neq 0$ . Thus $a$ is the uniquely nearest root to the non-root critical point $0$ . At one tenth of that segment, exact rational arithmetic gives

| Q (a / 10) | = \frac{9 a}{10} (\frac{a^{2}}{100} + p^{2}) (\frac{a^{2}}{100} + \frac{11 p^{2}}{10}) > 1.

For the cubic, take $g (z) = z^{3} - (99 / 100)^{3}$ . If $z, w$ are distinct roots, then $z + w$ is the negative of the third root, so

g (\frac{z + w}{2}) = - \frac{9}{8} {(\frac{99}{100})}^{3}, \frac{9}{8} {(\frac{99}{100})}^{3} = \frac{8732691}{8000000} > 1.

All three roots have modulus $99 / 100$ . ◻

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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