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

Will Cook

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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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Quintics with two missing coefficients

The missing cubic and quadratic terms of $z^{5} + a z^{4} + b z + c$ give three identities among the real and radial coordinates of its roots. We first use those identities to select two indices whose radial segments will be contained. For $0 < r < 2$ and $0 \leq s_{i} \leq 1$ , $x_{i}^{2} \leq s_{i}$ , put

E_{i} = s_{i}^{4} (s_{i} + r^{2} + 2 r x_{i}) .

Theorem 9.4 (a consequence of three moment identities). Suppose

\begin{matrix} (10) & \sum_{i = 0}^{4} x_{i} = - r, \sum_{i = 0}^{4} (2 x_{i}^{2} - s_{i}) = r^{2}, \sum_{i = 0}^{4} (4 x_{i}^{3} - 3 s_{i} x_{i}) = - r^{3} . \end{matrix}

Then at least one of the ten pairs $0 \leq i < j \leq 4$ satisfies $E_{i} < 1$ and $E_{j} < 1$ .

The theorem selects two distinct indices. It does not assert that the corresponding complex root values are distinct.

The polynomial must have no $z^{3}$ or $z^{2}$ term in the coordinates used here. For example, $z^{5} + b z + c$ is included, but a general quintic with all coefficients nonzero is not. The root-location assumption alone does not imply the three moment identities. Those identities are precisely what lets the finite inequality find two radial segments.

Theorem 9.5 (a quintic with two missing coefficients). Let

p (z) = z^{5} + a z^{4} + b z + c

and suppose its five zero occurrences $w_{0}, \dots, w_{4}$ lie in the closed unit disc. At least two distinct indices satisfy

\begin{matrix} (11) & | b w_{i} + c | \leq 1. \end{matrix}

If $a \neq 0$ , two indices can be chosen with strict inequalities. If $a = 0$ , every index satisfies (11), and equality holds exactly when $| w_{i} | = 1$ .

For open-disc zeros, two zero occurrences are joined inside ${| p | < 1}$ by a curve of length below $2$ : use the two radial spokes through $0$ when their values are distinct, and the constant path when the selected occurrences have the same value.

Proof. If $a = 0$ , the root equation gives $| b w_{i} + c | = | w_{i} |^{5}$ for every index, proving (11) and its equality clause. For $a \neq 0$ , write $a = r e^{i ϕ}$ with $r > 0$ and set $z_{i} = e^{- i ϕ} w_{i}$ . The missing $z^{3}$ and $z^{2}$ coefficients and Newton’s identities give

\begin{matrix} (12) & \sum z_{i} = - r, \sum z_{i}^{2} = r^{2}, \sum z_{i}^{3} = - r^{3} . \end{matrix}

At a zero,

| b w_{i} + c | = | w_{i} |^{4} | w_{i} + a | .

Thus, for $x_{i} = ℜ z_{i}$ and $s_{i} = | z_{i} |^{2}$ , the squared modulus $| b w_{i} + c |^{2}$ is exactly $E_{i}$ and (12) becomes (10). The third moment gives $r^{3} = | \sum_{i} z_{i}^{3} | \leq 5 < 8$ , so $0 < r < 2$ as required by the finite inequality.

We need to rule out four indices with squared modulus at least $1$ . The three known power sums determine the sum of any real harmonic polynomial of degree at most three over the roots. On the unit circle, $s^{4} (s + r^{2} + 2 r x) \geq 1$ reduces to $x \geq - r / 2$ . We therefore use the harmonic extension of $(- r / 2 - x) (1 - x)^{2}$ , which is nonpositive on that part of the circle. In the variables $x = ℜ z$ and $s = | z |^{2}$ , this extension is

H_{r} (x, s) = (- r / 2 - x) (1 - x)^{2} + (1 - s) (1 - \frac{r}{4} - \frac{3 x}{4}) .

The useful feature of this choice is the expansion

H_{r} (ℜ z, | z |^{2}) = 1 - \frac{3 r}{4} + ℜ [(r - \frac{7}{4}) z + (1 - \frac{r}{4}) z^{2} - \frac{z^{3}}{4}] .

It is harmonic, and summing this expression with the three moments in (12) gives

\begin{matrix} (13) & \sum_{i = 0}^{4} H_{r} (x_{i}, s_{i}) = 5 - 2 r . \end{matrix}

On the unit circle its value is $(- r / 2 - x) (1 - x)^{2}$ , and

4 - 2 r - (- r / 2 - x) (1 - x)^{2} = (x + 1) (\frac{(2 - r) (3 - x)}{2} + (1 - x)^{2}) \geq 0.

The maximum principle therefore gives $H_{r} \leq 4 - 2 r$ throughout the disc. For an index with $s^{4} (s + r^{2} + 2 r x) \geq 1$ , one has $s > 0$ and $s^{- 4} \geq 1 + 4 (1 - s)$ by convexity on $(0, 1]$ . Consequently

2 r (x + r / 2) \geq s^{- 4} - s \geq 5 (1 - s) .

Put $d = x + r / 2$ , $u = 1 - x$ , $A = 1 - r / 4 - 3 x / 4$ , and $P = - u^{2} + (2 r / 5) A$ . The assumption $s^{4} (s + r^{2} + 2 r x) \geq 1$ gives $0 \leq d < 2$ . Since $H_{r} = - d u^{2} + (1 - s) A$ , it is nonpositive when $A \leq 0$ ; when $A > 0$ , the preceding bound gives $H_{r} \leq d P$ . The remaining estimate is the exact sum-of-squares identity

\frac{1}{31} - P = \frac{2}{5} {(d - \frac{38}{31} + \frac{1}{4} (u - \frac{3}{31}))}^{2} + \frac{31}{40} {(u - \frac{3}{31})}^{2} \geq 0,

which gives $P \leq 1 / 31$ , and hence $H_{r} (x, s) \leq 2 / 31$ whether $P$ is positive or nonpositive. Thus $H_{r} \leq 2 / 31$ whenever the squared modulus is at least $1$ . If four indices had squared modulus at least $1$ , (13) would give

5 - 2 r \leq 4 - 2 r + \frac{8}{31} < 5 - 2 r,

a contradiction. This proves the two-index conclusion when $a \neq 0$ .

For either value of $a$ , at a zero $w$ ,

p (t w) = (1 - t) c + (t - t^{4}) (b w + c) - t^{4} (1 - t) w^{5} .

For $0 \leq t \leq 1$ the three nonnegative weights sum to $1 - t^{5}$ . The tail bound, $| w | \leq 1$ , and $| c | \leq 1$ therefore keep each selected radial segment in ${| p | \leq 1}$ . Under the open-disc hypothesis, $| c | < 1$ . Its coefficient $1 - t$ is positive for $0 \leq t < 1$ , so

| p (t w) | < (1 - t) + (t - t^{4}) + t^{4} (1 - t) = 1 - t^{5} \leq 1.

At $t = 1$ the value is zero. Thus both segments lie in ${| p | < 1}$ , and their total length is $| w_{i} | + | w_{j} | < 2$ . ◻

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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