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Claim. If $f(z)=(z-z_0)(z-z_1)(z-z_2)$ with $\lvert z_j\rvert<1$, then two roots of $f$, counted with multiplicity, can be joined inside $\{z:\lvert f(z)\rvert<1\}$ by a polygonal path of length less than $2$, which is the cubic case of [[problems/polynomials/E1041/_index|Problem 1041]]. Alexey Borisov registered the claim on the site's proof-claims tab on 2026-09-18 and published the preprint The cubic case of the Erdős–Herzog–Piranian problem on Zenodo the same day, explaining that Borisov lacks arXiv access. The argument has two steps. The local step: if $c$ is a critical point of a monic cubic $F$ with $\lvert F(c)\rvert<1$ and $a$ is a root with $d=a-c$ of modulus less than $1$, then $\lvert F\rvert<1$ along the whole segment $[c,a]$, because $F'(c)=0$ and $F(a)=0$ give the identity $F(c+td)=F(c)(1-t^2)-d^3t^2(1-t)$, so that $\lvert F(c+td)\rvert<1-t^3$ for $0<t<1$. The global step: some critical point $c$ with $\lvert F(c)\rvert<1$ lies at distance less than $1$ from two of the roots. For this the roots are translated by their centroid $g$ to $a_j=z_j-g$ and parametrized as $a_j=p\omega^j+q\omega^{-j}$ with $\omega=e^{2\pi i/3}$; since $\sum_j\lvert a_j\rvert^2=3(\lvert p\rvert^2+\lvert q\rvert^2)$ and $\sum_j\lvert z_j\rvert^2=3\lvert g\rvert^2+\sum_j\lvert a_j\rvert^2<3$, the hypothesis $\lvert z_j\rvert<1$ gives $\lvert p\rvert^2+\lvert q\rvert^2<1$. After a rotation, $p=Xe^{i\varphi}$ and $q=Ye^{-i\varphi}$ with $X,Y\ge0$, the translated cubic is $w^3-3XYw-(p^3+q^3)$ with critical points $\pm\sqrt{XY}$, the critical point $c=g+\delta\sqrt{XY}$ with the sign $\delta$ chosen so that $\delta\cos3\varphi\le0$ has $\lvert f(c)\rvert\le X^3+Y^3<1$, and an elementary trigonometric estimate shows that at least two roots lie at distance less than $1$ from $c$, which gives the two segments. The Zenodo abstract states that with the quartic case the result completes the affirmative answer for every degree at most four.

Submission note. Posted to erdosproblems.com as a proof claim by Borisov Alexey (account AlxBorisov17) on 18 September 2026:

Problem #1041 Claim: the cubic case admits a positive answer. If f(z) = (z−z₀)(z−z₁)(z−z₂) with |z_j| < 1, then two roots of f, counted with multiplicity, may be joined inside {|f| < 1} by a polygonal path of length < 2. The argument relies on a local fact. Take a critical point c of a monic cubic and a root a with |f(c)| < 1 and |a − c| < 1. Then the whole segment from c to a remains in {|f| < 1}, since near a critical point the polynomial behaves locally like a cube and its modulus grows too slowly to reach 1 before the endpoint. So it suffices to find one critical point lying within distance 1 of two roots. Such a point is obtained by centering the roots at their centroid and writing them as a_j = pω^j + qω^{−j} with ω = e^{2πi/3}. This yields |p|² + |q|² < 1, and the polynomial turns into a depressed cubic with explicit critical points. A brief trigonometric check produces the required path. Preprint: https://doi.org/10.5281/zenodo.22832490 Notes: Unfortunately I do not have access to arXiv, so the preprint is published on Zenodo instead: https://doi.org/10.5281/zenodo.22832490

Covers. Degree three only, with a path through a critical point. The quartic case is the separate claim on Pendyala 2026; the general question is claimed false in degree seven on ani 2026 (counterexample).

Depends on. No page of this wiki.

Standing. Claimed. The claim carries four comments on the tab. A reader wrote on 2026-09-19 and 2026-09-23 that Pendyala had proved the cubic case in June 2026 together with the quartic case and had sent the proof privately, saying that it does not follow from the quartic proof; Borisov replied on 2026-09-19 that the quartic preprint states no cubic result and that the cubic case does not follow from it, and added on 2026-09-26 that Borisov's path has a vertex at a critical point. A private communication gets no page. The site labels the problem FALSIFIABLE (page last edited 06 December 2025), and the preprint is not refereed.