Erdős Problem #1047

Erdős Problem 1047: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Statement and explicit counterexample

No. Let \[ f(z)=z^6-z,\qquad c=\frac{291}{500},\qquad r=\frac{137}{200}. \tag{1} \] The polynomial has six distinct roots. We prove that $\{|f|\le c\}$ has exactly six disjoint connected components and that the component containing zero is not convex. This uses the example attributed to the referee of Goodman's paper in https://www.erdosproblems.com/1047. The rational test points agree with the certificate in the linked Aristotle proof; all inequalities needed here are proved directly with rational bounds. No local Lean verification is claimed.

Six separate components

The roots are zero and the fifth roots of unity. At a critical point $\zeta$, the identity $6\zeta^5=1$ gives \[ |f(\zeta)|=\frac56|\zeta|=5\,6^{-6/5}=:\kappa. \] The exact rational inequality \[ \left(\frac{291}{500}\right)^5<\frac{5^5}{6^6} \tag{2} \] shows that $c<\kappa$. Choose $d$ strictly between them. There are no critical values over the disk $D(0,d)$. A polynomial is a proper map, so its restriction over this disk is a six-sheeted covering without branch points. Since the disk is simply connected, there are six single-valued holomorphic inverse branches $g_j$ on it, with disjoint images. One can obtain these by continuing the six local inverses at zero: the absence of critical values permits continuation along every path, properness prevents escape to infinity, and the monodromy theorem makes the continuation independent of the path.

Each set $g_j(\overline{D(0,c)})$ is compact and connected. They are pairwise disjoint and their union is exactly $\{|f|\le c\}$. A finite family of disjoint compact sets has positive mutual distances, so these are precisely its six connected components. This also checks the closed-sublevel-set convention of the question; components are not merely separated before taking closures.

Two points in the central component

Let $\omega=e^{2\pi i/5}$. For $0\le t\le r$, \[ f(t)=t^6-t,\qquad f(t\omega)=\omega(t^6-t). \] The function $t-t^6$ is increasing on this interval because $6r^5<1$. Exact rational arithmetic gives \[ 6\left(\frac{137}{200}\right)^5<1, \qquad \frac{137}{200}-\left(\frac{137}{200}\right)^6 <\frac{291}{500}. \tag{3} \] Thus both segments $[0,r]$ and $[0,r\omega]$ lie in $\{|f|<c\}$. Their endpoints $r,r\omega$ belong to the same component as zero.

Their midpoint is outside the entire sublevel set

Write \[ m=\frac r2(1+\omega)=R e^{i\pi/5},\qquad R=r\cos(\pi/5)=\frac r4(1+\sqrt5). \] Since $m^5=-R^5$, we have $|f(m)|=R+R^6$. The rational lower bound $\sqrt5>559/250$ follows by squaring positive numbers. It gives \[ R>R_0:=\frac{110833}{200000},\qquad R_0+R_0^6>\frac{291}{500}.\tag{4} \] The last inequality is again an exact integer comparison after multiplication by $200000^6$. For an easily checked margin it is enough to note that $R_0>554/1000$ and \[ \frac{554}{1000}+\left(\frac{554}{1000}\right)^6 >\frac{291}{500}; \] its left side is greater than $0.5829$. Therefore $|f(m)|>c$. The midpoint of two points in the central component is outside that component, so it is not convex.

Assessment

The maximal number of components, distinctness of the roots, strict separation of the closed components, and nonconvexity witness are all verified. The condition in the question means a level with as many components as distinct roots; our level is strictly below the first critical-value modulus and meets that condition. The standard inverse-function and monodromy principles are the only general complex-analytic inputs.

Completion rate estimate: 100%.