Erdős Problem #114

Erdős Problem 114: lengths of polynomial lemniscates

Statement and scope

For a monic degree-$n$ polynomial $p$, is the length of $\{z:|p(z)|=1\}$ maximal at $p(z)=z^n-1$? The assertion for every $n$ is unresolved in this write-up. We prove the maximization within the entire translated binomial family and compute its value, including the constant term in its asymptotic expansion.

This develops the valid length integral from the earlier GPT-5.2 attempt. Current statement: https://www.erdosproblems.com/114. Tao proves the full assertion for all sufficiently large degrees in https://arxiv.org/abs/2512.12455; that theorem is context, not an input to the restricted proof below. No novelty is claimed.

The inverse-image length formula

Translation of the variable does not affect length, so consider $p(z)=z^n-c$. For $n>1$ put $s=1-1/n\in(0,1)$. On the level curve, $z^n=c+e^{it}$. Away from the finitely many singular parameters, each of the $n$ inverse branches has speed \[ \frac1n |c+e^{it}|^{1/n-1}. \] Summing branch lengths gives \[ L_n(c)=\int_0^{2\pi}|c+e^{it}|^{-s}\,dt. \tag{1} \] This counts each smooth arc once; the branch singularity when $|c|=1$ is integrable because $s<1$. Individual branches need not return to their initial values after one circuit; summing all branches is what justifies $1$.

Rotation of $t$ shows that $1$ depends only on $r=|c|$. For $0\le r<1$, the binomial series and termwise integration, justified by absolute uniform convergence for each fixed $r<1$, give \[ L_n(r)=2\pi\sum_{j=0}^\infty \left(\frac{(s/2)_j}{j!}\right)^2 r^{2j}. \tag{2} \] Indeed expand $(1+r e^{it})^{-s/2}$ and its conjugate; only equal indices survive integration. All coefficients are positive, so $L_n(r)$ strictly increases on $[0,1)$.

For $r>1$, factoring out $r$ gives \[ L_n(r)=r^{-s}L_n(1/r) =2\pi\sum_{j=0}^\infty \left(\frac{(s/2)_j}{j!}\right)^2 r^{-s-2j}, \tag{3} \] which strictly decreases. Continuity at $r=1$ follows by dominated convergence: for $r\in[1/2,3/2]$, \[ |r+e^{it}|^2=(r-1)^2+4r\cos^2(t/2)\ge2\cos^2(t/2), \] and $|\cos(t/2)|^{-s}$ is integrable. Equations $2$--$3$ prove that the unique maximizing modulus is $|c|=1$. Thus the proposed extremizer is correct throughout the family $(z-a)^n-c$.

Exact length and a justified asymptotic

At $c=1$, $1$ yields \[ L_n(1)=2^{1/n}B\left(\frac1{2n},\frac12\right), \tag{4} \] where $B$ is the beta integral. For $\epsilon\downarrow0$, \[ B(\epsilon,\tfrac12)-\frac1\epsilon =\int_0^1 t^{\epsilon-1}\big((1-t)^{-1/2}-1\big)\,dt =2\log2+O(\epsilon). \tag{5} \] The limit integral is $2\log2$, by $u=\sqrt{1-t}$. The error estimate follows from $|t^\epsilon-1|\le\epsilon|\log t|$ and the integrability of $t^{-1}|\log t|((1-t)^{-1/2}-1)$ at both endpoints. Combining $4$--$5$ with $2^{1/n}=1+(\log2)/n+O(n^{-2})$ gives \[ L_n(1)=2n+4\log2+O(n^{-1}). \] For $n=1$, every monic linear polynomial has a unit-circle lemniscate of length $2\pi$.

Remaining gap

The positive-coefficient comparison uses the two-term form of the polynomial. It supplies no upper bound for a general monic polynomial in the degrees not covered by the cited large-degree theorem. That is the missing comparison for the full question.

COMPLETION ESTIMATE

COMPLETION: 45%

Erdős Problem #114

1) FORMAL RESTATEMENT

Let $p(z)\in\mathbb C[z]$ be a monic polynomial of degree $n\ge1$. Consider the lemniscate \[ L(p):=\{z\in\mathbb C:\ |p(z)|=1\}. \] Assume $L(p)$ is a rectifiable curve (for polynomials it is a finite union of real-analytic arcs except at finitely many singular points). Let $\ell(p)$ denote its total Euclidean arclength. Define \[ f(n):=\sup\{\ell(p):\ p\text{ monic of degree }n\}. \] Question: is $\ell(p)$ maximized by $p(z)=z^n-1$? (Up to the natural symmetries: rotation and translation as stated in the problem text.)

2) QUICK LITERATURE/CONTEXT CHECK

The problem text itself records a sequence of bounds and partial results, culminating in the statement that Tao $2025$ proved $z^n-1$ is the unique maximizer for all sufficiently large $n$. I do not re-prove those results here. I only compute the exact arclength for $p(z)=z^n-1$ and verify the claimed $2n+O(1)$ behavior.

3) ATTACK PLAN

• Compute $\ell(z^n-1)$ in a closed form, to at least have the conjectured extremal value explicitly.
• For small $n$ do a sanity check with explicit parametrizations.

4) WORK

Lemma 114.1 (the case $n=1$ is trivial).

If $n=1$ and $p(z)=z-a$ is monic linear, then $L(p)$ is the circle $|z-a|=1$ of length $2\pi$. Hence $f(1)=2\pi$, and every monic linear polynomial is a maximizer.

Proof. $|p(z)|=1$ is exactly $|z-a|=1$. The Euclidean circumference is $2\pi$. $\square$

Lemma 114.2 (exact formula for $\ell(z^n-1)$).

Let $p_n(z)=z^n-1$. Then the arclength of $L(p_n)$ is \[ \ell(p_n)=2^{1/n}\,\mathrm B\Big(\frac{1}{2n},\frac12\Big) \] where $\mathrm B$ is the Beta function. In particular, $\ell(p_n)=2n+O(1)$ as $n\to\infty$.

Proof. On the unit circle, write $z=e^{it}$. Then \[ |p_n(e^{it})|=|e^{int}-1|=2\,\big|\sin\frac{nt}{2}\big|. \] The lemniscate $|z^n-1|=1$ is the preimage under the map $w=z^n$ of the circle $|w-1|=1$. A convenient parametrization is obtained by parametrizing that circle: \[ w(\theta)=1+e^{i\theta},\qquad 0\le\theta<2\pi. \] Then $|w(\theta)-1|=1$, and $|w(\theta)|=|1+e^{i\theta}|=2|\cos(\theta/2)|$. Choose a branch of the $n$th root and set \[ z(\theta)=w(\theta)^{1/n}. \] As $\theta$ runs $0\to2\pi$, $w(\theta)$ traverses the circle once and $z(\theta)$ traces all $n$ components of the lemniscate (the $n$ choices of the $n$th root correspond to rotations by $2\pi/n$, contributing equally to length). Differentiating, \[ z'(\theta)=\frac1n w(\theta)^{1/n-1} w'(\theta)=\frac1n w(\theta)^{1/n-1} \cdot i e^{i\theta}. \] Hence \[ |z'(\theta)|=\frac1n |w(\theta)|^{1/n-1}. \] Therefore the length contributed by one branch over $\theta\in[0,2\pi]$ is \[ \int_0^{2\pi} |z'(\theta)|\,d\theta=\frac1n\int_0^{2\pi} |w(\theta)|^{1/n-1}\,d\theta. \] There are $n$ congruent branches (rotations), so the total length is \[ \ell(p_n)=n\cdot \frac1n\int_0^{2\pi} |w(\theta)|^{1/n-1}\,d\theta=\int_0^{2\pi} \big(2|\cos(\theta/2)|\big)^{1/n-1}\,d\theta. \] Using symmetry $|\cos(\theta/2)|$ is even about $\theta=\pi$, we get \[ \ell(p_n)=2\int_0^{\pi} \big(2\cos(\theta/2)\big)^{1/n-1}\,d\theta. \] Substitute $u=\theta/2$: \[ \ell(p_n)=4\int_0^{\pi/2} (2\cos u)^{1/n-1}\,du = 4\,2^{1/n-1}\int_0^{\pi/2} (\cos u)^{1/n-1}\,du. \] Now apply the standard Beta integral \[ \int_0^{\pi/2} (\cos u)^{\alpha-1}(\sin u)^{\beta-1}\,du = \frac12\,\mathrm B\Big(\frac\alpha2,\frac\beta2\Big) \] with $\alpha=1/n$ and $\beta=1$. This yields \[ \int_0^{\pi/2} (\cos u)^{1/n-1}\,du = \frac12\,\mathrm B\Big(\frac{1}{2n},\frac12\Big). \] Plugging in gives \[ \ell(p_n)=4\,2^{1/n-1}\cdot \frac12\,\mathrm B\Big(\frac{1}{2n},\frac12\Big)=2^{1/n}\,\mathrm B\Big(\frac{1}{2n},\frac12\Big). \]

For the asymptotic: using $\mathrm B(x,1/2)=\Gamma(x)\Gamma(1/2)/\Gamma(x+1/2)$ and $\Gamma(x)\sim 1/x$ as $x\to0^+$, we have \[ \mathrm B\Big(\frac{1}{2n},\frac12\Big)\sim \frac{\Gamma(1/2)}{\Gamma(1/2)}\cdot 2n = 2n, \] and $2^{1/n}=1+O(1/n)$. Thus $\ell(p_n)=2n+O(1)$. $\square$

FAST REALITY CHECK (numerical values for $\ell(z^n-1)$).

Using the closed form in Lemma 114.2, the following values were computed (rounded): \[ \begin{array}{c|c} n & \ell(z^n-1)\\\hline 1 & 6.283185307\ (\approx 2\pi)\\ 2 & 7.416298709\\ 3 & 9.179724222\\ 4 & 11.07002052\\ 5 & 13.00681138\\ 6 & 14.96573219 \end{array} \] These are consistent with $\ell(z^n-1)\approx 2n+\text{constant}$.

5) VERIFICATION

• Lemma 114.2 differentiates $z(\theta)=w(\theta)^{1/n}$; branch issues do not affect arclength because we integrate $|z'|$ and then sum over the $n$ rotationally congruent branches.
• Endpoint singularities occur when $w(\theta)=0$ (i.e. $\theta=\pi$), but the integrand behaves like $|\cos(\theta/2)|^{1/n-1}$, which is integrable since $1/n-1>-1$ for all $n\ge1$.

6) FINAL

UNRESOLVED.

(i) Strongest fully proved partial result obtained here. An exact closed form for the candidate extremal value $\ell(z^n-1)$ and its $2n+O(1)$ asymptotic (Lemma 114.2).

(ii) Exact first gap. A general upper bound $\ell(p)\le \ell(z^n-1)$ for every monic degree-$n$ polynomial $p$.

(iii) Top 3 next moves.

1. Prove a variational inequality showing $z^n-1$ is a global maximizer among monic polynomials (beyond the local/asymptotic results cited in the problem text).
2. Establish sharp universal bounds on $\ell(p)$ in terms of $n$ with the correct leading constant $2$.
3. For small $n$ (e.g. $n=2,3$), attempt classification of maximizers by reducing to a finite-dimensional optimization problem.

(iv) Minimal counterexample structure. A counterexample would be a monic polynomial $p$ of degree $n$ with $\ell(p)>\ell(z^n-1)$. Such a $p$ would necessarily have a lemniscate with unusually large total curvature/length relative to the $n$-fold covering map $z\mapsto z^n$.