gpt pro 5.2 attempt on Math-5435

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 $z\mapsto z^n$.