Erdős Problem #116

FORMAL RESTATEMENT

Erdős #116; reconstruction with an explicit harmonic-analysis input, 2026-09-23. Let $p(z)=\prod_{j=1}^n(z-w_j)$ be monic, with $|w_j|\leq1$. Must the planar area of $\Lambda_p=\{z:|p(z)|<1\}$ be bounded below by a constant times $1/\log n$? We prove \[ \operatorname{Area}(\Lambda_p)\geq\frac{c}{\log(2n)}\qquad(n\geq1) \tag{1} \] for an absolute $c>0$.

QUICK LITERATURE AND CONTEXT CHECK

Krishnapur, Lundberg, and Ramachandran proved this in 2025 by combining a zero-pushing construction with the Nazarov--Polterovich--Sodin theorem for harmonic functions. We reconstruct that route. For this lower bound alone, pushing zeros while working on the half-disc suffices; it gives an explicit degree-$4n$ auxiliary polynomial. The harmonic sign-area theorem is the sole deep external input. This is a reconstruction of the known method, not a novelty claim.

OBJECTIVE AND ATTACK PLAN

First dominate $|p|^4$ on $|z|\leq1/2$ by the modulus of a monic polynomial $Q$ all of whose roots lie on the unit circle. Then $\log|Q|$ is harmonic and vanishes at the origin. Its number of boundary sign changes is controlled by its degree, allowing a harmonic sign-area estimate.

WORK

The stated external theorem

We use Theorem 1.8 of Nazarov--Polterovich--Sodin: there is $c_0>0$ such that any nonzero real harmonic function $h$, harmonic in a neighborhood of the closed unit disc, with $h(0)=0$ and at most $d\geq2$ sign changes on the unit circle, satisfies \[ \operatorname{Area}(\{h<0\}\cap\mathbb D)\geq c_0/\log d. \tag{2} \] Their theorem is usually written for $h>0$; applying it to $-h$ gives $2$. Its proof is not reproduced here.

An explicit replacement for each root

Fix $w$ with $|w|\leq1$. If $|w|>1/2$, put $\zeta=w/|w|$ and $q_w(z)=(z-\zeta)^4$. For $|z|\leq1/2$, writing $w=r\zeta$ gives \[ |z-\zeta|^2-|z-w|^2 =(1-r)(1+r-2\operatorname{Re}(z\overline\zeta))\geq0. \] Thus $|q_w(z)|\geq|z-w|^4$.

If $|w|\leq1/2$, define \[ q_w(z)=\frac{(z-w)^4-(1-\overline wz)^4}{1-(\overline w)^4}. \tag{3} \] Its denominator is nonzero, and the leading coefficient is one. Its four zeros solve $B_w(z)^4=1$, where $B_w(z)=(z-w)/(1-\overline wz)$. The inverse transformation $z=(w+\eta)/(1+\overline w\eta)$ sends each fourth root of unity $\eta$ to a distinct point of the unit circle. Hence all zeros of $3$ lie on that circle.

For $|z|,|w|\leq1/2$ the identity \[ |1-\overline wz|^2-|z-w|^2=(1-|w|^2)(1-|z|^2) \] and the estimates $|1-\overline wz|^2\leq25/16$, $(1-|w|^2)(1-|z|^2)\geq9/16$ imply $|B_w(z)|\leq4/5$. When $z\ne w$, the reverse triangle inequality in $3$ consequently gives \[ \frac{|q_w(z)|}{|z-w|^4} \geq\frac{(5/4)^4-1}{1+(1/2)^4} =\frac{369}{272}>1. \tag{4} \] When $z=w$ the desired inequality has right-hand side zero and holds directly. Thus in both root cases $q_w$ is monic of degree four, has all roots on the unit circle, and satisfies $|q_w(z)|\geq|z-w|^4$ throughout the half-disc.

Area comparison and sign changes

Set $Q=\prod_{j=1}^nq_{w_j}$. Then $Q$ is monic of degree $m=4n$, all its roots have modulus one, and \[ |Q(z)|\geq|p(z)|^4\quad(|z|\leq1/2). \tag{5} \] Define $h(\xi)=\log|Q(\xi/2)|$. The polynomial $Q$ is zero-free on a neighborhood of the closed half-disc, so $h$ is harmonic on a neighborhood of $\overline{\mathbb D}$. Also $|Q(0)|=1$, because it is monic with unit-modulus roots, and therefore $h(0)=0$. It is not identically zero: otherwise the analytic function $Q$ would have constant modulus on an open disc and hence would be constant, contrary to $\deg Q=4n$.

On the unit circle the sign of $h(e^{it})$ is the sign of \[ |Q(e^{it}/2)|^2-1. \] This is a real trigonometric polynomial of degree at most $m$. It is not identically zero, since the maximum principle for the harmonic function $h$ would otherwise give $h\equiv0$. Multiplication by $e^{imt}$ turns it into a nonzero polynomial in $e^{it}$ of degree at most $2m$. Hence it has at most $2m=8n$ zeros, and in particular at most $8n$ sign changes. Apply $2$ with $d=8n$ and then scale by $z=\xi/2$: \[ \operatorname{Area}\{z:|z|<1/2,\ |Q(z)|<1\} \geq\frac{c_0}{4\log(8n)}. \tag{6} \] By $5$, every point counted on the left has $|p(z)|<1$. Since $\log(8n)\leq3\log(2n)$ for $n\geq1$, $6$ proves $1$ with $c=c_0/12$.

VERIFICATION AND SCOPE

The root replacement, degree bound, domination direction, sign-change count, and area scaling are proved explicitly. The offline exact checker verifies the rational constants $16/25$ and $369/272$ in the replacement estimate. No inference from sampled polynomial values is used. The input $2$ is a known theorem with all relevant hypotheses checked here. The result does not assert that the exact minimum area has order $1/\log n$: the current upper and lower bounds need not match.

SOURCES

• Current problem: https://www.erdosproblems.com/116.
• M. Krishnapur, E. Lundberg, and K. Ramachandran, On the area of polynomial lemniscates, particularly Section 6.1 and Lemma 21: https://arxiv.org/html/2503.18270v1.
• F. Nazarov, L. Polterovich, and M. Sodin, Sign and area in nodal geometry of Laplace eigenfunctions, Theorem 1.8: https://arxiv.org/html/math/0402412.

CONCLUSION

Attempt status: solved using the explicit harmonic sign-area theorem. The complete deduction above gives the requested universal logarithmic lower bound. No novelty is claimed. Completion rate estimate: 100%.