Erdős Problem #509

Erdős Problem 509 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/509

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/509.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/509.tex.

FORMAL RESTATEMENT

Let $f\in\mathbb{C}[z]$ be monic, non-constant, of degree $d\ge 1$. Define the (closed) lemniscate \[ E(f):=\{z\in\mathbb{C}: |f(z)|\le 1\}. \] Interpretation issue: the phrase "covered by circles" is ambiguous.

• Literal (1-dimensional) interpretation: circles mean circumferences. Then $E(f)$ typically has nonempty interior (e.g. for $f(z)=z$), so it cannot be covered by countably many circles of finite total radius. Hence that literal reading is impossible.
• Standard complex-analysis interpretation (Cartan/Pommerenke context): "circles" means closed discs (each disc determined by its boundary circle). We adopt this corrected interpretation: we ask whether there exist closed discs $\overline{D}(c_j,r_j)$ such that \[ E(f)\subseteq \bigcup_j \overline{D}(c_j,r_j) \qquad\text{and}\qquad \sum_j r_j\le 2. \] The covering family is understood to be finite or countable so that $\sum_j r_j$ is defined.

Edge cases: $d=1$ gives a single closed unit disc, covered with total radius $1$. For $d=2$ the trivial cover by unit discs about the roots already has total radius $2$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement itself records: Cartan proved a universal covering with total radius $\le 2e$; Pommerenke improved the universal constant to $2.59$; and Pommerenke proved the sharp constant $2$ holds when $E(f)$ is connected. (I do not reproduce those proofs here.)

ATTACK PLAN

• Proof-track idea: relate $E(f)$ to potential theory (logarithmic capacity $=1$ for monic polynomials) and compare capacity to a one-dimensional outer content given by disc coverings; then try to push constants down to $2$.
• Disproof-track idea: search for high-degree polynomials where $E(f)$ has many separated "thick" components forcing large total covering radius. A plausible candidate would be polynomials with many near-multiple roots arranged to make components fat.

I did not find either a proof with constant $2$ or a counterexample.

WORK

Fast reality check (small degrees).

• If $d=1$, $f(z)=z-a$, then $E(f)=\overline{D}(a,1)$, so one disc of radius $1$ covers it.
• If $d=2$, writing $f(z)=(z-\alpha)(z-\beta)$ (monic), then \[ E(f)\subseteq \overline{D}(\alpha,1)\cup \overline{D}(\beta,1) \] (because if $|z-\alpha|>1$ and $|z-\beta|>1$ then $|f(z)|>1$). Hence total radius $2$ suffices for $d=2$.

Lemma 1 (trivial disc cover by unit discs around roots). Let $f(z)=\prod_{j=1}^d (z-\alpha_j)$ be monic of degree $d\ge 1$ (roots counted with multiplicity). Then \[ E(f)\subseteq \bigcup_{j=1}^d \overline{D}(\alpha_j,1). \]

Proof. Take $z\in\mathbb{C}$ that lies outside every $\overline{D}(\alpha_j,1)$. Then $|z-\alpha_j|>1$ for each $j$, so \[ |f(z)|=\prod_{j=1}^d |z-\alpha_j| > 1. \] Thus $z\notin E(f)$. Contrapositively, every $z\in E(f)$ must belong to at least one $\overline{D}(\alpha_j,1)$. $\square$

Lemma 2 (each component of $\{|f|<1\}$ contains a zero of $f$). Let \[ \Omega(f):=\{z\in\mathbb{C}: |f(z)|<1\} \] (which is open). Then every connected component $U$ of $\Omega(f)$ contains at least one zero of $f$.

Proof. Because $f$ is a non-constant polynomial, $|f(z)|\to\infty$ as $|z|\to\infty$. Hence there exists $R>0$ such that $|f(z)|>1$ whenever $|z|\ge R$, so $\Omega(f)\subset \overline{D}(0,R)$ is bounded. In particular, every connected component $U$ of $\Omega(f)$ is a bounded open set.

Fix such a component $U$ and suppose, for contradiction, that $f$ has no zeros in $U$. Then $f$ is nonzero on $\overline{U}$ as well: if $f(\zeta)=0$ for some $\zeta\in\overline{U}$, then $|f(\zeta)|=0<1$ and continuity would imply $\zeta\in\Omega(f)$, hence $\zeta\in U$, contradicting the assumption.

Define $g(z):=1/f(z)$. Then $g$ is holomorphic on $U$ and continuous on $\overline{U}$. Moreover, for $z\in U$ we have $|f(z)|<1$, so $|g(z)|>1$.

Next, we claim that for every boundary point $\zeta\in\partial U$ we have $|f(\zeta)|=1$, hence $|g(\zeta)|=1$. Indeed, since $\zeta$ is a limit of points $z_j\in U$ with $|f(z_j)|<1$, continuity gives $|f(\zeta)|\le 1$. If $|f(\zeta)|<1$, then by continuity $|f|<1$ in a neighborhood of $\zeta$, implying $\zeta$ is an interior point of $\Omega(f)$, contradicting $\zeta\in\partial U$. Thus $|f(\zeta)|=1$.

Now apply the maximum modulus principle to $g$ on the bounded domain $U$: the maximum of $|g|$ on $\overline{U}$ is attained on $\partial U$. But $|g|=1$ on $\partial U$ and $|g|>1$ on $U$, a contradiction. Therefore $U$ must contain a zero of $f$. $\square$

VERIFICATION

• Lemma 1: checked the contrapositive carefully; it uses only the factorisation of monic polynomials and that if all factors exceed $1$ in modulus, the product exceeds $1$.
• Lemma 2: uses that $\Omega(f)=\{|f|<1\}$ is bounded for a polynomial $f$, that $|f|=1$ on $\partial U$ for any component $U$, and then applies the maximum modulus principle to $g=1/f$ (which has $|g|>1$ on $U$ but $|g|=1$ on $\partial U$), yielding a contradiction unless $U$ contains a zero.
• These lemmas do not approach the sharp constant $2$ except in degree $2$; they are consistency checks and structural facts.

FINAL

UNRESOLVED

(i) Strongest proved partial result: For degree $d=2$ the desired bound $\sum r_j\le 2$ holds by the trivial cover by unit discs about the two roots (and Lemma 1 gives a degree-$d$ cover with total radius $d$ in general).

(ii) First gap (crisp): Prove or disprove that for every monic polynomial $f$ there is a disc cover of $E(f)$ with total radii $\le 2$.

(iii) Top 3 next moves:

1. Attempt a potential-theoretic reduction: prove an inequality of the form "for polynomial lemniscates $E(f)$, the one-dimensional outer content by discs is $\le 2\,\mathrm{cap}(E(f))$" and then show $\mathrm{cap}(E(f))=1$.
2. Search numerically for extremisers: sample high-degree polynomials, approximate $E(f)$ by level sets, and attempt computational covering optimisations to see whether total radius $>2$ is ever forced.
3. Try a reduction to connected components: show one can cover each component of $\Omega(f)=\{|f|<1\}$ by a disc with radius controlled by something additive over components, and then sum over components.

(iv) Minimal counterexample structure (if false): would require a monic $f$ for which $E(f)$ has many "thick" separated components so that any disc cover needs total radius $>2$ despite being able to place many small discs; heuristically this would demand components whose diameters do not shrink fast enough as the number of components grows.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.

FORMAL RESTATEMENT

Let $f\in\mathbb{C}[z]$ be monic, non-constant, of degree $d\ge 1$. Define the (closed) lemniscate \[ E(f):=\{z\in\mathbb{C}: |f(z)|\le 1\}. \] Interpretation issue: the phrase "covered by circles" is ambiguous.

• Literal (1-dimensional) interpretation: circles mean circumferences. Then $E(f)$ typically has nonempty interior (e.g. for $f(z)=z$), so it cannot be covered by countably many circles of finite total radius. Hence that literal reading is impossible.
• Standard complex-analysis interpretation (Cartan/Pommerenke context): "circles" means closed discs (each disc determined by its boundary circle). We adopt this corrected interpretation: we ask whether there exist closed discs $\overline{D}(c_j,r_j)$ such that \[ E(f)\subseteq \bigcup_j \overline{D}(c_j,r_j) \qquad\text{and}\qquad \sum_j r_j\le 2. \] The covering family is understood to be finite or countable so that $\sum_j r_j$ is defined.

Edge cases: $d=1$ gives a single closed unit disc, covered with total radius $1$. For $d=2$ the trivial cover by unit discs about the roots already has total radius $2$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement itself records: Cartan proved a universal covering with total radius $\le 2e$; Pommerenke improved the universal constant to $2.59$; and Pommerenke proved the sharp constant $2$ holds when $E(f)$ is connected. (I do not reproduce those proofs here.)

ATTACK PLAN

• Proof-track idea: relate $E(f)$ to potential theory (logarithmic capacity $=1$ for monic polynomials) and compare capacity to a one-dimensional outer content given by disc coverings; then try to push constants down to $2$.
• Disproof-track idea: search for high-degree polynomials where $E(f)$ has many separated "thick" components forcing large total covering radius. A plausible candidate would be polynomials with many near-multiple roots arranged to make components fat.

I did not find either a proof with constant $2$ or a counterexample.

WORK

Fast reality check (small degrees).

• If $d=1$, $f(z)=z-a$, then $E(f)=\overline{D}(a,1)$, so one disc of radius $1$ covers it.
• If $d=2$, writing $f(z)=(z-\alpha)(z-\beta)$ (monic), then \[ E(f)\subseteq \overline{D}(\alpha,1)\cup \overline{D}(\beta,1) \] (because if $|z-\alpha|>1$ and $|z-\beta|>1$ then $|f(z)|>1$). Hence total radius $2$ suffices for $d=2$.

Lemma 1 (trivial disc cover by unit discs around roots). Let $f(z)=\prod_{j=1}^d (z-\alpha_j)$ be monic of degree $d\ge 1$ (roots counted with multiplicity). Then \[ E(f)\subseteq \bigcup_{j=1}^d \overline{D}(\alpha_j,1). \]

Proof. Take $z\in\mathbb{C}$ that lies outside every $\overline{D}(\alpha_j,1)$. Then $|z-\alpha_j|>1$ for each $j$, so \[ |f(z)|=\prod_{j=1}^d |z-\alpha_j| > 1. \] Thus $z\notin E(f)$. Contrapositively, every $z\in E(f)$ must belong to at least one $\overline{D}(\alpha_j,1)$. $\square$

Lemma 2 (each component of $\{|f|<1\}$ contains a zero of $f$). Let \[ \Omega(f):=\{z\in\mathbb{C}: |f(z)|<1\} \] (which is open). Then every connected component $U$ of $\Omega(f)$ contains at least one zero of $f$.

Proof. Because $f$ is a non-constant polynomial, $|f(z)|\to\infty$ as $|z|\to\infty$. Hence there exists $R>0$ such that $|f(z)|>1$ whenever $|z|\ge R$, so $\Omega(f)\subset \overline{D}(0,R)$ is bounded. In particular, every connected component $U$ of $\Omega(f)$ is a bounded open set.

Fix such a component $U$ and suppose, for contradiction, that $f$ has no zeros in $U$. Then $f$ is nonzero on $\overline{U}$ as well: if $f(\zeta)=0$ for some $\zeta\in\overline{U}$, then $|f(\zeta)|=0<1$ and continuity would imply $\zeta\in\Omega(f)$, hence $\zeta\in U$, contradicting the assumption.

Define $g(z):=1/f(z)$. Then $g$ is holomorphic on $U$ and continuous on $\overline{U}$. Moreover, for $z\in U$ we have $|f(z)|<1$, so $|g(z)|>1$.

Next, we claim that for every boundary point $\zeta\in\partial U$ we have $|f(\zeta)|=1$, hence $|g(\zeta)|=1$. Indeed, since $\zeta$ is a limit of points $z_j\in U$ with $|f(z_j)|<1$, continuity gives $|f(\zeta)|\le 1$. If $|f(\zeta)|<1$, then by continuity $|f|<1$ in a neighborhood of $\zeta$, implying $\zeta$ is an interior point of $\Omega(f)$, contradicting $\zeta\in\partial U$. Thus $|f(\zeta)|=1$.

Now apply the maximum modulus principle to $g$ on the bounded domain $U$: the maximum of $|g|$ on $\overline{U}$ is attained on $\partial U$. But $|g|=1$ on $\partial U$ and $|g|>1$ on $U$, a contradiction. Therefore $U$ must contain a zero of $f$. $\square$

VERIFICATION

• Lemma 1: checked the contrapositive carefully; it uses only the factorisation of monic polynomials and that if all factors exceed $1$ in modulus, the product exceeds $1$.
• Lemma 2: uses that $\Omega(f)=\{|f|<1\}$ is bounded for a polynomial $f$, that $|f|=1$ on $\partial U$ for any component $U$, and then applies the maximum modulus principle to $g=1/f$ (which has $|g|>1$ on $U$ but $|g|=1$ on $\partial U$), yielding a contradiction unless $U$ contains a zero.
• These lemmas do not approach the sharp constant $2$ except in degree $2$; they are consistency checks and structural facts.

FINAL

UNRESOLVED

(i) Strongest proved partial result: For degree $d=2$ the desired bound $\sum r_j\le 2$ holds by the trivial cover by unit discs about the two roots (and Lemma 1 gives a degree-$d$ cover with total radius $d$ in general).

(ii) First gap (crisp): Prove or disprove that for every monic polynomial $f$ there is a disc cover of $E(f)$ with total radii $\le 2$.

(iii) Top 3 next moves:

1. Attempt a potential-theoretic reduction: prove an inequality of the form "for polynomial lemniscates $E(f)$, the one-dimensional outer content by discs is $\le 2\,\mathrm{cap}(E(f))$" and then show $\mathrm{cap}(E(f))=1$.
2. Search numerically for extremisers: sample high-degree polynomials, approximate $E(f)$ by level sets, and attempt computational covering optimisations to see whether total radius $>2$ is ever forced.
3. Try a reduction to connected components: show one can cover each component of $\Omega(f)=\{|f|<1\}$ by a disc with radius controlled by something additive over components, and then sum over components.

(iv) Minimal counterexample structure (if false): would require a monic $f$ for which $E(f)$ has many "thick" separated components so that any disc cover needs total radius $>2$ despite being able to place many small discs; heuristically this would demand components whose diameters do not shrink fast enough as the number of components grows.