Erdős Problem #1041

Erdős Problem 1041 --- collection record

PROVENANCE AND STATUS

Saved collaborative-database status: falsifiable. 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_5.2/1041.tex.

REUSED SOURCE 1

Let $f(z)=\prod_{i=1}^n(z-z_i)\in \mathbb{C}[z]$ with $\lvert z_i\rvert < 1$ for all $i$. Must there always exist a path of length less than $2$ in\[\{z: \lvert f(z)\rvert < 1\}\]which connects two of the roots of $f$? A problem of Erdős, Herzog, and Piranian [EHP58], who proved that this set always has a connected component containing at least two of the roots. References [EHP58] Erdős, P. and Herzog, F. and Piranian, G., Metric properties of polynomials . J. Analyse Math. $1958$, 125-148.

FORMAL RESTATEMENT Let f(z)=$\prod_{i=1}^n$ $z-z_i$ with all roots satisfying |z_i|<1, and let L(f) := { z in C : |f(z)| < 1 }. Question: must there exist indices i!=j and a rectifiable path gamma in L(f) connecting z_i to z_j such that the Euclidean length of gamma is strictly less than 2? (For n=1 the question is vacuous; the interesting case is n>=2.)

QUICK LITERATURE/CONTEXT CHECK Only facts explicitly present in the problem statement are treated as established here. In particular, it is stated that L(f) always has a connected component containing at least two roots. The additional length <2 requirement is the new constraint.

ATTACK PLAN $1$ Prove the statement for n=2 exactly (this is the smallest nontrivial case). $2$ Find a clean sufficient condition guaranteeing that the straight segment between two roots stays inside L(f). $3$ Attempt to argue that such a sufficient condition must occur for some pair, or else try to build counterexamples by spreading roots.

WORK FAST REALITY CHECK. * n=2 is fully solved below: there is always a path of length |z_1-z_2|<2 (Lemma 1041.1). * For n=3,4,5,6 a random experiment (roots uniform in the unit disc) shows that the sufficient condition in Lemma 1041.2 holds for at least one pair with frequencies: n=3: about 0.641 n=4: about 0.619 n=5: about 0.540 n=6: about 0.448 This does not prove the conjecture but is a sanity check that the condition is not extremely rare for small n.

Lemma 1041.1 (the conjecture holds for n=2). Let f(z)=$z-a$$z-b$ with |a|<1 and |b|<1 and a!=b. Then there is a path in L(f) from a to b of length |a-b|<2.

Proof. Let d:=|a-b|. Since |a|,|b|<1, the triangle inequality gives d <= |a|+|b| < 2, hence d<2. Let m=$a+b$/2 be the midpoint. Consider the straight segment from a to m, parameterized by z(t) = a + t$m-a$ = a + t$b-a$/2, 0<=t<=1. Then |z(t)-a| = t d/2. Also, z(t)-b = a-b + t$b-a$/2 = -$1-t/2$$b-a$, so |z(t)-b| = $1-t/2$ d. Therefore, along this segment we have |f(z(t))| = |z(t)-a||z(t)-b| = $t d/2$ * $(1-t/2)d$ = $t(1-t/2)$ * d^2/2. For t in [0,1], the function t$1-t/2$ is increasing (its derivative is 1-t>=0), so its maximum on [0,1] is attained at t=1 and equals 1/2. Hence for all t in [0,1], |f(z(t))| <= $1/2$ * d^2/2 = d^2/4 < 1. Thus the entire segment from a to m lies in L(f). By symmetry, the same argument shows the segment from m to b also lies in L(f). Concatenating these two segments gives a path from a to b inside L(f) of total length |a-m| + |m-b| = d/2 + d/2 = d < 2.

Lemma 1041.2 (a sufficient condition: close pair implies a short connecting segment). Let n>=2 and f(z)=$\prod_{k=1}^n$ $z-z_k$ with all |z_k|<1. Suppose there exist i!=j with d:=|z_i-z_j| such that d^2 $2+d$^{n-2} < 1. Then the straight line segment [z_i,z_j] is contained in L(f). In particular, z_i and z_j can be connected inside L(f) by a path of length d $<2$.

Proof. Let z be any point on the straight segment from z_i to z_j. Then by convexity, |z-z_i| <= d and |z-z_j| <= d. Now fix any other root z_k with k not in {i,j}. Using the triangle inequality, |z-z_k| <= |z-z_i| + |z_i-z_k|. Since |z_i|<1 and |z_k|<1, we have |z_i-z_k| <= |z_i|+|z_k| < 2. Thus |z-z_k| < d+2. Combining these bounds over all factors, |f(z)| = $\prod_{k=1}^n$ |z-z_k| < d * d * $2+d$^{n-2} = d^2$2+d$^{n-2} < 1. Therefore every z on [z_i,z_j] lies in L(f), and the segment length is d<2.

VERIFICATION * Lemma 1041.1 is a direct computation along a segment and is fully rigorous. * Lemma 1041.2 is a direct product bound using only triangle inequality and the root constraints. * The random experiment numbers were computed by a minimal script; they are not used as proof.

FINAL UNRESOLVED (i) Strongest proved partial result here: the statement is true for n=2 (Lemma 1041.1). More generally, if two roots are sufficiently close in the sense d^2$2+d$^{n-2}<1, then the straight segment between them lies in {|f|<1} and yields a path of length <2 (Lemma 1041.2). (ii) First gap: show that for every n>=2 and every choice of roots with |z_i|<1, there exists some pair i!=j satisfying a condition strong enough to force a path of length <2 inside L(f); or else construct an explicit counterexample polynomial. (iii) Top 3 next moves: 1. Try to strengthen Lemma 1041.2 by replacing the crude bound |z-z_k|<2+d with something depending on where the segment lies (e.g. if the segment stays well inside the unit disc, improve the 2 constant). 2. Combine a packing argument (some pair of roots must be close) with a lower bound on how small the other factors can be along a connecting corridor; attempt to force |f|<1 along a short path. 3. Computational: for small degrees (say n<=6) attempt an explicit search over root configurations to maximize the minimal in-domain path length between roots (e.g. discretize the plane and do shortest-path computations in the sublevel set) to look for counterexamples or patterns. (iv) Minimal counterexample structure (if false): roots arranged near the unit circle and spread so that (a) the component containing multiple roots exists but (b) any connecting path must detour around regions where |f|>=1, forcing length >=2.

COMPLETION ESTIMATE

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

Let $f(z)=\prod_{i=1}^n(z-z_i)\in \mathbb{C}[z]$ with $\lvert z_i\rvert < 1$ for all $i$. Must there always exist a path of length less than $2$ in\[\{z: \lvert f(z)\rvert < 1\}\]which connects two of the roots of $f$? A problem of Erdős, Herzog, and Piranian [EHP58], who proved that this set always has a connected component containing at least two of the roots. References [EHP58] Erdős, P. and Herzog, F. and Piranian, G., Metric properties of polynomials . J. Analyse Math. $1958$, 125-148.

FORMAL RESTATEMENT Let f(z)=$\prod_{i=1}^n$ $z-z_i$ with all roots satisfying |z_i|<1, and let L(f) := { z in C : |f(z)| < 1 }. Question: must there exist indices i!=j and a rectifiable path gamma in L(f) connecting z_i to z_j such that the Euclidean length of gamma is strictly less than 2? (For n=1 the question is vacuous; the interesting case is n>=2.)

QUICK LITERATURE/CONTEXT CHECK Only facts explicitly present in the problem statement are treated as established here. In particular, it is stated that L(f) always has a connected component containing at least two roots. The additional length <2 requirement is the new constraint.

ATTACK PLAN $1$ Prove the statement for n=2 exactly (this is the smallest nontrivial case). $2$ Find a clean sufficient condition guaranteeing that the straight segment between two roots stays inside L(f). $3$ Attempt to argue that such a sufficient condition must occur for some pair, or else try to build counterexamples by spreading roots.

WORK FAST REALITY CHECK. * n=2 is fully solved below: there is always a path of length |z_1-z_2|<2 (Lemma 1041.1). * For n=3,4,5,6 a random experiment (roots uniform in the unit disc) shows that the sufficient condition in Lemma 1041.2 holds for at least one pair with frequencies: n=3: about 0.641 n=4: about 0.619 n=5: about 0.540 n=6: about 0.448 This does not prove the conjecture but is a sanity check that the condition is not extremely rare for small n.

Lemma 1041.1 (the conjecture holds for n=2). Let f(z)=$z-a$$z-b$ with |a|<1 and |b|<1 and a!=b. Then there is a path in L(f) from a to b of length |a-b|<2.

Proof. Let d:=|a-b|. Since |a|,|b|<1, the triangle inequality gives d <= |a|+|b| < 2, hence d<2. Let m=$a+b$/2 be the midpoint. Consider the straight segment from a to m, parameterized by z(t) = a + t$m-a$ = a + t$b-a$/2, 0<=t<=1. Then |z(t)-a| = t d/2. Also, z(t)-b = a-b + t$b-a$/2 = -$1-t/2$$b-a$, so |z(t)-b| = $1-t/2$ d. Therefore, along this segment we have |f(z(t))| = |z(t)-a||z(t)-b| = $t d/2$ * $(1-t/2)d$ = $t(1-t/2)$ * d^2/2. For t in [0,1], the function t$1-t/2$ is increasing (its derivative is 1-t>=0), so its maximum on [0,1] is attained at t=1 and equals 1/2. Hence for all t in [0,1], |f(z(t))| <= $1/2$ * d^2/2 = d^2/4 < 1. Thus the entire segment from a to m lies in L(f). By symmetry, the same argument shows the segment from m to b also lies in L(f). Concatenating these two segments gives a path from a to b inside L(f) of total length |a-m| + |m-b| = d/2 + d/2 = d < 2.

Lemma 1041.2 (a sufficient condition: close pair implies a short connecting segment). Let n>=2 and f(z)=$\prod_{k=1}^n$ $z-z_k$ with all |z_k|<1. Suppose there exist i!=j with d:=|z_i-z_j| such that d^2 $2+d$^{n-2} < 1. Then the straight line segment [z_i,z_j] is contained in L(f). In particular, z_i and z_j can be connected inside L(f) by a path of length d $<2$.

Proof. Let z be any point on the straight segment from z_i to z_j. Then by convexity, |z-z_i| <= d and |z-z_j| <= d. Now fix any other root z_k with k not in {i,j}. Using the triangle inequality, |z-z_k| <= |z-z_i| + |z_i-z_k|. Since |z_i|<1 and |z_k|<1, we have |z_i-z_k| <= |z_i|+|z_k| < 2. Thus |z-z_k| < d+2. Combining these bounds over all factors, |f(z)| = $\prod_{k=1}^n$ |z-z_k| < d * d * $2+d$^{n-2} = d^2$2+d$^{n-2} < 1. Therefore every z on [z_i,z_j] lies in L(f), and the segment length is d<2.

VERIFICATION * Lemma 1041.1 is a direct computation along a segment and is fully rigorous. * Lemma 1041.2 is a direct product bound using only triangle inequality and the root constraints. * The random experiment numbers were computed by a minimal script; they are not used as proof.

FINAL UNRESOLVED (i) Strongest proved partial result here: the statement is true for n=2 (Lemma 1041.1). More generally, if two roots are sufficiently close in the sense d^2$2+d$^{n-2}<1, then the straight segment between them lies in {|f|<1} and yields a path of length <2 (Lemma 1041.2). (ii) First gap: show that for every n>=2 and every choice of roots with |z_i|<1, there exists some pair i!=j satisfying a condition strong enough to force a path of length <2 inside L(f); or else construct an explicit counterexample polynomial. (iii) Top 3 next moves: 1. Try to strengthen Lemma 1041.2 by replacing the crude bound |z-z_k|<2+d with something depending on where the segment lies (e.g. if the segment stays well inside the unit disc, improve the 2 constant). 2. Combine a packing argument (some pair of roots must be close) with a lower bound on how small the other factors can be along a connecting corridor; attempt to force |f|<1 along a short path. 3. Computational: for small degrees (say n<=6) attempt an explicit search over root configurations to maximize the minimal in-domain path length between roots (e.g. discretize the plane and do shortest-path computations in the sublevel set) to look for counterexamples or patterns. (iv) Minimal counterexample structure (if false): roots arranged near the unit circle and spread so that (a) the component containing multiple roots exists but (b) any connecting path must detour around regions where |f|>=1, forcing length >=2.