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Statement

Notation (pp. 125, 143): f(z)=∏ν=1n(z−zν)f(z)=\prod_{\nu=1}^n(z-z_\nu) and D(f)\mathcal D(f) its discriminant, so that ∣D(f)∣=∏ν<μ∣zν−zμ∣2|\mathcal D(f)|=\prod_{\nu<\mu}|z_\nu-z_\mu|^2.

Problem 13 (p. 143). "For a fixed value of nn, what is the maximum value of ∣D(f)∣|\mathcal D(f)| in the space of polynomials (1) with ∣zμ−zν∣≤2|z_\mu-z_\nu|\leq2 (1≤μ<ν≤n)(1\leq\mu<\nu\leq n)? Is the maximum achieved if the zνz_\nu are the vertices of a regular nn-gon whose greatest diagonal has length 2?"

The problem follows Theorem 10 and its Remark on the same page. The paper states the regular-polygon sentence as a question and gives no proof, bound or construction for the diameter-constrained class.

Source. P. Erdős, F. Herzog, G. Piranian, Metric properties of polynomials, J. Analyse Math. 6 (1958), 125--148, doi:10.1007/BF02790232; Problem 13 on p. 143. The copy read is identified on the source card.

Read depth. Claims checked: the problem was read clause by clause on the page image of p. 143 on 2026-10-08. Nothing here is independently reviewed.

Dependencies

None. Theorem 10 bounds the same quantity over a different class, defined by the critical values of ff.

Bears on

  • #1045: the problem's Δ=∏i≠j∣zi−zj∣\Delta=\prod_{i\ne j}|z_i-z_j| equals $\prod_{i<j}|z_i-z_j|^2=|\mathcal D(f)|$, and its constraint ∣zi−zj∣≤2|z_i-z_j|\le2 is the paper's, so the two ask the same maximization; the problem's "regular polygon" is the paper's regular nn-gon whose greatest diagonal has length 22.