Newton-polytope saturation of Vandermonde powers

A polynomial has a saturated Newton polytope if every lattice point in that polytope occurs as a monomial with nonzero coefficient. For every fixed k≥1, do Vandermonde k-th powers eventually fail this property as the variable count grows?

Why it matters and what counts as progress

Why it matters: An absent interior lattice monomial distinguishes geometric support from actual polynomial support.

What counts as progress: An explicit missing-monomial seed for each remaining power, together with a propagation proof to all sufficiently large variable counts.

How the problem developed

7 Mar 2017 · Conjecture [1]
Cara Monical, Neriman Tokcan and Alexander Yong Conjectured that every fixed positive power of the Vandermonde determinant eventually fails to have saturated Newton polytope as the number of variables grows.
26 Jul 2026 · Proof [2]
Thien Le and Melanie Weber; they credit OpenAI Codex GPT Sol 5.6 Extra High with the key even-power construction and proof strategy Proved the conjecture for every fixed positive power; for even k ≥ 4 they exhibit missing lattice monomials in sufficiently many variables, and the odd case follows by alternation.

AI contribution

Powers of the Vandermonde determinant are eventually non-SNP

Thien Le, Melanie Weber

For every fixed power k≥1, sufficiently many variables give a lattice monomial in the Newton polytope of the k-th Vandermonde power with zero coefficient. The new step is an explicit even-power seed for k≥4; earlier cases and propagation complete the result.

AI involvementAI-centered

ImportanceNotable

Technical verificationFormal (Lean 4 even-power seed theorem (Ward recurrence; Jack route conditional on stated Macdonald facts) + human)

Human understandingFull