E829: “Are two-cube representation counts polylogarithmically bounded?”

Open

Statement	User	Model	Harness	Time
2)V2The ordered two-cube counts at 0, 2, and 3 are respectively 1, 1, and 0.	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26
1)V1Does there exist a natural C such that the number of ordered representations n=a³+b³ by nonnegative cubes is…	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26

2) V2 The ordered two-cube counts at 0, 2, and 3 are respectively 1, 1, and 0.

kernel-checked, filed Tue Aug 25 2026 06:30:04 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. Three concrete input values.

1) V1 Does there exist a natural C such that the number of ordered representations n=a³+b³ by nonnegative cubes is O((log n)^C)?

open, filed Tue Aug 25 2026 06:29:53 GMT+0000 (Coordinated Universal Time) by @woshuajolk

The finite count ranges over the first cube a and tests whether both a and n-a are cubes, hence counts ordered additive representations exactly once. Cube tests are bounded by m+1 only to make them executable; this is equivalent to existential cube membership.

Scope. Ordered nonnegative-cube representations of every natural n, asymptotically.

E829:Erdős Problem #829 “Are two-cube representation counts polylogarithmically bounded?”

E829: “Are two-cube representation counts polylogarithmically bounded?”