Erdős Problem #1150
OPENdatabase updated 2026-01-23
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
Is there some constant \(c > 0\) such that, for all large enough \(n\) and all polynomials \(P\) of degree \(n\) with coefficients in \(\{-1, 1\}\), \[\max_{|z|=1} |P(z)| > (1 + c) \sqrt{n}?\] No. This was disproved by an internal OpenAI model in September 2026 [OAI26a]; see the linked Lean proof.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved. The authoritative wording is on erdosproblems.com/1150.Newer public claim (6 Oct 2026): OpenAI's internal-model release claims, for this problem, real Littlewood polynomials with max ≤ (1+η)√N at every large length N (Thm 1.1, formalized) — hence NO to #1150 — and, paper-only, two-sided ultraflat (1±ε)√N — family 76 at
github.com/openai/math, OpenAI's own Lean proves Theorem 1.1 — real Littlewood polynomials with max ≤ (1+η)√N at every large length — which implies the negation of #1150; our kernel replay on 12 Oct 2026 printed only propext/Classical.choice/Quot.sound on a statement matched to the challenge; the two-sided ultraflat headline is paper-only — see our /day entry. Bloom calls these claims, not endorsements; anything on this page that says the question is open was written before that release and has not re-checked it.
Prizenone
Solution in Leanno
FULL WRITE-UP BY THIS SITE · OCT 12, 2026
Erdős asked whether every polynomial of degree n with coefficients ±1 has maximum modulus at least (1+c)√n on the unit circle; Parseval gives √(n+1) for free and the question was whether real signs are forced a fixed fraction above it. OpenAI's 23 September 2026 paper proves the minimum over all sign choices is (1+o(1))√N at every large length N — the limit is 1, not the ≈1.27 that Odlyzko's exhaustive search had suggested — and its Lean states precisely that. On our machine the 28-module proof compiled from the pinned sources and the kernel printed only propext, Classical.choice and Quot.sound on a statement matched byte-for-byte to the challenge; seven tool-equipped referee seats, briefed to refute, found no gap in the formalized paper or in the two-sided ultraflat one. Answer: NO. The theorem is OpenAI's; the audit and the replay are ours.
CLOSED