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Squarefree values of irreducible quartic polynomials, and (d-2)-power-free values in every degree d at least 4

Let f∈Z[x]f\in\mathbb Z[x] be irreducible of degree dd and k≥2k\ge2. If no prime kkth power divides every value of ff, one expects f(n)f(n) to be kk-free for a proportion cf,k=∏p(1−ρf(pk)/pk)>0c_{f,k}=\prod_p(1-\rho_f(p^k)/p^k)>0 of n≤Xn\le X, where ρf(q)\rho_f(q) counts roots mod qq. Ricci proved this for k≥dk\ge d; Erdos (1953) and Hooley (1967) handled k=d−1k=d-1, which includes squarefree values of cubics; Nair, Heath-Brown and Browning reached k=d−2k=d-2 only for d≥9d\ge9. Erdos singled out the squarefreeness of n4+2n^4+2 in 1953 and returned to the k=d−2k=d-2 obstacle in 1965. Does every irreducible integer quartic with no fixed prime-square divisor take squarefree values with the predicted density, and more generally does the (d−2)(d-2)-free density hold in every degree d≥4d\ge4?

Result
Proved(see note)
Status
Candidate (review pending)
AI contribution
AI-discovered
Method
Argument
Field
Analytic number theory, sieve methods
Posed by
Paul Erdos (the case n^4 + 2 and the exponent d - 2 barrier)
Year posed
1953
Years open
73y
Solved
2026-09-24
Model
Unreleased internal OpenAI model
Vendor
OpenAI
Collaborators
—
Verification
Lean-checked, statement unaudited
Publication
Announced
Collection
OpenAI math release (October 2026), version adc7f12
Significance
48 / 100
Disclosed cost
—
Wikipedia
No dedicated article

What was actually shown

Theorem 1.1: for f∈Z[x]f\in\mathbb Z[x] irreducible over Q\mathbb Q of degree 4≤d≤84\le d\le8 with no fixed prime (d−2)(d-2)th-power divisor, the number of n≤Xn\le X with f(n)f(n) (d−2)(d-2)-free is cf,d−2X+of(X)c_{f,d-2}X+o_f(X) with the Euler-product constant, which is positive; for d=4d=4 this is squarefree values of quartics, e.g. n4+2n^4+2 and n4+1n^4+1. With Browning's theorem for d≥9d\ge9 (Corollary 1.2) the (d−2)(d-2)-free density holds for all d≥4d\ge4; Corollary 1.3 treats separable products. Not shown: a power-saving error term, uniformity in ff, or squarefree values in degree five and above. Priority: Carella's preprint claims n4+1n^4+1 and n4+2n^4+2, and Zapata Ceballos-Jalalvand claim positive density for a class including n4+1n^4+1; the manuscript uses neither.

What the AI did

The release README says the results were produced by an unreleased internal OpenAI model with a fixed procedure, on average about three hours of ChatGPT Pro thinking compute per result, and that some outputs build on earlier model results. This result is not among the README's exceptions (the Hodge conjecture for CM abelian varieties and the Re(s) > 11/12 zero-free region). The manuscripts are authored 'OpenAI' and name no human author. The single manuscript (September 24, 2026) is the whole family.

Verification

No independent mathematician has checked this yet. Checked here: the abstract, introduction, Theorem 1.1 and Corollaries 1.2-1.3 were read against the problem as the manuscript states it from Erdos's papers. The proof (number-field factorization, determinant estimates with adaptive auxiliary primes, low-degree geometry and an exact parameter certificate) was not refereed. lean/formalization.yaml lists comparator PowerFreeValues, declaration OAI.QuarticPowerFree.allDegrees (file OAI/NumberTheory/PowerFree/Main.lean). Its statement was read here: for f∈Z[x]f\in\mathbb Z[x] irreducible over Q\mathbb Q with deg⁡f≥4\deg f\ge4 and ρf(pd−2)<pd−2\rho_f(p^{d-2})<p^{d-2} for every prime pp, the local factors are multipliable, their product cc is positive, and the number of 1≤n≤X1\le n\le X with f(n)f(n) (d−2)(d-2)-free is cX+o(X)cX+o(X). That states the headline, including squarefree values of quartics, and in Lean also the degrees d≥9d\ge9 that the paper takes from Browning. Not rebuilt here. Permitted axioms: propext, Quot.sound, Classical.choice.

Sources

Changelog1 change

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