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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. T. D. Browning, Power-free values of polynomials, Arch. Math. (Basel) 96 (2011), no. 2, 139--150. Let f∈Z[x]f\in\mathbb{Z}[x] be irreducible of degree d≥3d\ge3, let ρf(q)\rho_f(q) count the residues aa modulo qq with f(a)≡0(modq)f(a)\equiv0\pmod q, and let r≥(3d+1)/4r\ge(3d+1)/4. Then the number of n≤xn\le x with f(n)f(n) rr-power-free is cf,r x+o(x)c_{f,r}\,x+o(x), where cf,r=∏p(1−ρf(pr)/pr)c_{f,r}=\prod_p\bigl(1-\rho_f(p^r)/p^r\bigr) is the Euler product of the local factors. The exponent r=d−2r=d-2 satisfies d−2≥(3d+1)/4d-2\ge(3d+1)/4 exactly when d≥9d\ge9. So every irreducible ff of degree at least 99 with no prime pp such that pd−2p^{d-2} divides every value takes (d−2)(d-2)-power-free values on a set of nn of positive density cf,d−2c_{f,d-2}, hence infinitely often. The theorem needs neither the exclusion of degrees that are powers of 22 nor the sign condition on the leading coefficient. Browning obtains the range by combining Heath-Brown's determinant method, which gave r≥(3d+2)/4r\ge(3d+2)/4 (its claim page), with Salberger's global determinant estimates.

Covers. The second question of Problem 978 for every polynomial of degree k≥9k\ge9, answered yes, with a positive density in place of infinitude. Not covered: the degrees 4≤k≤84\le k\le8 and the third question, both settled by OpenAI's density theorem, and the first question.

Read depth. The statement is checked against B. Z. Moroz's zbMATH review of the paper (Zbl 1252.11070), which states the asymptotic for irreducible ff of degree d≥3d\ge3 and k≥(3d+1)/4k\ge(3d+1)/4 and the consequence for d≥9d\ge9 under the local condition f(Z/pd−2Z)≠{0}f(\mathbb{Z}/p^{d-2}\mathbb{Z})\ne\{0\} for every prime pp; the review gives no theorem number, and the paper is not held. The sentence on the method follows Section 1.1 of the release manuscript carded as OpenAI 2026. The proof is not checked.

Depends on. No page of this wiki; the claim rests on the paper above.

Acceptance. Refereed: Arch. Math. (Basel) 96 (2011), 139--150. The site's curator credits Browning with the case k≥9k\ge9 in the problem's remarks, but the site labels the problem OPEN, so that credit is commentary and is not counted as review.