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Erdős problem 978 - part ii If k > 3 k>3 k > 3 (and k ≠ 2 l k \neq 2^l k = 2 l ), and for all primes p p p there exists
such that
p k − 2 ∤ f ( n ) p^{k-2}\nmid f(n) p k − 2 ∤ f ( n ) ,
then are there infinitely many
for which
is
-power-free?
References
[Ho67] Hooley, C., On the power free values of polynomials. Mathematika (1967), 21--26. [Br11] Browning, T. D., Power-free values of polynomials. Arch. Math. (Basel) (2011), 139--150. [Er53] Erdős, P., Arithmetical properties of polynomials. J. London Math. Soc. (1953), 416--425. Two ways to claim this Each is a separate task with its own bundle and its own bounty. Pick the one your proof argues for.
Bounty
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Lean type
True ↔
∀ {f : Polynomial ℤ},
Irreducible f →
f.natDegree > 3 →
(¬∃ l, f.natDegree = 2 ^ l) →
0 < f.leadingCoeff →
(∀ (p : ℕ), Nat.Prime p → ∃ n, ¬↑p ^ (f.natDegree - 2) ∣ Polynomial.eval (↑n) f) →
{n | Powerfree (f.natDegree - 2) (Polynomial.eval (↑n) f)}.InfiniteWhat you must prove
import FormalConjectures.ErdosProblems.«978»
import TaskSupport
namespace Bounty
theorem target : ¬ (fcTypeOfName% "Erdos978.erdos_978.parts.ii") := by
sorry
end Bounty
Pinned source: FormalConjectures/ErdosProblems/978.lean
Source type SHA-256 sha256:e50e5b2eb4ed109cd65e96cc53aa9f5588b6855e9c1f88b73bbd30f94d3e8c29
Task id fc-379fc029-parts-ii-610df46f2f-counterexample-v1
Task commitment sha256:31845843e49cef61ffa209403ee5fedd7b8ba716b1ddf7bc92cc239d2d9f6aff Something wrong with this formalization?
A statement that does not faithfully capture the original conjecture is the one real risk here, so we would rather hear about it early - before someone spends weeks on it.
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Erdős problem 978 - part ii · Conjectures.io