Field theory and polynomials
Erdős 477 - X pow three
Probably there is no such for the polynomial .References
- Sek59 Milan Sekanina, Замечания к фактoризации беcкoнечнoй цикличеcкoй группы, Czechoslovak Mathematical Journal, Vol. 9 (1959), No. 4, 485–495
No one has attempted this yet.
Formal statement
Lean type
∀ (A : Set ℤ), ∃ z, ¬∃! a, (a ∈ A ×ˢ Set.range fun x => Polynomial.eval x (Polynomial.X ^ 3)) ∧ z = a.1 + a.2What you must prove
import FormalConjectures.ErdosProblems.«477»
import TaskSupport
namespace Bounty
theorem target : fcTypeOfName% "Erdos477.erdos_477.variants.X_pow_three" := by
sorry
end Bounty
Pinned source: FormalConjectures/ErdosProblems/477.lean
- Source type SHA-256
- sha256:6b92749cc0b4a99fc7d402a7918889949b99f7e50709bd13e5546219d03dea46
- Task id
- fc-379fc029-variants-x-pow-three-f045d570a6-formalized-v1
- Task commitment
- sha256:f278c242bb7374eee87d73a878784a6b1e1e0fac3bdb25131040a70b03828e62
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.