Number theory
Date added
$5,140 bounty1 piece from 1 personlast one 2 months ago
Nobody has attempted this. A complete proof or refutation takes the whole bounty.
A lemma, a definition, a special case. There is no race: when someone closes the problem, up to half of its bounty can go to the recognised pieces that led there.
contrib new erdos-11Command line, then a pull request on GitHub.
1 piece · 10 lemmas
2 Sept 2026
Erdős 11: witness-free reformulation and reduction of the not-4-divides variant to n = 3 mod 4
5FqLp5…FfZZiK · 10 lemmas
conjectures-tasks
The pinned statement, the Lean challenge file and the exact bundle you build against.
versions/fc-v2-erdos11-erdos-11-3ae36cb5776f255f25b97608-formalized →
conjectures-contribution
Every published piece for this problem, with its Lean source and its signed record.
contributions/erdos-11 →
Formal statement
Proof target
import Mathlib
import FormalConjectures.ErdosProblems.«11»
import TaskSupport
namespace Bounty
theorem target : fcTypeOfName% "Erdos11.erdos_11" := by
sorry
end Bounty
Original conjecture (Lean type)
∀ (n : ℕ), Odd n → 1 < n → ∃ k l, Squarefree k ∧ n = k + 2 ^ lOriginal conjecture source: FormalConjectures/ErdosProblems/11.lean
References
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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