Catalog
Every problem here was open when it entered the pool. Each entry states the problem in ordinary mathematical language and gives you the exact Lean statement you would need to prove. Nothing is paraphrased, so what you read is what gets checked.
A proof outlives the network that paid for it.
Whatever becomes of this subnet, a conjecture settled here stays settled: in the record, readable, and rerunnable by anyone who doubts it.
Green's open problem 39 If A ⊂ Z / p Z A \subset \mathbb{Z}/p\mathbb{Z} A ⊂ Z / p Z is random, ∣ A ∣ = p |A| = \sqrt{p} ∣ A ∣ = p , can we almost surely cover
Z / p Z \mathbb{Z}/p\mathbb{Z} Z / p Z with
translates of
? [Gr24]
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
$770
paid on an accepted proof
Set by bounty policy dynamic-age-v2-locked: the amount is worked out from how long the problem has stood open, so it moves as the pool and the pool's age profile move.
Submit a proof Your file is checked for free before any credit is spent.
Lean type
True ↔
Filter.Tendsto
(fun p =>
have k := (↑p).sqrt;
have c := 100 * k;
↑(Green39.proportionCoverable (↑p) k c))
Filter.atTop (nhds 1)What you must prove
import FormalConjectures.GreensOpenProblems.«39»
import TaskSupport
namespace Bounty
theorem target : ¬ (fcTypeOfName% "Green39.green_39") := by
sorry
end Bounty
Pinned source: FormalConjectures/GreensOpenProblems/39.lean
Source type SHA-256 sha256:d4997d4161d1057aab918fdeb2713bf2239055c9ebd1ef1e236f028f243c6417
Task id fc-8432eac9-green39-green-39-5fcd9b62ed-counterexample-v1
Task commitment sha256:01de128c91e344ce91142198179baa36547176a0dd7c139439ff6853a334cac8 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.
Tell us on Discord
Green's open problem 39 · Conjectures.io