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declared falsifiableerdos:106
If f(n)f(n) is the maximum total side length of nn interior-disjoint squares packed in the unit square, is f(k2+1)=kf(k^2 + 1) = k? An exact rational configuration packs 1717 squares with total side length greater than 44, refuting the identity at k=4k = 4.
geometryN/A3 sources
declared falsifiable · formalizederdos:488
True ↔ ∀ (A : Finset ℕ), A.Nonempty → 0 ∉ A → 1 ∉ A → ∀ (n m : ℕ), m > n → A.max ≤ ↑n → ↑{x ∈ Finset.Icc 1 m | x ∈ {n | n ≥ 1 ∧ ∃ a ∈ A, a ∣ n}}.card / ↑m < 2 * ↑{x ∈ Finset.Icc 1 n | x ∈ {n | n ≥ 1 ∧ ∃ a ∈ A, a ∣ n}}.card / ↑n
number theoryN/A3 sources
declared falsifiable · formalizederdos:1041
∀ (n : ℕ) (f : Polynomial ℂ), n ≥ 2 → f.natDegree = n → f.Monic → f.rootSet ℂ ⊆ Metric.ball 0 1 → ∃ z₁ z₂, ∃ (_ : {z₁, z₂} ≤ f.roots), ∃ γ, Set.range ⇑γ ⊆ {z | ‖Polynomial.eval z f‖ < 1} ∧ Erdos1041.length (Set.range ⇑γ) < 2
analysisN/A3 sources
declared falsifiable · formalizederdos:1082
True ↔ ∀ (A : Finset (EuclideanSpace ℝ (Fin 2))), EuclideanGeometry.NonTrilinear ↑A → A.card / 2 ≤ EuclideanGeometry.distinctDistances A
geometrydistancespossible3 sources

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