1. FormalConjectures
  2. ErdosProblems
  3. «364»
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Erdős Problem 364

Reference: erdosproblems.com/364

open Natnamespace Erdos364

There is no consecutive triple of powerful numbers.

@[category research open, AMS 11] theorem erdos_364 : ¬ ∃ (n : ℕ), Powerful n ∧ Powerful (n + 1) ∧ Powerful (n + 2) := ⊢ ¬∃ n, n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful All goals completed! 🐙

Erdős [Er76d] conjectured a stronger statement: if $n_k$ is the $k$th powerful number, then $n_{k+2} - n_k > n_k^c$ for some constant $c > 0$.

[Er76d] Erdős, P., Problems and results on number theoretic properties of consecutive integers and related questions. Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1975) (1976), 25-44.

@[category research open, AMS 11] theorem erdos_364.variants.strong : ∃ (c : ℝ) (h : c > 0), ∀ (k : ℕ), Nat.nth Powerful (k + 2) - Nat.nth Powerful k > (Nat.nth Powerful k : ℝ) ^ c := ⊢ ∃ c, ∃ (_ : c > 0), ∀ (k : ℕ), ↑(nth Powerful (k + 2)) - ↑(nth Powerful k) > ↑(nth Powerful k) ^ c All goals completed! 🐙

There is no quadruple of powerful numbers, since at least one of the four numbers must be $2 \pmod{4}$, which cannot be powerful (since $2$ divides it, but $2^2$ does not).

n:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh2mod4:n % 4 = 2 ∨ (n + 1) % 4 = 2 ∨ (n + 2) % 4 = 2 ∨ (n + 3) % 4 = 2⊢ False n:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:n % 4 = 2⊢ Falsen:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:(n + 1) % 4 = 2⊢ Falsen:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:(n + 2) % 4 = 2⊢ Falsen:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:(n + 3) % 4 = 2⊢ False n:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:n % 4 = 2⊢ Falsen:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:(n + 1) % 4 = 2⊢ Falsen:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:(n + 2) % 4 = 2⊢ Falsen:ℕhn:n.Powerful ∧ (n + 1).Powerful ∧ (n + 2).Powerful ∧ (n + 3).Powerfulh✝:(n + 3) % 4 = 2⊢ False All goals completed! 🐙end Erdos364