1. FormalConjectures
  2. ErdosProblems
  3. «366»
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Erdős Problem 366

Reference: erdosproblems.com/366

@[expose] public sectionnamespace Erdos366

Are there any $2$-full $n$ such that $n+1$ is $3$-full?

@[category research open, AMS 11] theorem erdos_366 : answer(sorry) ↔ ∃ n > 0, (2).Full n ∧ (3).Full (n + 1) := ⊢ True ↔ ∃ n > 0, Nat.Full 2 n ∧ Nat.Full 3 (n + 1) All goals completed! 🐙

Note that $8$ is $3$-full and $9$ is 2-full.

@[category test, AMS 11] theorem exists_three_full_then_two_full : ∃ n > 0, (3).Full n ∧ (2).Full (n + 1) := ⊢ ∃ n > 0, Nat.Full 3 n ∧ Nat.Full 2 (n + 1) ⊢ 8 > 0 ∧ Nat.Full 3 8 ∧ Nat.Full 2 (8 + 1) All goals completed! 🐙

Erdős and Graham asked whether $(8, 9)$ is the only pair of consecutive integers $n$, $n+1$ with $n$ $3$-full and $n+1$ $2$-full. The answer is no: $12167 = 23^3$ and $12168 = 2^3 3^2 13^2$ is another such pair, already known to Golomb [Go70].

@[category research solved, AMS 11] theorem erdos_366.variants.three_two : answer(False) ↔ ∀ n > 0, (3).Full n ∧ (2).Full (n + 1) → n = 8 := ⊢ False ↔ ∀ n > 0, Nat.Full 3 n ∧ Nat.Full 2 (n + 1) → n = 8 refine ⟨False.elim, fun h ↦ absurd (h 12167 (h:∀ n > 0, Nat.Full 3 n ∧ Nat.Full 2 (n + 1) → n = 8⊢ 12167 > 0 All goals completed! 🐙) ?_) (h:∀ n > 0, Nat.Full 3 n ∧ Nat.Full 2 (n + 1) → n = 8⊢ ¬12167 = 8 All goals completed! 🐙)⟩ All goals completed! 🐙

Are there any consecutive pairs of $3$-full integers?

@[category research open, AMS 11] theorem erdos_366.variants.weaker : answer(sorry) ↔ ∃ n > 0, (3).Full n ∧ (3).Full (n + 1) := ⊢ True ↔ ∃ n > 0, Nat.Full 3 n ∧ Nat.Full 3 (n + 1) All goals completed! 🐙end Erdos366