1. FormalConjectures
  2. ErdosProblems
  3. «937»
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Erdős Problem 937

Reference: erdosproblems.com/937

References:

    [BBC24] Bajpai, P., Bennett, M. A. and Chan, T. H., Arithmetic progressions in squarefull / powerful numbers, Int. J. Number Theory 20 (2024), 19-45.

namespace Erdos937 open Nat

The four numbers $a, a+d, a+2d, a+3d$ form a four-term arithmetic progression ($d > 0$) of pairwise coprime powerful numbers.

def IsCoprimePowerfulAP4 (a d : ℕ) : Prop := 0 < d ∧ a.Powerful ∧ (a + d).Powerful ∧ (a + 2 * d).Powerful ∧ (a + 3 * d).Powerful ∧ a.Coprime (a + d) ∧ a.Coprime (a + 2 * d) ∧ a.Coprime (a + 3 * d) ∧ (a + d).Coprime (a + 2 * d) ∧ (a + d).Coprime (a + 3 * d) ∧ (a + 2 * d).Coprime (a + 3 * d)

Are there infinitely many four-term arithmetic progressions of coprime powerful numbers? (A number $n$ is powerful if $p \mid n \to p^2 \mid n$; Nat.Powerful.)

Erdős [Er76d] asked this; the answer is yes: Bajpai, Bennett and Chan [BBC24] proved that there are infinitely many four-term arithmetic progressions of pairwise coprime powerful numbers. (Without coprimality this is easy, and by a theorem of Fermat there are no four squares in arithmetic progression.)

@[category research solved, AMS 11] theorem declaration uses 'sorry'erdos_937 : answer(True) ↔ {p : ℕ × ℕ | IsCoprimePowerfulAP4 p.1 p.2}.Infinite := ⊢ True ↔ {p | IsCoprimePowerfulAP4 p.1 p.2}.Infinite All goals completed! 🐙

Sanity check for IsCoprimePowerfulAP4: the progression $0, 1, 2, 3$ is not a valid example, since $2$ is not powerful.

@[category test, AMS 11] theorem not_isCoprimePowerfulAP4_zero_one : ¬ IsCoprimePowerfulAP4 0 1 := ⊢ ¬IsCoprimePowerfulAP4 0 1 h:IsCoprimePowerfulAP4 0 1⊢ False h2:(0 + 2 * 1).Powerful⊢ False exact Nat.not_full_of_prime_mod_prime_sq 2 1 Nat.prime_two (h2:(0 + 2 * 1).Powerful⊢ 2 % 2 ^ (1 + 1) = 2 All goals completed! 🐙) h2 end Erdos937