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module
public import FormalConjecturesUtil
@[expose] public sectionnamespace Erdos137We say that $N$ is powerful if whenever $p\mid N$ we also have $p^2\mid N$.
Let $k\geq 3$. Can the product of any $k$ consecutive positive integers ever be powerful?
@[category research open, AMS 11]
theorem erdos_137 : answer(sorry) ↔ ∃ k ≥ 3, ∃ n, (∏ x ∈ Finset.Ioc n (n + k), x).Powerful := ⊢ True ↔ ∃ k ≥ 3, ∃ n, (∏ x ∈ Finset.Ioc n (n + k), x).Powerful
All goals completed! 🐙Let $k\geq 2$. Erdős and Selfridge [ES75] proved that the product of any $k$ consecutive integers $N$ cannot be a perfect power.
[ES75] P. Erdös, J. L. Selfridge, "The product of consecutive integers is never a power", Illinois J. Math. 19(2): 292-301, 1975
@[category research solved, AMS 11]
theorem erdos_137.variants.perfect_power (k : ℕ) (hk : k ≥ 2) (n : ℕ) (x l : ℕ) (hl : 2 ≤ l) :
(∏ x ∈ Finset.Ioc n (n + k), x) ≠ x ^ l := k:ℕhk:k ≥ 2n:ℕx:ℕl:ℕhl:2 ≤ l⊢ ∏ x ∈ Finset.Ioc n (n + k), x ≠ x ^ l
All goals completed! 🐙Erdős [Er82c] conjectures that, if $k$ is fixed, then for all $n$ sufficiently large and all positive integers $m$, there must be at least $k$ distinct primes $p$ such that $p\mid m(m+1)\cdots (m+n)$ and yet $p^2$ does not divide the right hand side.
[Er82c] Erdős, Paul, "Miscellaneous problems in number theory". Congr. Numer. (1982), 25-45.,
@[category research open, AMS 11]
theorem erdos_137.variants.multiple_powerful_factors (k : ℕ) : ∀ᶠ n in Filter.atTop,
∀ (m : ℕ) (hm : 0 < m),
letI N := ∏ x ∈ Finset.Ioc m (m + n), x
∃ P : Finset ℕ, P.card = k ∧ ∀ p ∈ P, p.Prime ∧
p ∣ N ∧ ¬ p ^ 2 ∣ N := k:ℕ⊢ ∀ᶠ (n : ℕ) in Filter.atTop,
∀ (m : ℕ),
0 < m →
∃ P,
P.card = k ∧ ∀ p ∈ P, Nat.Prime p ∧ p ∣ ∏ x ∈ Finset.Ioc m (m + n), x ∧ ¬p ^ 2 ∣ ∏ x ∈ Finset.Ioc m (m + n), x
All goals completed! 🐙end Erdos137