364.lean`

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3 declarations, 2 contain sorry, 0 declared as axioms

Erdős Problem 364

Reference: erdosproblems.com/364

Imports (1)

FormalConjecturesUtil

theorem Erdos364.erdos_364 There is no consecutive triple of powerful numbers sorry line 34 source
Source documentation
There is no consecutive triple of powerful numbers.

depends on sorry what this means
Statement uses: Not, Exists, Nat, And, Nat.Powerful, HAdd.hAdd, OfNat.ofNat
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
6 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_364 :
    ¬ ∃ (n : ℕ), Powerful n ∧ Powerful (n + 1) ∧ Powerful (n + 2) := by
…
```
theorem Erdos364.erdos_364.variants.strong 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 $c > 0$ sorry line 45 source
Source documentation
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.

depends on sorry what this means
Statement uses: Exists, Real, GT.gt, OfNat.ofNat, Nat, HSub.hSub, Nat.nth, Nat.Powerful, HAdd.hAdd, HPow.hPow
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
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 := by
…
```
Mentions: c
theorem Erdos364.erdos_364.variants.weak 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) line 55 source
Source documentation
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).

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Exists, Nat, And, Nat.Powerful, HAdd.hAdd, OfNat.ofNat
Proof uses: False, Or, Eq, HMod.hMod, Decidable.byContradiction, False.elim, Eq.mp, Eq.trans, congrFun', congrArg, HPow.hPow, eq_false, Nat.Full, Nat.not_full_of_prime_mod_prime_sq, Nat.prime_two, of_eq_true, True, eq_self, false_and, and_false
17 instances, coercions and auxiliary constants left out.

Tactic trace: 6 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_364.variants.weak :
    ¬ ∃ (n : ℕ), Powerful n ∧ Powerful (n + 1) ∧ Powerful (n + 2) ∧ Powerful (n + 3) := by
…
```