FormalConjectures/ErdosProblems/366.lean

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L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL
Lean4 Domain Explanation Language

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

4 declarations, 2 contain sorry, 0 declared as axioms

Erdős Problem 366

Reference: erdosproblems.com/366

Imports (1)
  1. theorem Erdos366.erdos_366 Are there any $2$-full $n$ such that $n+1$ is $3$-full? sorry line 34 source

    Source documentation

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

    depends on sorry what this means

    Statement uses: Iff, True, Exists, Nat, And, GT.gt, OfNat.ofNat, Nat.Full, HAdd.hAdd

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    8 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_366 : answer(sorry) ↔ ∃ n > 0, (2).Full n ∧ (3).Full (n + 1) := by
    …
  2. theorem Erdos366.exists_three_full_then_two_full Note that $8$ is $3$-full and $9$ is 2-full line 41 source

    Source documentation

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

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: Exists, Nat, And, GT.gt, OfNat.ofNat, Nat.Full, HAdd.hAdd

    Proof uses: Exists.intro, of_eq_true, Eq.trans, True, congr, congrArg, eq_true, Mathlib.Meta.NormNum.isNat_lt_true, Mathlib.Meta.NormNum.isNat_ofNat, Eq.refl, Bool, Bool.false, Membership.mem, Finset, Nat.primeFactors, Dvd.dvd, HPow.hPow, forall_congr, Eq, implies_congr_ctx and 34 more

    37 instances, coercions and auxiliary constants left out.

    Tactic trace: 2 steps

    Open to load the checked before-and-after goal states.

    theorem exists_three_full_then_two_full : ∃ n > 0, (3).Full n ∧ (2).Full (n + 1) := by
    …
  3. theorem Erdos366.erdos_366.variants.three_two 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 line 51 source

    Source documentation

    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].

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: Iff, False, Nat, GT.gt, OfNat.ofNat, And, Nat.Full, HAdd.hAdd, Eq

    Proof uses: Iff.intro, False.elim, absurd, Mathlib.Meta.NormNum.isNat_lt_true, Mathlib.Meta.NormNum.isNat_ofNat, Eq.refl, Bool, Bool.false, of_eq_true, Eq.trans, True, congr, congrArg, Membership.mem, Finset, Nat.primeFactors, Dvd.dvd, HPow.hPow, forall_congr, implies_congr_ctx and 44 more

    39 instances, coercions and auxiliary constants left out.

    Tactic trace: 4 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_366.variants.three_two :
        answer(False) ↔ ∀ n > 0, (3).Full n ∧ (2).Full (n + 1) → n = 8 := by
    …
  4. theorem Erdos366.erdos_366.variants.weaker Are there any consecutive pairs of $3$-full integers? sorry line 60 source

    Source documentation

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

    depends on sorry what this means

    Statement uses: Iff, True, Exists, Nat, And, GT.gt, OfNat.ofNat, Nat.Full, HAdd.hAdd

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    8 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_366.variants.weaker : answer(sorry) ↔
        ∃ n > 0, (3).Full n ∧ (3).Full (n + 1) := by
    …