FormalConjectures/ErdosProblems/723.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

5 declarations, 5 contain sorry, 0 declared as axioms

Erdős Problem 723: The prime power conjecture.

Reference: erdosproblems.com/723

Imports (1)
  1. theorem Erdos723.erdos_723 If there is a finite projective plane of order $n$ then must $n$ be a prime power? sorry line 36 source

    Source documentation

    If there is a finite projective plane of order $n$ then must $n$ be a prime power?

    depends on sorry what this means

    Statement uses: Iff, True, Membership, Fintype, Configuration.ProjectivePlane, IsPrimePow, Nat, Configuration.ProjectivePlane.order

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

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_723 :
        answer(sorry) ↔ ∀ {P L : Type} (_: Membership P L) (_ : Fintype P) (_ : Fintype L),
          ∀ pp : ProjectivePlane P L, IsPrimePow pp.order := by
    …
  2. theorem Erdos723.erdos_723.variants.prime_power_is_projplane_order These always exist if $n$ is a prime power sorry line 45 source

    Source documentation

    These always exist if $n$ is a prime power.

    depends on sorry what this means

    Statement uses: Nat, IsPrimePow, Exists, Membership, Fintype, Configuration.ProjectivePlane, Eq, Configuration.ProjectivePlane.order

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

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_723.variants.prime_power_is_projplane_order :
        ∀ n, IsPrimePow n → ∃ (P L : Type) (_ : Membership P L) (_ : Fintype P) (_ : Fintype L)
          (pp : ProjectivePlane P L), pp.order = n := by
    …
  3. theorem Erdos723.erdos_723.variants.leq_11 This conjecture has been proved for $n \leq 11$ sorry line 54 source

    Source documentation

    This conjecture has been proved for $n \leq 11$.

    depends on sorry what this means

    Statement uses: Membership, Fintype, Configuration.ProjectivePlane, LE.le, Nat, Configuration.ProjectivePlane.order, OfNat.ofNat, IsPrimePow

    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_723.variants.leq_11 {P L : Type} [Membership P L] [Fintype P] [Fintype L] :
        ∀ pp : ProjectivePlane P L, pp.order ≤ 11 → IsPrimePow pp.order := by
    …
  4. theorem Erdos723.erdos_723.variants.eq_12 It is open whether there exists a projective plane of order 12 sorry line 62 source

    Source documentation

    It is open whether there exists a projective plane of order 12.

    depends on sorry what this means

    Statement uses: Iff, True, Exists, Membership, Fintype, Configuration.ProjectivePlane, Eq, Nat, Configuration.ProjectivePlane.order, OfNat.ofNat

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

    2 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_723.variants.eq_12 : answer(sorry) ↔
        ∃ (P L : Type) (_ : Membership P L) (_ : Fintype P) (_ : Fintype L) (pp : ProjectivePlane P L),
          pp.order = 12 := by
    …
  5. theorem Erdos723.erdos_723.variants.bruck_ryser Bruck and Ryser have proved that if $n \equiv 1 (\mod 4)$ or $n \equiv 2 (\mod 4)$ then $n$ must be the sum of two squares sorry line 72 source

    Source documentation

    Bruck and Ryser have proved that if $n \equiv 1 (\mod 4)$ or $n \equiv 2 (\mod 4)$ then $n$ must be the sum of two squares.

    depends on sorry what this means

    Statement uses: Membership, Fintype, Nat, Configuration.ProjectivePlane, Eq, Configuration.ProjectivePlane.order, Or, Nat.ModEq, OfNat.ofNat, Exists, HAdd.hAdd, HPow.hPow

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

    14 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_723.variants.bruck_ryser {P L : Type} [Membership P L] [Fintype P] [Fintype L]
        (n : ℕ) (pp : ProjectivePlane P L) (hpp : pp.order = n) :
        (n ≡ 1 [MOD 4] ∨ n ≡ 2 [MOD 4]) → ∃ a b, n = a ^ 2 + b ^ 2 := by
    …