FormalConjectures/ErdosProblems/405.lean

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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

3 declarations, 2 contain sorry, 0 declared as axioms

Erdős Problem 405

References:

Imports (1)
  1. theorem Erdos405.erdos_405 Let $p$ be an odd prime sorry line 48 source

    Source documentation

    Let $p$ be an odd prime. Is it true that the equation $(p-1)!+a^{p-1}=p^k$ has only finitely many solutions?

    Originally proposed by Erdős and Graham [ErGr80]. Brindza and Erdős [BrEr91] proved that there are finitely many such solutions.

    depends on sorry what this means

    Statement uses: Set.Finite, Prod, Nat, Set.ofPred, Erdos405.erdos_405.match_1, And, Nat.Prime, Odd, Eq, HAdd.hAdd, Nat.factorial, HSub.hSub, OfNat.ofNat, HPow.hPow

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

    20 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_405 :
        Set.Finite { x : ℕ × ℕ × ℕ | let (a, k, p) := x; p.Prime ∧ Odd p ∧
    …
  2. theorem Erdos405.erdos_405.variants.nonodd_p Erdős and Graham [ErGr80] ask this allowing $p=2$, but this is presumably an oversight, since clearly there are infinitely many solutions when $p=2$ line 60 source

    Source documentation

    Erdős and Graham [ErGr80] ask this allowing $p=2$, but this is presumably an oversight, since clearly there are infinitely many solutions when $p=2$.

    Observe that this creates $1! + a = 2^k$. For all $k$, fix $a = 2^k - 1$.

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

    Statement uses: Set.Infinite, Prod, Nat, Set.ofPred, Erdos405.erdos_405.match_1, And, Nat.Prime, Eq, HAdd.hAdd, Nat.factorial, HSub.hSub, OfNat.ofNat, HPow.hPow

    Proof uses: Set.infinite_of_injective_forall_mem, Prod.mk, And.left, True, And.right, Eq.mp, Eq.trans, Prod.mk.injEq, congrArg, eq_self, id, Membership.mem, Set, And.intro, Nat.prime_two, Eq.mpr, pow_one, LT.lt, pow_pos, Mathlib.Meta.Positivity.pos_of_isNat and 6 more

    35 instances, coercions and auxiliary constants left out.

    Tactic trace: 14 steps

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

    theorem erdos_405.variants.nonodd_p :
        Set.Infinite { x : ℕ × ℕ × ℕ | let (a, k, p) := x; p.Prime ∧
    …
  3. theorem Erdos405.erdos_405.variants.yu_liu Yu and Liu [YuLi96] showed that the only solutions to $(p-1)! + a^(p-1) = p^k$ for an odd prime p are: $$ 2! + 1^2 = 3\\ 2! + 5^2 = 3^3\\ 4! + 1^4 = 5^2 $$ sorry line 94 source

    Source documentation

    Yu and Liu [YuLi96] showed that the only solutions to $(p-1)! + a^(p-1) = p^k$ for an odd prime p are: $$ 2! + 1^2 = 3\\ 2! + 5^2 = 3^3\\ 4! + 1^4 = 5^2 $$

    depends on sorry what this means

    Statement uses: Eq, Set, Prod, Nat, Set.ofPred, Erdos405.erdos_405.match_1, And, Nat.Prime, Odd, HAdd.hAdd, Nat.factorial, HSub.hSub, OfNat.ofNat, HPow.hPow, Insert.insert, Prod.mk, Singleton.singleton

    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_405.variants.yu_liu :
        { x : ℕ × ℕ × ℕ | let (a, k, p) := x; p.Prime ∧ Odd p ∧
    …