FormalConjectures/ErdosProblems/403.lean

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Explanation legend Lean4 Domain Explanation Language (L4DEL)

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

1 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 403

References:

Imports (1)
  1. theorem Erdos403.erdos_403 Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1 0$ sorry line 58 source

    Source documentation

    Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1<a_2<\cdots <a_k$ have only finitely many solutions?

    Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is $$2^7=2!+3!+5!.$$ In fact Lin showed that the largest power of $2$ which can divide a sum of distinct factorials containing $2$ is $2^{254}$, and that there are only 5 solutions to $3^m=a_1!+\cdots+a_k!$ (when $m=0,1,2,3,6$).

    See also [404].

    A solution is encoded below as a pair $(m, s)$ where $s$ is the finite set $\{a_1 < a_2 < \cdots < a_k\}$ of positive integers, so the distinctness of the $a_i$ is given by set membership. The empty set contributes no solutions since $2^m \geq 1 > 0$.

    The linked proof gives more than finiteness: it classifies the solutions outright, as $(0,\{1\})$, $(1,\{2\})$, $(3,\{2,3\})$, $(5,\{2,3,4\})$ and $(7,\{2,3,5\})$, so the set below has exactly five elements.

    L4DEL explanation

    Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1 0$. The linked proof gives more than finiteness: it classifies the solutions outright, as $(0,\{1\})$, $(1,\{2\})$, $(3,\{2,3\})$, $(5,\{2,3,4\})$ and $(7,\{2,3,5\})$, so the set below has exactly five elements.

    Sources and attribution: Pinned Lean source (1646ca16afd6)

    depends on sorry what this means

    Statement uses: Iff, True, Set.Finite, Prod, Nat, Finset, Set.ofPred, And, Membership.mem, Prod.snd, LT.lt, OfNat.ofNat, Eq, HPow.hPow, Prod.fst, Finset.sum, Nat.factorial

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

    18 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_403 : answer(True) ↔
        {p : ℕ × Finset ℕ | (∀ a ∈ p.2, 0 < a) ∧
          2 ^ p.1 = ∑ a ∈ p.2, a.factorial}.Finite := by
    …