FormalConjectures.ErdosProblems.«397»

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import FormalConjecturesUtil

Erdős Problem 397

References:

erdosproblems.com/397

[MathOverflow] (https://mathoverflow.net/questions/138209/product-of-central-binomial-coefficients)


open Nat


namespace Erdos397

Are there only finitely many solutions to $$ \prod_i \binom{2m_i}{m_i}=\prod_j \binom{2n_j}{n_j} $$ with the $m_i,n_j$ distinct?

Somani, using ChatGPT, has given a negative answer. In fact, for any $a\geq 2$, if $c=8a^2+8a+1$, $\binom{2a}{a}\binom{4a+4}{2a+2}\binom{2c}{c}= \binom{2a+2}{a+1}\binom{4a}{2a}\binom{2c+2}{c+1}.$ Further families of solutions are given in the comments by SharkyKesa.

This was earlier asked about in a [MathOverflow] question, in response to which Elkies also gave an alternative construction which produces solutions - at the moment it is not clear whether Elkies' argument gives infinitely many solutions (although Bloom believes that it can).

This was formalized in Lean by Wu using Aristotle.

@[category research solved, AMS 11,
formal_proof using lean4 at "https://gist.github.com/llllvvuu/40d68cfa9de9f43eece07ff4fdc3b0ef",
formal_proof using formal_conjectures at "https://github.com/XC0R/formal-conjectures/blob/3c356a50a21bcbf3543f960b0c92d7fb26228cb6/FormalConjectures/ErdosProblems/397.lean#L147"]
theorem declaration uses 'sorry'erdos_397 :
    answer(False) ↔
      {(M, N) : Finset ℕ × Finset ℕ | Disjoint M N ∧
      ∏ i ∈ M, centralBinom i = ∏ j ∈ N, centralBinom j}.Finite := by⊢ False ↔ {(M, N) | Disjoint M N ∧ ∏ i ∈ M, i.centralBinom = ∏ j ∈ N, j.centralBinom}.Finite
  sorryAll goals completed! 🐙


end Erdos397