erdos_397

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).

theorem erdos_397 :
    answer(False) ↔
      {(M, N) : Finset ℕ × Finset ℕ | Disjoint M N ∧
      ∏ i ∈ M, centralBinom i = ∏ j ∈ N, centralBinom j}.Finite

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