Erdős Problem #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). This was formalized in Lean by Wu using Aristotle.