DISPROVED (LEAN)
This has been solved in the negative and the proof verified in Lean.
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 I believe it can).
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This page was last edited 12 January 2026. View history
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Formalised statement?
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Additional thanks to: SharkyKesa, Neel Somani, Nat Sothanaphan, and Terence Tao
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #397, https://www.erdosproblems.com/397, accessed 2026-04-11
