Problem Statement
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?
Categories:
Number Theory Binomial Coefficients
Progress
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).
Source: erdosproblems.com/397 | Last verified: January 14, 2026