Erdős Problem #397

declared status 'disproved'. Formalized: yes. 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? Current best: 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. Prize: no. Tags: binomial coefficients, number theory.

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