OPEN
This is open, and cannot be resolved with a finite computation.
Are there infinitely many pairs of integers $n\neq m$ such that $\binom{2n}{n}$ and $\binom{2m}{m}$ have the same set of prime divisors?
A problem of Erdős, Graham, Ruzsa, and Straus
[EGRS75], who believed there is 'no doubt' that the answer is yes.
For example $(87,88)$ and $(607,608)$. Those $n$ such that there exists some suitable $m>n$ are listed as
A129515 in the OEIS.
A triple of such $n$ for which $\binom{2n}{n}$ all share the same set of prime divisors is $(10003,10004,10005)$. It is not known whether there are such pairs of the shape $(n,n+k)$ for every $k\geq 1$.
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This page was last edited 27 December 2025.
Additional thanks to: AlphaProof, Moritz Firsching, and Salvatore Mercuri
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 #730, https://www.erdosproblems.com/730, accessed 2026-01-16
