Problem Statement
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?
Categories:
Number Theory Binomial Coefficients Base Representations
Progress
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 $k\geq 1$.
Source: erdosproblems.com/730 | Last verified: January 16, 2026