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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 $(607,608)$.

Kummer's theorem implies that, for all odd primes $p$, $p\mid \binom{2n}{n}$ if and only some base $p$ digit of $n$ is $>p/2$, and hence $(n,n+1)$ has the required property if for all primes $p\leq n$ we have $n\not\in \{\frac{p-1}{2},p-1\}\pmod{p}$. Standard heuristics then predict there should be \[\gg \frac{x}{(\log x)^2}\] many such $n\leq x$.