It is easy to see that $4\mid \binom{2n}{n}$ except when $n=2^k$, and hence it suffices to prove this when $n$ is a power of $2$.

Proved by Sárközy [Sa85] for all sufficiently large $n$, and independently by Granville and Ramaré [GrRa96] and Velammal [Ve95] for all $n\geq 5$.

More generally, if $f(n)$ is the largest integer such that, for some prime $p$, we have $p^{f(n)}$ dividing $\binom{2n}{n}$, then $f(n)$ should tend to infinity with $n$. Can one even disprove that $f(n)\gg \log n$?

Sander [Sa92] proved that $f(n)\to \infty$, and later [Sa95] quantified this to $f(n) \gg (\log n)^{1/10-o(1)}$. Sander [Sa95] also proved that $f(n)\ll \log n$ for all $n$, and that $f(n) \gg \log n$ for almost all $n$. The proofs of the latter two facts are very easy using Kummer's theorem -- indeed, this immediately implies that any $p$ divides $\binom{2n}{n}$ with multiplicity $\ll \log_p n$, and that the multiplicity with which $2$ divides $\binom{2n}{n}$ is equal to the number of $1$s in the binary expansion of $n$, which is $\gg \log n$ for almost all $n$.

Erdős and Kolesnik [ErKo99] improved the lower bound for all $n$ to\[f(n) \gg (\log n)^{1/4-o(1)}.\]It remains open whether $f(n) \gg \log n$ for all $n$.

Sander [Sa92b] proved that, for all $0<\epsilon<1$, if $n$ is sufficiently large and $\lvert d\rvert\leq n^{1-\epsilon}$ then $\binom{2n+d}{n}$ is not squarefree.

The largest $n$ known for which $\binom{2n}{n}$ is not divisible by the square of an odd prime is $n=786$ (found by Levine). Guy [Gu04] reports that Erdős 'feels sure that there are no larger such $n$'.

See also [379].

This is mentioned in problem B33 of Guy's collection [Gu04].

View the LaTeX source

This page was last edited 08 February 2026.