In the paper, P Erdos, R Graham, I Ruzsa, E Straus, *On the prime factors of $\binom{2n}n$*, Math. Comp., 29:83–92, 1975, it was conjectured that the central binomials are never square-free for $n>4$. The proof was given in A Granville, O Ramare, *Explicit bounds on exponential sums and the scarcity
of squarefree binomial coefficients,* Mathematika, 43:73–107, 1996.


Cute fact: $\frac12\binom{2n}n\in\mathbb{Z}$, and it is odd iff $n=2^k$. So, remove powers of $2$ to write $\frac1{2^n}\binom{2n}n:=\frac{U\_n}{V\_n}$ as reduced fractions, and ask:



> 
> **Question.** Is this true? There exists $n\*\in\mathbb{N}$ such that
>  $U\_n$ is never square-free whenever $n>n\*$.
> 
> 
> 



Central binomial coefficients deprived of $2$'s: not radicals?
The real thrust of the work of Granville and Ramare was to get explicit bounds, so that one could get a complete resolution of the Erdos problem. Earlier work of [Sarkozy](http://www.sciencedirect.com/science/article/pii/0022314X85900174) already gave asymptotic results that are quite a bit sharper. Thus, Sarkozy showed that if one writes 
$$ 
\binom{2n}{n} = (s(n))^2 q(n), 
$$ 
where $q(n)$ is square-free, then 
$$ 
\exp((c-\epsilon)\sqrt{n}) \le s(n)^2 \le \exp((c+\epsilon)\sqrt{n})
$$ 
for large $n$. Here 
$$ 
c= \sqrt{2} \sum\_{k=1}^{\infty} \Big( \frac{1}{\sqrt{2k-1}} -\frac{1}{\sqrt{2k}}\Big) 
$$ 
is a positive constant. Note that the power of $2$ dividing $\binom{2n}{n}$ is no more than $\log\_2 (2n)$, and so the square-factor in Sarkozy's result certainly doesn't just come from powers of $2$. 


