TITLE: Central binomial coefficients deprived of $2$'s: not radicals?
QUESTION [5 upvotes]: 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 $n>n*$.

REPLY [14 votes]: 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 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 $2$.