# divisibility of central binomial coefficient

In this entry, we shall prove two results about the divisibility of central binomial coefficients which were stated in the main entry.

###### Theorem 1.

If $n\geq 3$ is an integer and $p$ is a prime number  such that $n, then $p$ divides ${2n\choose n}$.

###### Proof.
 ${2n\choose n}={2n(2n-1)\cdots(n+2)(n+1)\over n(n-1)\cdots 3\cdot 2\cdot 1}.$

Since $n, we find $p$ appearing in the numerator. However, $p$ cannot appear in the denominator because the terms there are all smaller than $n$. Hence, $p$ cannot be cancelled, so it must divide ${2n\choose n}$. ∎

###### Theorem 2.

If $n\geq 3$ is an integer and $p$ is a prime number such that $2n/3, then $p$ does not divide ${2n\choose n}$.

###### Proof.

We will again examine our expression for our binomial coefficient:

 ${2n\choose n}={2n(2n-1)\cdots(n+2)(n+1)\over n(n-1)\cdots 3\cdot 2\cdot 1}.$

This time, because $2n/3, we find $p$ appearing in the denominator and $2p$ appearing in the numerator. No other multiples  will appear because, if $m>2$, then $mp>2n$. The two occurrences of $p$ noted above cancel, hence $p$ is not a prime factor  of ${2n\choose n}$. ∎

Title divisibility of central binomial coefficient DivisibilityOfCentralBinomialCoefficient 2013-03-22 17:41:22 2013-03-22 17:41:22 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Proof msc 05A10 msc 11B65