divisibility of central binomial coefficient

In this entry, we shall prove two results about thedivisibility of central binomial coefficientswhich were stated in the main entry.

Theorem 1.

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

Proof.

We will examine the following 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}.

Since $n<p<2n$ , we find $p$ appearing in the numerator. However, $p$ cannot appear in the denominator because the terms there are allsmaller 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 $2n/3<p\\leq n$ , 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 $2n/3<p\\leq n$ , we find $p$ appearing in the denominatorand $2p$ 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 ${2n\\choose n}$ .∎