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 $n<p<2n$ , 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 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 $2n/3<p\leq n$ , 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 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
Canonical name	DivisibilityOfCentralBinomialCoefficient
Date of creation	2013-03-22 17:41:22
Last modified on	2013-03-22 17:41:22
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	8
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 05A10
Classification	msc 11B65