# 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 $n\mathrm{\ge}\mathrm{3}$ is an integer and $p$ is a prime number^{} such that $$, then
$p$ divides $\mathrm{\left(}\genfrac{}{}{0pt}{}{\mathrm{2}\mathit{}n}{n}\mathrm{\right)}$.

###### Proof.

We will examine the following expression for our binomial coefficient^{}:

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\frac{2n(2n-1)\mathrm{\cdots}(n+2)(n+1)}{n(n-1)\mathrm{\cdots}3\cdot 2\cdot 1}.$$ |

Since $$, 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 $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$. ∎

###### Theorem 2.

If $n\mathrm{\ge}\mathrm{3}$ is an integer and $p$ is a prime number such that $$, then $p$ does not divide $\mathrm{\left(}\genfrac{}{}{0pt}{}{\mathrm{2}\mathit{}n}{n}\mathrm{\right)}$.

###### Proof.

We will again examine our expression for our binomial coefficient:

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\frac{2n(2n-1)\mathrm{\cdots}(n+2)(n+1)}{n(n-1)\mathrm{\cdots}3\cdot 2\cdot 1}.$$ |

This time, because $$, 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 $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$.
∎

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 |