Lucas' Theorem/Corollary

From ProofWiki
Jump to navigation Jump to search

Corollary to Lucas' Theorem

Let $p$ be a prime number.

Let $n, k \in \Z$.


Let the representations of $n$ and $k$ to the base $p$ be given by:

$n = a_r p^r + \cdots + a_1 p + a_0$
$k = b_r p^r + \cdots + b_1 p + b_0$

Then:

$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$

where $\dbinom n k$ denotes a binomial coefficient.


Proof

Consider the representations of $n$ and $k$ to the base $p$:

$n = a_r p^r + \cdots + a_1 p + a_0$
$k = b_r p^r + \cdots + b_1 p + b_0$

Let:

$n_1 = \floor {n / p}$
$k_1 = \floor {k / p}$

We have that:

$n \bmod p = a_0$
$k \bmod p = b_0$
$n_1 = a_r p^{r - 1} + a_{r - 1} p^{r - 2} \cdots + a_1$
$k_1 = b_r p^{r - 1} + b_{r - 1} p^{r - 2} \cdots + b_1$

It follows from Lucas' Theorem that:

$\dbinom n k \equiv \dbinom {n_1} {k_1} \dbinom {a_0} {b_0} \pmod p$


Now we do the same again to the representation to the base $p$ of $n_1$ and $n_2$.

Thus:

$\dbinom n k \equiv \dbinom {\floor {n_1 / p}} {\floor {k_1 / p} } \dbinom {a_1} {b_1} \dbinom {a_0} {b_0} \pmod p$

and so on until:

$\floor {n_r / p}$

and:

$\floor {k_r / p}$

Hence the result.

$\blacksquare$


Source of Name

This entry was named for Édouard Lucas.


Sources