Lucas's theorem

For the theorem in complex analysis, see Gauss–Lucas theorem.

In number theory, Lucas's theorem expresses the remainder of division of the binomial coefficient ${\tbinom {m}{n}}$ by a prime number p in terms of the base p expansions of the integers m and n.

Lucas's theorem first appeared in 1878 in papers by Édouard Lucas.

Formulation

For non-negative integers m and n and a prime p, the following congruence relation holds:

${\binom {m}{n}}\equiv \prod _{i=0}^{k}{\binom {m_{i}}{n_{i}}}{\pmod {p}},$

where

$m=m_{k}p^{k}+m_{k-1}p^{k-1}+\cdots +m_{1}p+m_{0},$

and

$n=n_{k}p^{k}+n_{k-1}p^{k-1}+\cdots +n_{1}p+n_{0}$

are the base p expansions of m and n respectively. This uses the convention that ${\tbinom {m}{n}}$ = 0 if m < n.

Consequence

• A binomial coefficient ${\tbinom {m}{n}}$ is divisible by a prime p if and only if at least one of the base p digits of n is greater than the corresponding digit of m.

Proofs

There are several ways to prove Lucas's theorem. We first give a combinatorial argument and then a proof based on generating functions.

Combinatorial argument

Let M be a set with m elements, and divide it into m_i cycles of length pⁱ for the various values of i. Then each of these cycles can be rotated separately, so that a group G which is the Cartesian product of cyclic groups C_pⁱ acts on M. It thus also acts on subsets N of size n. Since the number of elements in G is a power of p, the same is true of any of its orbits. Thus in order to compute ${\tbinom {m}{n}}$ modulo p, we only need to consider fixed points of this group action. The fixed points are those subsets N that are a union of some of the cycles. More precisely one can show by induction on k-i, that N must have exactly n_i cycles of size pⁱ. Thus the number of choices for N is exactly $\prod _{i=0}^{k}{\binom {m_{i}}{n_{i}}}{\pmod {p}}$.

Proof based on generating functions

This proof is due to Nathan Fine.

If p is a prime and n is an integer with 1≤n≤p-1, then the numerator of the binomial coefficient

${\binom {p}{n}}={\frac {p\cdot (p-1)\cdots (p-n+1)}{n\cdot (n-1)\cdots 1}}$

is divisible by p but the denominator is not. Hence p divides ${\tbinom {p}{n}}$. In terms of ordinary generating functions, this means that

$(1+X)^{p}\equiv 1+X^{p}{\pmod {p}}.$

Continuing by induction, we have for every nonnegative integer i that

$(1+X)^{p^{i}}\equiv 1+X^{p^{i}}{\pmod {p}}.$

Now let m be a nonnegative integer, and let p be a prime. Write m in base p, so that $m=\sum _{i=0}^{k}m_{i}p^{i}$ for some nonnegative integer k and integers m_i with 0 ≤ m_i ≤ p-1. Then

${\begin{aligned}\sum _{n=0}^{m}{\binom {m}{n}}X^{n}&=(1+X)^{m}=\prod _{i=0}^{k}\left((1+X)^{p^{i}}\right)^{m_{i}}\\&\equiv \prod _{i=0}^{k}\left(1+X^{p^{i}}\right)^{m_{i}}=\prod _{i=0}^{k}\left(\sum _{n_{i}=0}^{m_{i}}{\binom {m_{i}}{n_{i}}}X^{n_{i}p^{i}}\right)\\&=\prod _{i=0}^{k}\left(\sum _{n_{i}=0}^{p-1}{\binom {m_{i}}{n_{i}}}X^{n_{i}p^{i}}\right)=\sum _{n=0}^{m}\left(\prod _{i=0}^{k}{\binom {m_{i}}{n_{i}}}\right)X^{n}{\pmod {p}},\end{aligned}}$

where in the final product, n_i is the ith digit in the base p representation of n. This proves Lucas's theorem.

Variations and generalizations

• The largest integer k such that p^k divides the binomial coefficient ${\tbinom {m}{n}}$ (or in other words, the valuation of the binomial coefficient with respect to the prime p) is equal to the number of carries that occur when n and m − n are added in the base p. (This result is known as Kummer's theorem.)
• Andrew Granville has given a generalization of Lucas's theorem to the case of p being a power of prime.

External links

• Lucas's Theorem at PlanetMath.
• Alternate Proof of Lucas'Theorem