Lucas' theorem

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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.[1]

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.

Proof

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

Let M be a set with m elements, and divide it into mi 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 Cpi 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 ni cycles of size pi. Thus the number of choices for N is exactly \prod _{{i=0}}^{k}{\binom  {m_{i}}{n_{i}}}{\pmod  {p}}.

Here is a proof based on generating functions, due to Nathan Fine.[2]

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

{\binom  pn}={\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}{\text{ mod }}p.

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

(1+X)^{{p^{i}}}\equiv 1+X^{{p^{i}}}{\text{ mod }}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 mi with 0 ≤ mip-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)\\&=\sum _{{n=0}}^{{m}}\left(\prod _{{i=0}}^{{k}}{\binom  {m_{i}}{n_{i}}}\right)X^{n}{\text{ mod }}p,\end{aligned}}

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

Variations and generalizations

  • The largest integer k such that pk 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.[3]

References

  1. Fine, Nathan (1947). "Binomial coefficients modulo a prime". American Mathematical Monthly 54: 589–592. 
  2. Andrew Granville (1997). "Arithmetic Properties of Binomial Coefficients I: Binomial coefficients modulo prime powers". Canadian Mathematical Society Conference Proceedings 20: 253–275. MR 1483922. 

External links

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