proof of Lucas’s theorem by binomial expansion


We work with polynomials in x over the integers modulo p.
By the binomial theoremMathworldPlanetmath we have (1+x)p=1+xp. More generally, by inductionMathworldPlanetmath on i we have (1+x)pi=1+xpi.

Hence the following holds:

(1+x)n=(1+x)[∑i=0kai⁢pi]=∏i=0k(1+xpi)ai=∏i=0k∑b=0ai(aib)⁢xb⁢pi

Then the coefficient on xm on the left hand side is (nm).

As m is uniquely base p, the coefficient on xm on the right hand side is ∏i=0k(aibi).

Equating the coefficients on xm on either therefore yields the result.

Title proof of Lucas’s theorem by binomial expansion
Canonical name ProofOfLucassTheoremByBinomialExpansion
Date of creation 2013-03-22 18:19:59
Last modified on 2013-03-22 18:19:59
Owner whm22 (2009)
Last modified by whm22 (2009)
Numerical id 4
Author whm22 (2009)
Entry type Proof
Classification msc 11B65