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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)[ki=0aipi]=ki=0(1+xpi)ai=ki=0aib=0(aib)xbpi

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 ki=0(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