proof of Lucas’s theorem by binomial expansion
We work with polynomials in x over the integers modulo p.
By
the binomial theorem we have (1+x)p=1+xp. More
generally, by induction
on i we have (1+x)pi=1+xpi.
Hence the following holds:
(1+x)n=(1+x)[∑ki=0aipi]=k∏i=0(1+xpi)ai=k∏i=0ai∑b=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 |
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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 |