proof of Lucas’s theorem


Let n≥m∈ℕ. Let a0,b0 be the least non-negative residues of n,m(modp), respectively. (Additionally, we set r=n-m, and r0 is the least non-negative residue of r modulo p.) Then the statement follows from

(nm)≡(a0b0)⁢(⌊np⌋⌊mp⌋)(modp).

We define the ’carry indicators’ ci for all i≥0 as

ci=(1ifbi+ri≥p0otherwise,

and additionally c-1=0.

The special case s=1 of Anton’s congruenceMathworldPlanetmath is:

(n!)p≡(-1)⌊np⌋⋅a0!(modp), (1)

where a0 as defined above, and (n!)p is the product of numbers ≤n not divisible by p. So we have

n!⌊np⌋!⁢p⌊np⌋=(n!)p≡(-1)⌊np⌋⋅a0!(modp)

When dividing by the left-hand terms of the congruences for m and r, we see that the power of p is

⌊np⌋-⌊mp⌋-⌊rp⌋=c0

So we get the congruence

(nm)(⌊np⌋⌊mp⌋)⋅pc0≡(-1)c0⋅(a0b0),

or equivalently

(-1p)c0⋅(nm)≡(a0b0)⁢(⌊np⌋⌊mp⌋)(modp). (2)

Now we consider c0=1. Since

a0=b0+r0-p⁢c0=p⏟,

b0+r0<b0↔c0=1↔b0-(p-r0)=a0<b0↔(a0b0)=0. So both congruences–the one in the statement and (2)– produce the same results.

Title proof of Lucas’s theorem
Canonical name ProofOfLucassTheorem
Date of creation 2013-03-22 13:22:56
Last modified on 2013-03-22 13:22:56
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 7
Author mathcam (2727)
Entry type Proof
Classification msc 11A07
Related topic BinomialCoefficient