Binomial Coefficient of Prime/Proof 2

Theorem

Let $p$ be a prime number.


Then:

$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$

where $\dbinom p k$ is defined as a binomial coefficient.


Proof

Lucas' Theorem gives:

$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$

So, substituting $p$ for $n$:

$\dbinom p k \equiv \dbinom {\floor {p / p} } {\floor {k / p} } \dbinom {p \bmod p} {k \bmod p} \pmod p$


But $p \bmod p = 0$ by definition.

Hence, if $0 < k < p$, we have that:

$k \bmod p \ne 0$

and so:

$\dbinom {p \bmod p} {k \bmod p} = \dbinom 0 {k \bmod p} = 0$

by definition of binomial coefficients.

The result follows immediately.

$\blacksquare$


Sources

  • 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(b)}$
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