Binomial Coefficient involving Power of Prime
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Theorem
Let $p$ be a prime number.
Let $k \in \Z$.
Let $n \in \Z_{>0}$.
Then:
- $\dbinom {p^n k} {p^n} \equiv k \pmod p$
where $\dbinom {p^n k} {p^n}$ is a binomial coefficient.
Proof 1
From Prime Power of Sum Modulo Prime we have:
- $(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$
We can write this:
- $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$
By $(1)$ and Congruence of Powers, we therefore have:
- $\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} + b^{p^n} }^k \pmod p$
The coefficient $\dbinom {p^n k} {p^n}$ is the binomial coefficient of $b^{p^n}$ in $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$.
Expanding $\paren {a^{p^n} + b^{p^n} }^k$ using the Binomial Theorem, we find that the coefficient of $b^{p^n}$, the second term, is $\dbinom k 1 = k$.
So:
- $\dbinom {p^n k} {p^n} \equiv k \pmod p$
$\blacksquare$
Proof 2
Lucas' Theorem states that for $n, k, p \in \Z$ and $p$ be a prime number, such that:
- $n = a_r p^r + \cdots + a_1 p + a_0$
- $k = b_r p^r + \cdots + b_1 p + b_0$
then:
- $\ds \binom n k \equiv \prod_{j \mathop = 0}^r \binom {a_j}{b_j} \pmod p$
Therefore:
\(\ds \binom {p^n k} {p^n}\) | \(\equiv\) | \(\ds \binom k 1 \prod_{j \mathop = 0}^{n - 1} \binom 0 0 \pmod p\) | Lucas' Theorem | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds k \prod_{j \mathop = 0}^{n - 1} 1 \pmod p\) | Binomial Coefficient with One, Binomial Coefficient with Zero | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds k \pmod p\) |
$\blacksquare$