Lucas' Theorem

Let $p$ be a prime. If $(\overline{n_mn_{m-1}\cdots n_0})_p$ is the base $p$ representation of $n$ and $(\overline{i_mi_{m-1}\cdots i_0})_p$ is the base $p$ representation of $i$ , where $n\geq i$ , Lucas' Theorem states that $\binom{n}{i}\equiv \prod_{j=0}^{m}\binom{n_j}{i_j}\pmod{p}$

Proof

Lemma

For $p$ prime and $x,r\in\mathbb{Z}$ , $(1+x)^{p^r}\equiv 1+x^{p^r}\pmod{p}$

Proof

For all $1\leq k \leq p-1$ , $\binom{p}{k}\equiv 0 \pmod{p}$ . Then we have that

$\begin{eqnarray*}(1+x)^p&\equiv &\binom{p}{0}+\binom{p}{1}x+\binom{p}{2}x^2+\cdots+\binom{p}{p-1}x^{p-1}+\binom{p}{p}x^p\\ &\equiv& 1+x^p\pmod{p}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

Assume we have $(1+x)^{p^k}\equiv 1+x^{p^k}\pmod{p}$ . Then

$\begin{eqnarray*}(1+x)^{p^{k+1}}

&\equiv&\left((1+x)^{p^k}\right)^p\\ &\equiv&\left(1+x^{p^k}\right)^p\\ &\equiv&\binom{p}{0}+\binom{p}{1}x^{p^k}+\binom{p}{2}x^{2p^k}+\cdots+\binom{p}{p-1}x^{(p-1)p^k}+\binom{p}{p}x^{p^{k+1}}\\

&\equiv&1+x^{p^{k+1}}\pmod{p}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

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Lucas' Theorem

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Proof

Lemma

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