Difference between revisions of "Lucas' Theorem"

Revision as of 21:31, 21 April 2008

Lucas' Theorem states that for any prime $p$ , 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 $n\geq i$ , then $\binom{n}{i}\equiv \prod_{j=0}^{m}\binom{n_j}{i_j}\pmod{p}$ .

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

$

\begin{array}{rcl} (1 + x)^{p} & \equiv & (\binom{p}{0}) + (\binom{p}{1}) x + (\binom{p}{2}) x^{2} + \dots + (\binom{p}{p - 1}) x^{p - 1} + (\binom{p}{p}) x^{p} \\ \equiv & 1 + x^{p} (\mod p) \end{array}

$ (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)

Proof

Consider $(1+x)^n$ . If $(\overline{n_mn_{m-1}\cdots n_0})_p$ is the base $p$ representation of $n$ , then $0\leq n_k \leq p-1$ for all $0\leq k \leq m$ and $n=n_mp^m+n_{m-1}p^{m-1}+\cdots+n_1p+n_0$ . We then have

$\begin{eqnarray*}(1+x)^n&=&(1+x)^{n_mp^m+n_{m-1}p^{m-1}+\cdots+n_1p+n_0}\

&=&[(1+x)^{p^m}]^{n_m}[(1+x)^{p^{m-1}}]^{n_{m-1}}\cdots[(1+x)^p]^{n_1}(1+x)^{n_0}\ &\equiv&(1+x^{p^m})^{n_m}(1+x^{p^{m-1}})^{n_{m-1}}\cdots(1+x^p)^{n_1}(1+x)^{n_0}\pmod{p}

\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

We want the coefficient of $x^i$ in $(1+x)^n$ . Since $i=i_mp^m+i_{m-1}p^{m-1}+\cdots+i_1p+i_0$ , we want the coefficient of $(x^{p^{m}})^{i_{m}}(x^{p^{m-1}})^{i_{m-1}}\cdots (x^p)^{i_1}x^{i_0}$ .

The coefficient of each $(x^{p^{k}})^{i_{k}}$ comes from the binomial expansion of $(1+x^{p^k})^{n_k}$ , which is $\binom{n_k}{i_k}$ . Therefore we take the product of all such $\binom{n_k}{i_k}$ , and thus we have

$\binom{n}{i}\equiv\prod_{k=1}^{n}\binom{n_k}{i_k}\pmod{p}$

Note that $n_k<i_k\Longrightarrow\binom{n_k}{i_k}=0\Longrightarrow\binom{n}{i}\equiv 0 \pmod{p}$ .

This is equivalent to saying that there is no $x^i$ term in the expansion of $(1+x)^n=(1+x^{p^m})^{n_m}(1+x^{p^{m-1}})^{n_{m-1}}\cdots(1+x^p)^{n_1}(1+x)^{n_0}$ .

@@ Line 34: / Line 34: @@
 * [[Binomial Theorem]]
+[[Category:Theorems]]
 [[Category:Number theory]]
-[[Category:Theorems]]

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Difference between revisions of "Lucas' Theorem"

Revision as of 21:31, 21 April 2008

Contents

Lemma

Proof

Proof

See also