Difference between revisions of "Lucas' Theorem"

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=== Proof ===
 
=== Proof ===
 
For all <math>1\leq k \leq p-1</math>, <math>\binom{p}{k}\equiv 0 \pmod{p}</math>. Then we have
 
For all <math>1\leq k \leq p-1</math>, <math>\binom{p}{k}\equiv 0 \pmod{p}</math>. Then we have
<center><math>\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\\
+
<cmath>\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*}</math></center>
+
&\equiv& 1+x^p\pmod{p}\end{eqnarray*}</cmath>
 
Assume we have <math>(1+x)^{p^k}\equiv 1+x^{p^k}\pmod{p}</math>. Then
 
Assume we have <math>(1+x)^{p^k}\equiv 1+x^{p^k}\pmod{p}</math>. Then
<center><math>\begin{eqnarray*}(1+x)^{p^{k+1}}
+
<cmath>\begin{eqnarray*}(1+x)^{p^{k+1}}
 
&\equiv&\left((1+x)^{p^k}\right)^p\\
 
&\equiv&\left((1+x)^{p^k}\right)^p\\
 
&\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&\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*}</math></center>
+
&\equiv&1+x^{p^{k+1}}\pmod{p}\end{eqnarray*}</cmath>
  
 
== Proof ==
 
== Proof ==

Revision as of 19:10, 10 March 2015

Lucas' Theorem states that for any prime $p$ and any positive integers $n\geq i$, if $(\overline{n_mn_{m-1}\cdots n_0})_p$ is the representation of $n$ in base $p$ and $(\overline{i_mi_{m-1}\cdots i_0})_p$ is the representation of $i$ in base $p$ (possibly with some leading $0$s) 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{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*} 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*}

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. ! Missing \endgroup inserted.)

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 $(1+x)^n=(1+x^{p^m})^{n_m}(1+x^{p^{m-1}})^{n_{m-1}}\cdots(1+x^p)^{n_1}(1+x)^{n_0}$.

See also

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