Difference between revisions of "Lucas' Theorem"

m (Proof)
(Proof)
 
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&\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*}</cmath>
 
&\equiv&1+x^{p^{k+1}}\pmod{p}\end{eqnarray*}</cmath>
 
== Proof ==
 
Consider <math>(1+x)^n</math>. If <math>(\overline{n_mn_{m-1}\cdots n_0})_p</math> is the base <math>p</math> representation of <math>n</math>, then <math>0\leq n_k \leq p-1</math> for all <math>0\leq k \leq m</math> and <math>n=n_mp^m+n_{m-1}p^{m-1}+\cdots+n_1p+n_0</math>. We then have
 
<cmath>\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*}</cmath>
 
We want the coefficient of <math>x^i</math> in <math>(1+x)^n</math>. Since <math>i=i_mp^m+i_{m-1}p^{m-1}+\cdots+i_1p+i_0</math>, we want the coefficient of <math>(x^{p^{m}})^{i_{m}}(x^{p^{m-1}})^{i_{m-1}}\cdots (x^p)^{i_1}x^{i_0}</math>.
 
 
The coefficient of each <math>(x^{p^{k}})^{i_{k}}</math> comes from the binomial expansion of <math>(1+x^{p^k})^{n_k}</math>, which is <math>\binom{n_k}{i_k}</math>. Therefore we take the product of all such <math>\binom{n_k}{i_k}</math>, and thus we have
 
<center><math>\binom{n}{i}\equiv\prod_{k=1}^{n}\binom{n_k}{i_k}\pmod{p}</math></center>
 
Note that <math>n_k<i_k\Longrightarrow\binom{n_k}{i_k}=0\Longrightarrow\binom{n}{i}\equiv 0 \pmod{p}</math>.
 
 
This is equivalent to saying that there is no <math>x^i</math> term in the expansion of <math>(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}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 15:13, 11 August 2020

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*}

See also

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