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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> |
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− | == Proof ==
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− | 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
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− | <cmath>\begin{eqnarray*}(1+x)^n&=&(1+x)^{n_mp^m+n_{m-1}p^{m-1}+\cdots+n_1p+n_0}\\
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− | &=&[(1+x)^{p^m}]^{n_m}[(1+x)^{p^{m-1}}]^{n_{m-1}}\cdots[(1+x)^p]^{n_1}(1+x)^{n_0}\\
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− | &\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}
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− | \end{eqnarray*}</cmath>
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− | 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>.
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− | 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
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− | <center><math>\binom{n}{i}\equiv\prod_{k=1}^{n}\binom{n_k}{i_k}\pmod{p}</math></center>
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− | Note that <math>n_k<i_k\Longrightarrow\binom{n_k}{i_k}=0\Longrightarrow\binom{n}{i}\equiv 0 \pmod{p}</math>.
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− | 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>.
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| == See also == | | == See also == |
Latest revision as of 14:13, 11 August 2020