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| {{For|the theorem in complex analysis|Gauss–Lucas theorem}}
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| In [[number theory]], '''Lucas's theorem''' expresses the [[remainder]] of division of the [[binomial coefficient]] <math>\tbinom{m}{n}</math> by a [[prime number]] ''p'' in terms of the [[radix|base]] ''p'' expansions of the integers ''m'' and ''n''.
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| Lucas's theorem first appeared in 1878 in papers by [[Édouard Lucas]].<ref>
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| *{{cite journal| author=Edouard Lucas |title=Théorie des Fonctions Numériques Simplement Périodiques| jstor=2369308 |journal=[[American Journal of Mathematics]] |year=1878 |volume=1 |issue=2 |pages=184–196 |doi=10.2307/2369308| mr=1505161}} (part 1);
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| *{{cite journal| author=Edouard Lucas |title=Théorie des Fonctions Numériques Simplement Périodiques| jstor=2369311 |journal=[[American Journal of Mathematics]] |year=1878 |volume=1 |issue=3 |pages=197–240 |doi=10.2307/2369311| mr=1505164}} (part 2);
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| *{{cite journal| author=Edouard Lucas |title=Théorie des Fonctions Numériques Simplement Périodiques| jstor=2369373 |journal=[[American Journal of Mathematics]] |year=1878 |volume=1 |issue=4 |pages=289–321 |doi=10.2307/2369373| mr=1505176}} (part 3)</ref>
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| == Formulation ==
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| For non-negative integers ''m'' and ''n'' and a prime ''p'', the following [[modular arithmetic|congruence relation]] holds:
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| :<math>\binom{m}{n}\equiv\prod_{i=0}^k\binom{m_i}{n_i}\pmod p,</math>
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| where
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| :<math>m=m_kp^k+m_{k-1}p^{k-1}+\cdots +m_1p+m_0,</math>
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| and
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| :<math>n=n_kp^k+n_{k-1}p^{k-1}+\cdots +n_1p+n_0</math>
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| are the base ''p'' expansions of ''m'' and ''n'' respectively. This uses the convention that <math>\tbinom{m}{n}</math> = 0 if ''m'' < ''n''.
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| == Consequence ==
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| * A binomial coefficient <math>\tbinom{m}{n}</math> is divisible by a prime ''p'' if and only if at least one of the base ''p'' digits of ''n'' is greater than the corresponding digit of ''m''.
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| == Proof ==
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| There are several ways to prove Lucas's theorem. We first give a combinatorial argument and then a proof based on generating functions.
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| Let ''M'' be a set with ''m'' elements, and divide it into ''m<sub>i</sub>'' cycles of length ''p<sup>i</sup>'' for the various values of ''i''. Then each of these cycles can be rotated separately, so that a group ''G'' which is the Cartesian product of cyclic groups ''C<sub>p<sup>i</sup></sub>'' acts on ''M''. It thus also acts on subsets ''N'' of size ''n''. Since the number of elements in ''G'' is a power of ''p'', the same is true of any of its orbits. Thus in order to compute <math>\tbinom{m}{n}</math> modulo ''p'', we only need to consider fixed points of this group action. The fixed points are those subsets ''N'' that are a union of some of the cycles. More precisely one can show by induction on ''k''-''i'', that ''N'' must have exactly ''n<sub>i</sub>'' cycles of size ''p<sup>i</sup>''. Thus the number of choices for ''N'' is exactly
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| <math>\prod_{i=0}^k\binom{m_i}{n_i}\pmod{p}</math>.
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| Here is a proof based on generating functions, due to Nathan Fine.<ref>{{cite journal|last=Fine|first=Nathan|title=Binomial coefficients modulo a prime|journal=American Mathematical Monthly|year=1947|volume=54|pages=589–592}}</ref>
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| If ''p'' is a prime and ''n'' is an integer with 1≤''n''≤''p''-1, then the numerator of the binomial coefficient
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| :<math> \binom p n = \frac{p \cdot (p-1) \cdots (p-n+1)}{n \cdot (n-1) \cdots 1} </math>
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| is divisible by ''p'' but the denominator is not. Hence ''p'' divides <math>\tbinom{p}{n}</math>. In terms of ordinary generating functions, this means that
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| :<math>(1+X)^p\equiv1+X^p\text{ mod }p.</math>
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| Continuing by induction, we have for every nonnegative integer ''i'' that
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| :<math>(1+X)^{p^i}\equiv1+X^{p^i}\text{ mod }p.</math>
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| Now let ''m'' be a nonnegative integer, and let ''p'' be a prime. Write ''m'' in base ''p'', so that <math>m=\sum_{i=0}^{k}m_ip^i</math> for some nonnegative integer ''k'' and integers ''m''<sub>''i''</sub> with 0 ≤ ''m''<sub>''i''</sub> ≤ ''p''-1. Then
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| :<math>\begin{align}
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| \sum_{n=0}^{m}\binom{m}{n}X^n &
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| =(1+X)^m=\prod_{i=0}^{k}\left((1+X)^{p^i}\right)^{m_i}\\
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| & \equiv \prod_{i=0}^{k}\left(1+X^{p^i}\right)^{m_i}
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| =\prod_{i=0}^{k}\left(\sum_{n_i=0}^{m_i}\binom{m_i}{n_i}X^{n_ip^i}\right)\\
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| & =\sum_{n=0}^{m}\left(\prod_{i=0}^{k}\binom{m_i}{n_i}\right)X^n
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| \text{ mod } p,
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| \end{align}</math>
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| where in the final product, ''n''<sub>''i''</sub> is digit ''i'' in the base ''p'' representation of ''n''. This proves Lucas's theorem.
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| == Variations and generalizations ==
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| * The largest integer ''k'' such that ''p''<sup>''k''</sup> divides the binomial coefficient <math>\tbinom{m}{n}</math> (or in other words, the [[p-adic order|valuation]] of the binomial coefficient with respect to the prime ''p'') is equal to the number of [[Carry (arithmetic)|carries]] that occur when ''n'' and ''m'' − ''n'' are added in the [[Positional notation#Base of the numeral system|base ''p'']]. (This result is known as [[Kummer's theorem]].)
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| * [[Andrew Granville]] has given a generalization of Lucas's theorem to the case of ''p'' being a power of prime.<ref>{{cite journal |author=[[Andrew Granville]] |title=Arithmetic Properties of Binomial Coefficients I: Binomial coefficients modulo prime powers |journal=Canadian Mathematical Society Conference Proceedings |volume=20 |pages=253–275 |year=1997 |url=http://www.dms.umontreal.ca/%7Eandrew/PDF/BinCoeff.pdf |mr=1483922}}</ref>
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| ==References==
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| {{reflist}}
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| ==External links==
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| *{{PlanetMath|urlname=LucassTheorem|title=Lucas's Theorem}}
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| *[http://arxiv.org/abs/1301.4250 Alternate Proof of Lucas'Theorem]
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| [[Category:Articles containing proofs]]
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| [[Category:Theorems about prime numbers]]
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