Prove that $$\binom{2p}{p} \equiv 2\pmod{p^3},$$ where $p\ge 5$ is a prime number.
Is there a combinatorial proof of this congruence identity?
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$\begingroup$
number-theory
combinatorics
binomial-coefficients
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0It's easy to prove it by algebraic method, but I am very interested to find a combinatorial interpretation of it. – 2011-01-15
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0This congruence identity can be generalized as follows$${ap \choose bp} \equiv {a \choose b} \pmod{p^3},$$ where $p$ is a prime number and $a,b$ are positive integers. The combinatorial proof of it can be reduce to the case $a=2,b=1$. – 2011-01-16
1 Answers
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A combinatorial proof for the congruence $\bmod p^2$ is given at this MO question but the answerer suggests the congruence $\bmod p^3$ does not have a natural combinatorial proof.
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0Have you read the bottom of that answer? – 2011-01-15
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3Yes. What about it? – 2011-01-15
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0Thank to QiaoChu Yuan very much. Indeed, there is a more general result, which have a beautiful combinatorial proof, for the congruence mod $p^2$ as follows $${ ap \choose bp} \equiv { a \choose b } \pmod {p^2}, $$ wherer $p$ is a prime number and $a, b$ are positive integers. – 2011-01-16