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Prove that $$\binom{2p}{p} \equiv 2\pmod{p^3},$$ where $p\ge 5$ is a prime number.

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It's easy to prove it by algebraic method, but I am very interested to find a combinatorial interpretation of it. –  Daniel Jan 15 '11 at 14:33
    
This 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 $a=2,b=1$. –  Daniel Jan 16 '11 at 1:44

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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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Have you read the bottom of that answer? –  Yuval Filmus Jan 15 '11 at 16:13
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Yes. What about it? –  Qiaochu Yuan Jan 15 '11 at 17:34
    
Thank 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. –  Daniel Jan 16 '11 at 1:33

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