5
$\begingroup$

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

  • 0
    It's easy to prove it by algebraic method, but I am very interested to find a combinatorial interpretation of it.2011-01-15
  • 0
    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 $a=2,b=1$.2011-01-16

1 Answers 1