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How can I prove the following, where $p$ is a prime and $x$ a positive integer?

$$\dfrac{(2px)!}{((px)!)^2}\equiv\dfrac{(2x)!}{((x)!)^2}\pmod{p^2}$$

I'm not sure if it is actually true, but I tested for small numbers and it checked.

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    Are you aware that the expressions you've written are [binomial coefficients](http://en.wikipedia.org/wiki/Binomial_coefficient)?2012-07-29
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    Also, [see this question](http://math.stackexchange.com/q/78533/264) for the case of $p=2$.2012-07-29
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    This is a special case of Babbage's theorem (http://en.wikipedia.org/wiki/Wolstenholme's_theorem#A_proof_of_the_theorem).2012-07-29
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    It seems to be true also for $\mod p^3$ at least when $p>3$, all primes up to 101.2012-07-29
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    I was not aware of Babbage's theorem, it does solve it indeed. Here's an algebraic proof of the theorem: http://answers.yahoo.com/question/index?qid=20100810104024AAhqaaJ2012-07-29
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    @i.m.soloveichik: It is. The $\bmod p^3$ version is Wolstenholme's theorem, which is also proved at the link in my previous comment.2012-07-29
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    Perhaps someone should write up these comments as an answer. Ricbit, it's OK if you do it, now that you have an answer to your question.2012-07-30

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