# Prove: $\frac{(2px)!}{((px)!)^2}\equiv\frac{(2x)!}{((x)!)^2}\pmod{p^2}$

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? – Zev Chonoles Jul 29 '12 at 15:38
Also, see this question for the case of $p=2$. – Zev Chonoles Jul 29 '12 at 15:40
This is a special case of Babbage's theorem (en.wikipedia.org/wiki/…). – Micah Jul 29 '12 at 15:52
It seems to be true also for $\mod p^3$ at least when $p>3$, all primes up to 101. – i. m. soloveichik Jul 29 '12 at 15:54
I was not aware of Babbage's theorem, it does solve it indeed. Here's an algebraic proof of the theorem: answers.yahoo.com/question/index?qid=20100810104024AAhqaaJ – Ricbit Jul 29 '12 at 16:03