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.
Asked
Active
Viewed 436 times
3
-
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 (http://en.wikipedia.org/wiki/Wolstenholme's_theorem#A_proof_of_the_theorem). – 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: http://answers.yahoo.com/question/index?qid=20100810104024AAhqaaJ – Ricbit Jul 29 '12 at 16:03
-
@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. – Micah Jul 29 '12 at 16:06
-
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. – Gerry Myerson Jul 30 '12 at 05:44