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January 2nd, 2017, 04:00 PM  #1 
Newbie Joined: Jan 2017 From: Deutschland Posts: 2 Thanks: 0  Divisibility of a binomial coefficient
I want to proof, why $\displaystyle \frac {\binom {2n}{n}  2}{n}$ always returns an integer if n is a prime, but I don't know how.. You can write it like the following but I don't see a beginning for a proof...: $\displaystyle \frac {\binom {2n}{n}2}{n}=\frac{\frac{(2n)!}{n!\cdot(2nn)!)}2}{n}=\frac{\frac{(2n)!)}{(n!)^2)}2}{n}$ I hope, that someone can help me! Thanks 
January 2nd, 2017, 05:43 PM  #2 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
Expand the factorials as products. How can you simplify the resulting fraction? Then recall Fermat's little Theorem. 
January 3rd, 2017, 04:31 AM  #3 
Newbie Joined: Jan 2017 From: Deutschland Posts: 2 Thanks: 0 
I don't know excactly what u mean but I tried the first part: $\displaystyle \frac {\frac{(2n)!)}{(n!)^2)}2}{n}=\frac {\frac{1\cdot 2 \cdot 3\cdot ...\cdot 2n}{1^2\cdot 2^2 \cdot 3^2 \cdot ... \cdot n^2}2}{n}=\frac {\frac{(n+1)\cdot(n+2)\cdot...\cdot(2n)}{1\cdot 2\cdot 3 \cdot ... \cdot (n1))}2}{n}=\frac {\frac{\frac{(2n)!)}{n!}}{(n1)!)}2}{n}=\frac{\frac{(2n)!)}{(n1)!\cdot n!}2}{n}$ Of course I can simplify the fraction, but I can't see how this will help and also I can't the the connection to Fermat's little theorem :/ 
January 5th, 2017, 06:42 AM  #4 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
Forget Fermat's Little Theorem  my mistake. But it's easy. Look at the factors of the fraction in the numerator (the one with all factorials replaced by linear factors). Since n is prime, you can mod each factor out by n. Then you should see some cancellations. 
January 5th, 2017, 09:06 AM  #5 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
$$\frac {\frac{(n+1)\cdot(n+2)\cdots\cdot(2n)}{1\cdot 2\cdot 3\cdots\cdot (n1))}2}{n}$$ The above expression was one of your attempts. This one is on the right track, but you already made an algebraic error. How did the factor $2n$ survive intact? 

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binomial, coefficient, divisibility 
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