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 January 2nd, 2017, 03:00 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...: $\displaystyle \frac {\binom {2n}{n}-2}{n}=\frac{\frac{(2n)!}{n!\cdot(2n-n)!)}-2}{n}=\frac{\frac{(2n)!)}{(n!)^2)}-2}{n}$ I hope, that someone can help me! Thanks  January 2nd, 2017, 04:43 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, 03:31 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: $\displaystyle \frac {\frac{(2n)!)}{(n!)^2)}-2}{n}=\frac {\frac{1\cdot 2 \cdot 3\cdot ...\cdot 2n}{1^2\cdot 2^2 \cdot 3^2 \cdot ... \cdot n^2}-2}{n}=\frac {\frac{(n+1)\cdot(n+2)\cdot...\cdot(2n)}{1\cdot 2\cdot 3 \cdot ... \cdot (n-1))}-2}{n}=\frac {\frac{\frac{(2n)!)}{n!}}{(n-1)!)}-2}{n}=\frac{\frac{(2n)!)}{(n-1)!\cdot n!}-2}{n}$ 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, 05:42 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, 08:06 AM #5 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 $$\frac {\frac{(n+1)\cdot(n+2)\cdots\cdot(2n)}{1\cdot 2\cdot 3\cdots\cdot (n-1))}-2}{n}$$ 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? Tags binomial, coefficient, divisibility Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post panky Algebra 1 September 30th, 2016 05:31 PM Lux Calculus 4 April 17th, 2015 02:00 PM pikachu26134 Number Theory 3 July 26th, 2011 07:03 PM sunflower Number Theory 0 April 2nd, 2011 07:15 AM coolhandluke Applied Math 1 March 26th, 2010 04:55 PM

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