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January 2nd, 2017, 03:00 PM   #1
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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
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January 2nd, 2017, 04:43 PM   #2
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Expand the factorials as products.

How can you simplify the resulting fraction?

Then recall Fermat's little Theorem.
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January 3rd, 2017, 03:31 AM   #3
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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 :/
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January 5th, 2017, 05:42 AM   #4
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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.
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January 5th, 2017, 08:06 AM   #5
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$$\frac {\frac{(n+1)\cdot(n+2)\cdots\cdot(2n)}{1\cdot 2\cdot 3\cdots\cdot (n-1))}-2}{n}$$

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?
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