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I want to proof, why $\frac {\binom {2n}{n}-2}{n}$ always returns an integer if $n$ is a prime, but I don't know how..

You can rewrite it like this:

$\frac {\binom {2n}{n}-2}{n} = \frac{\frac{(2n)!}{(n!)^2}-2}{n}$

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    Someone (was it you?) asked the same question yesterday or the day before, and I posted a comment with some hints. – 2017-01-04

3 Answers 3

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In $\mathbb F_p[x]$ we have

$$(x+1)^{2p} = \Big( (x+1)^p \Big)^2 = (x^p+1)^2 = x^{2p} + 2x^p +1,$$

hence $\binom{2p}{p}=2$ by comparing the coefficient of $x^p$.

Without any further work, this proof of course immediately generalizes to $\binom{np}{p} \equiv n \mod p$ for any $n \geq 1$.

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    What do you mean with $\mathbb F_p[x]$? – 2017-01-04
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We have that:

$\binom {2n}{n}= \binom{n}{0}^2 + \binom{n}{1}^2 +...+\binom{n}{n-1}^2+\binom{n}{n}^2\Rightarrow \binom {2n}{n}-2= \binom{n}{1}^2 +...+\binom{n}{n-1}^2$

But when $n$ is a prime, then $n|\binom{n}{k}$ for $k\in \{2,...,n-1\}$. Thus then we get:

$n^2|\binom{n}{1}^2 +...+\binom{n}{n-1}^2\Rightarrow \exists k_n\ \text{integer}:k_nn^2=\binom {2n}{n}-2\Rightarrow\frac{\binom {2n}{n}-2}{n}=nk_n $

and you are done.

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    Moreover the integer is divisible by the prime number $n$. – 2017-01-04
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$$ \frac{2n(2n-1)...(n+1)}{n(n-1)...1}=\frac{2(2n-1)...(n+1)}{(n-1)...1}$$ if $ n>2 $ is a prime, $ \left(\{1...n-1\},mod(n)\right) $ is a multiplicative group. So, $$ (n-1)!\equiv-1(mod\:n) $$ Dtto $$ (2n-1)...(n+1)\equiv-1(mod\:n) $$ and $$ \frac{2(2n-1)...(n+1)}{(n-1)...1}\equiv2\:(mod\:n) $$ Q.E.D

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    No need for Wilson's Theorem. Just reduce each factor of the product $$(2n-1)\ldots (n+1)$$ (mod n), then match them up and cancel against the corresponding factors of $$(n-1)\ldots..1$$The cancellation works if n is an odd prime, since primality guarantees nonzero factors (mod n) and since n is odd, there are an even number negative signs in the numerator) and – 2017-01-04
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    Note that the matching is in reverse order. Also, for my argument, the case n = 2 needs to be done separately (so just compute it). – 2017-01-04
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    Thanks, two questions: 1. can I write: $$ \frac{2(2n-1)...(n+1)}{(n-1)...1} \equiv \frac{2 \cdot (-1)}{-1}\:(mod\:n) \equiv 2\:(mod\:n) $$?$$$$ 2. Also for Slepecky Mamut's solution I have to proof n=2 separately, haven't I? – 2017-01-04
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    2. Yes. $ (6-2)/2=2 $ But quasi solution is better. – 2017-01-04