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The problem is as follows: Let $p>3$ be a prime. Show that $\tbinom{2p}{p}-2$ is divisible by $p^3$. The only thing I can think of is that $(2p)!-2(p!)^2$ is divisible by $p^2$ which doesn't help me much. Can someone point me in the right direction? Is there a combinatorial approach to this problem? Thanks

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There's a straightforward combinatorial proof that it's divisible by $p$: $\Bbb{Z}/2p$ acts on the $p$-element subsets of $\{1,2,\dots,2p\}$ by addition mod $2p$, and this action has no fixed points and only one order-2 orbit (consisting of $\{1,3,5,\dots,2p-1\}$ and $\{2,4,6,\dots,2p\}$). Maybe you can extend this somehow? – Micah Jul 15 '12 at 16:18
2… discusses a combinatorial proof of divisibility by $p^2$ but the accepted answer suggests that there is no combinatorial proof for divisibility by $p^3$. – Qiaochu Yuan Jul 15 '12 at 16:25
Problem 14 in Chapter 1 of Stanley's "Enumerative Combinatorics", second edition, has several parts. One is to give a combinatorial proof that $\binom{pa}{pb} \equiv \binom{a}{b} \mod p^2$. The next part is to prove the same equivalence mod $p^3$, and that includes the question, "Is there a combinatorial proof?" – Graphth Jul 15 '12 at 16:51
Note that the problem is marked [3-]: "A few students may be capable of solving a [3–] problem, while almost none could solve a [3] in a reasonable period of time." – joriki Jul 16 '12 at 10:01
up vote 4 down vote accepted

$${2p\choose p}=\frac{(2p)(2p-1)\ldots (2p-(p-1))}{p!}=\frac{2(2p-1)\ldots (2p-(p-1))}{(p-1)!}=2{2p-1\choose p-1}$$ Now by Wolstenholme's theorem $${2p\choose p}\equiv 2\cdot1\equiv 2\mod p^3$$ ${} {} {}$

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Wolstenholme's Theorem tells us that

$$\binom {2p-1}{p-1}=1\pmod{p^3}$$ and from here...

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Thats not correct, it is $1$ not $0$ – pritam Jul 15 '12 at 16:59
Yup. Edited the typo alread. t – DonAntonio Jul 15 '12 at 17:00

$\tbinom{2p}{p} = \frac{(p+1)(p+2)...(p+p-1)(p+p)}{1.2...(p-1)p} = \frac{(p+1)(p+2)...(p+p-1)2}{1.2...(p-1)}$

$\tbinom{2p}{p} -2 $ will be divisible by $p^3$

iff $ \frac{(p+1)(p+2)...(p+p-1)2}{1.2...(p-1)} -2 $ is divisible by $p^3$

iff $ ((p+1)(p+2)...(p+p-1)) -(p-1)! $ is divisible by $p^3$ as (2,p)=1 and (p,(p-1)!)=1

Let f(x)= $\prod_{1≤r≤p-1}(x+r) = \sum _{0≤r≤p-1}a_rx^r$ . Then (x+1)f(x+1)=(x+p)f(x). Putting the values of f(x) and f(x+1) and comparing the coefficients of the different powers of x,

for $x^p$, 1=1

for $x^{p-1}, p+a_{p-1}=pC_1+a_{p-1}$

and so on.

Clearly,p |$a_r$ and by Wolstenholme's Theorem $p^2|a_1$

So in $\prod_{1≤r≤p-1}(p+r)$, $a_0$ is (p-1)!, the co-efficient of p($a_1$) is the sum of product of r taken 2 at a time, the co-efficient of $p^2$($a_2$) is the sum of product of r taken 3 at a time and the rest terms are divisible by $p^3$.

As $p|a_2$ and $p^2|a_1$, the result follows.

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