Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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  
mathoverflow.net/questions/26137/… 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
add comment

3 Answers

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 $${2p\choose p}\equiv 2.1\equiv 2\mod p^3$$ ${} {} {}$

share|improve this answer
add comment

Wolstenholme's Theorem tells us that

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

share|improve this answer
    
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
add comment

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

$\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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.