Prove that ${2p\choose p}\equiv 2\pmod {p^2}.$

Question

Prove that ${2p\choose p}\equiv 2\pmod {p^2}.$

My progress:

Consider any $S={(x,y),~~1\le x\le p,~~1\le y\le 2}.$ Note that $|S|=2p.$

Now consider $A\subseteq S$ such that $|A|=p.$

So there are a total of ${2p}\choose{p}$ possible value of $A.$

Then let $A=\{(a_1,b_1),......,(a_p,b_p)\}.$

Define $$f(A)=\{(a_2,b_2),.............(a_p,b_p),(a_1,b_1)\}$$ . . . $$f^{p-1}(A)=\{(a_p,b_p),\dots ,(a_{p-1},b_{p-1})\}$$

Now, define a new binary relation $R$ such $f^{m}(A)Rf^l(A)$ when $$b_{m+1}=b_{l+1},~~b_{m+2}=b_{l+2},~~\dots$$ Note that this is possible only when $b_i=c$ a constant where $c=1$ or $2.$

For the rest ${{2p}\choose{p}}-2$ sets, the orbit size is $p.$

Hence we get $p\mid{{2p}\choose{p}}-2.$

I want to go with the same essence, but I don't know how to get $p^2.$ Any hints?

once you know that $(x+y)^p\equiv x^p+y^p$ mod $p$ (meaning that each binomial coefficient there is $0$ mod $p$), just square it and you'll get it — user8268, Aug 12 '21 at 06:09
@JohnOmielan The first non-prime shows that $\binom{2p}{p}\equiv2\pmod{p}$ does not hold if $p$ is not prime. — Servaes, Aug 12 '21 at 06:09
@SunainaPati FYI, the Wolstenholme's theorem Wikipedia article mentions Babbage's theorem that states $\binom{ap}{bp} \equiv \binom{a}{b} \pmod{p^2}$. Using $a = 2$ and $b = 1$ gives what you're asking about. This is also mentioned as Theorem $1.12$ in The Lucas and Babbage congruences. — John Omielan, Aug 12 '21 at 06:13
@user8268 You get that $\binom{2p}{p}\equiv0\pmod{p}$, but how do you get the congruence mod $p^2$? — Servaes, Aug 12 '21 at 06:22
Does this answer your question? CHECK: Let p be a prime number. Show that $\binom{2p}{p}$ is congruent to $2$ mod $p$. — Sil, Aug 12 '21 at 08:21
See also Prove that the binomial coefficent $\binom{2p}{p} $ is $\equiv 2\pmod{p}$ using group actions. (answer uses $p^2$) and MO post https://mathoverflow.net/questions/26137. The duplicate voted also mentions $p$ in statement but answer works for $p^2$, maybe better stated question can be found... — Sil, Aug 12 '21 at 08:23

score 7 · Answer 1 · answered Aug 12 '21 at 07:31

7

Vandermonde's Identity says that $$ \binom{2p}{p}=\sum_{k=0}^p\binom{p}{k}\binom{p}{p-k}\tag1 $$ For $k\ne0\pmod{p}$, we have $\left.p\,\middle|\binom{p}{k}\right.$, thus $\left.p^2\middle|\binom{p}{k}\binom{p}{p-k}\right.$ for all $k$ except $0$ and $p$.

$\binom{p}{k}\binom{p}{p-k}=1$ for $k=0$ and $k=p$.

answered Aug 12 '21 at 07:31

robjohn

345,667

I was about to type up an answer expanding $(1+x)^{2p}$ in $\Bbb{Z}_{p^2}[x]$. Thankfully I spotted your equivalent way in time :-). +1 – Jyrki Lahtonen Aug 12 '21 at 08:35
1

Not surprisingly, this approach has been used before (as noted in a comment to the question). – robjohn Aug 12 '21 at 13:45

score 3 · Answer 2 · answered Aug 12 '21 at 07:09

$$\binom{2p}p=\frac{2p(2p-1)\cdots(p+1)}{p(p-1)\cdots 1}=2\frac{(p+1)(p+2)\cdots(2p-1)}{1\cdot 2\cdots (p-1)}$$

Now, note that $(p+k)(p-k)\equiv -k^2\pmod{p^2}$ for $1\le k\le p-1$; also $\gcd(k^2,p^2)=1$. Basically, for any $n$ such that $p\not\mid n$, we have $\gcd(n,p^2)=1$ and we can divide num/denom by it.

Hence,

$$\prod_1^{p-1}\frac{p+k}{p-k}\equiv\prod_1^{p-1}\frac{(p+k)(p-k)}{(p-k)^2}\equiv\prod_1^{p-1}\frac{-k^2}{(p-k)^2}\equiv (-1)^{p-1}\pmod{p^2}$$

whence we have $\binom{2p}p\equiv 2(-1)^{p-1}\pmod{p^2}$

For prime $p\gt 2$, $(-1)^{p-1}\equiv 1$ and the claim follows. For $p=2$, we have $-2\equiv 2\pmod{2^2}$, so the claim holds there as well.

Thus, $\binom{2p}p\equiv 2\pmod{p^2}$ for all primes $p$

How do you get $\prod_1^{p-1}\frac{-k^2}{(p-k)^2}\equiv (-1)^{p-1}\pmod{p^2}$? — Servaes, Aug 12 '21 at 07:27
@Servaes: Index changing. $\prod_1^{p-1}k=\prod_1^{p-1}(p-k)$; note that $$\prod_1^{p-1}\frac{-k^2}{(p-k)^2}=(-1)^{p-1}\frac{1^2\cdot 2^2\cdots (p-2)^2(p-1)^2}{(p-1)^2(p-2)^2\cdots 2^2\cdot 1^2}$$ — Prasun Biswas, Aug 12 '21 at 07:31

score 3 · Accepted Answer · answered Aug 12 '21 at 07:59

I'm assuming $p$ is prime.

To paraphrase your work so far, you have a $2\times n$ array of points and you're acting by rotating the $n$ columns cyclically. Instead, rotate the $n$ points in each row independently.

That is, let $f$ map $(x,1)$ to $(x+1\text{ mod }p,1)$ while fixing the points $(x,2)$, let $g$ map $(x,2)$ to $(x+1\text{ mod }p,2)$ while fixing the points $(x,1)$, and consider the action of the group $\langle f,g\rangle$ on $\{A\subset S : |A|=p\}$. Then for any $A$ other than $\{(x,1):1\le x\le p\}$ and $\{(x,2):1\le x\le p\}$, the orbit of $A$ has size $p^2$.

(This is closely related to robjohn's answer, but perhaps sticks closer to the spirit of actions and orbits.)

Yess this work! I worked with the details, like showing p^2|2pcp-2. Really nice! — Sunaina Pati, Aug 12 '21 at 10:47

Prove that ${2p\choose p}\equiv 2\pmod {p^2}.$

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