I found this $\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le ktoday,when I deal this problem,I found this result:
let $p>3$ be prime number, show that
$$\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le k<p^2$$
I found this when $k=1,2,3$ it is easy to prove it,because
$k=1$
$$\binom{2p^2}{k}=\binom{2p^2}{1}=2p^2\equiv 0=2\binom{p^2}{1}\pmod {p^2}$$
$k=2$
$$\binom{2p^2}{k}=\binom{2p^2}{2}=p^2(2p^2-1)\equiv 0=2\binom{p^2}{2}\pmod {p^2}$$
$k=3$
$$\binom{2p^2}{k}=\binom{2p^2}{3}=\dfrac{2p^2(2p^2-1)(2p^2-2)}{6}\equiv 0=2\binom{p^2}{3}\pmod {p^2}$$
but for all postive integer $k$,I think is also right.But I can't prove it

Let $n$ be a positive integer and $1\le k<n$. By a straightforward counting argument we have the identity
$$\binom{2n}{k}=\sum_{a,b\ge 0, a+b=k}\binom{n}a\binom{n}b.$$
Setting $n=p^2$ and rewriting the equation gives
$$\binom{2p^2}k=2\binom{p^2}{k}+\sum_{a,b\ge 1, a+b=k}\binom{p^2}a\binom{p^2}b.$$
It therefore suffices to show that $p\mid\binom{p^2}a$ for all $1\le a<p^2$ and this follows immediately by expanding $x^{p^2}+1=(x+1)^{p^2}$ modulo $p$ using the binomial theorem.
