Given a family $K$ of Binomial distributions $B(\cdot|m,p)$ with two unknown parameters $m\in N$ and $0<p<1$. Find a sufficient statistics for $K$.
My attempt: The joint distribution of i.i.d random variables $X_1,\ldots, X_n$ is $f(X_1,\ldots, X_n|m,p) = {m \choose x_1}{m \choose x_2}\ldots {m \choose x_n}p^{x_1+\ldots + x_n}(1-p)^{nm-x_1-\ldots - x_n}$. But I could not find another term, besides $x_1+\ldots + x_n$ that would be a sufficient statistics for the unknown parameter $m$. The term $x_1+\ldots + x_n$ is only a sufficient statistics for $p$.
My question: Could someone please help me complete the proof above? I would sincerely appreciate any help or inputs.
In case $mp\gg 1$ and $m(1-p)\gg 1$, the binomial distribution $B(m,p)$ approaches a Gaussian distribution $N(\mu=mp,\sigma^2=mp(1-p))$. Then the likelihood function
$$L(x_1,\cdots,x_n|\mu,\sigma^2)=\frac{1}{(2\pi)^{n/2}\sigma^n}\prod_{i=1}^n\exp\left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right).$$
The data groups into
$$\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2=\frac{1}{2\sigma^2}\left(\sum_{i=1}^nx_i^2\right)-\frac{\mu}{\sigma^2}\left(\sum_{i=1}^nx_i\right)+\frac{\mu^2}{2\sigma^2}.$$
So under the Gaussian approximation, the two sufficient statistics are the sum $\sum_ix_i$ and the sum of squares $\sum_ix_i^2$, which could help determine $\mu$ and $\sigma^2$ (or equivalently $m$ and $p$).
In case $m\gg 1$ but $mp\rightarrow\mbox{const}$, the binomial distribution $B(m,p)$ approaches a Poisson distribution $\mbox{Poisson}(\mu=mp)$. Then the likelihood function
$$L(x_1,\cdots,x_n|\mu)=e^{-n\lambda}\prod_{i=1}^n\frac{\lambda^{x_i}}{x_i!}=\prod_{i=1}^n\frac{1}{x_i!}\times\lambda^{\sum_ix_i}\times e^{-n\lambda}.$$
Only one statistic, the sum $\sum_ix_i$ is needed and only one parameter $\lambda=mp$ can be determined. It is not possible to know $m$ and $p$ individually from a nearly Poisson distribution.
In case $m\gg 1$ does not hold, the binomial distribution $B(m,p)$ does not allow any approximation. Then the likelihood function
$$L(x_1,\cdots,x_n|m,p)=\frac{m!}{x_1!(m-x_1)!}\times\cdots\times\frac{m!}{x_n!(m-x_n)!}p^{\sum_ix_i}(1-p)^{mn-\sum_ix_i}.$$
The sum $\sum_ix_i$ is the sufficient statistic for $p$ if $m$ is known. If $m$ is not known, since the factorials $(m-x_1)!(m-x_2)!\cdots(m-x_n)!$ in the denomenator all factorize, there's no way to aggregate $x_1,\cdots,x_n$ to some statistic while still allowing the factorial product to be calculable. So I think the sufficient statistic is the entire data set $x_1, x_2,\ldots,x_n$.