Poisson characterization by independent binomial splitting

Let $N$ be a nonnegative integer-valued random variable with finite expected value $\mu$, and independently label each of its items with either of two equally likely labels. If the two resulting counts are independent random variables, then $N$ has a Poisson distribution. Indeed, its probability generating function obeys $G(s)=G((1+s)/2{)}^2$. With $H(t)=G(1-t)$, iteration gives $H(t)=H(t/2^k{)}^{2^k}$. Finite expected value gives $H(h)=1-\mu h+o(h)$, hence $H(t)=e^{-\mu t}$ and $G(s)=e^{\mu (s-1)}$. Conversely, Poisson thinning produces independent Poisson distributions for the two counts.