= Solution
With equal type probabilities, $F$ and $S$ have the same marginal <probability generating function>, namely $G_N((1+s)/2)$. Their assumed <independence>, together with $N=F+S$, gives
$$
\boxed{G_N(s)=G_F(s)G_S(s)=G_N\left(\frac{1+s}{2}\right)^2.}
$$
Set $H(t)=G_N(1-t)$ for $0\le t\le1$. The identity becomes $H(t)=H(t/2)^2$ and, after $k$ iterations,
$$
H(t)=H(t/2^k)^{2^k}.
$$
The finite <expected value> gives $G_N(1-h)=1-\mu h+o(h)$ as $h\downarrow0$. To justify this without any analyticity assumption at the endpoint, note that $(1-(1-h)^N)/h\to N$ and is bounded above by $N$; <dominated convergence> gives the derivative $G_N'(1-)=\mu$. Substituting the expansion into the iterated identity yields
$$
H(t)=\left(1-\frac{\mu t}{2^k}+o(2^{-k})\right)^{2^k}\longrightarrow e^{-\mu t}.
$$
The left side is independent of $k$, so it equals the limit. Thus $G_N(s)=e^{\mu(s-1)}$ on $[0,1]$ and uniqueness of the power-series coefficients proves \b[$N\sim\operatorname{Poisson}(\mu)$]. Only finite <expected value> was needed; the given finite <variance> is stronger. This proves the <Poisson characterization by independent binomial splitting> rather than assuming the original count distribution.
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