I am having trouble understanding the original statement of Raikov's theorem and its implementation in terms of characteristic functions. Raikov's theorem states that a shifted Poisson distributed random variable $X \sim Pois(\lambda)$ has only shifted Poisson distributed independent factors $X_1 \sim Pois(\mu)$,$X_2 \sim Pois(\nu)$, where $\mu + \nu = \lambda$. In terms of distribution functions, the result is stated as $F(x - m,\lambda) = \sum_{k \leq x - m} e^{-\lambda} \frac{\lambda^k}{k!}$,
is only factorisable into the form
$F(x-m,\lambda)= F_1(x- m +n,\mu) * F_2(x - m - n,\nu)$.
Note that in terms of the original result [D. Raikov (1937) On the decomposition of Poisson laws] I have taken $\alpha = m$, $\sigma = 1$ and $\beta = n$, where $m$ is some positive integer. In terms of characteristic functions, the characteristic function for a shifted Poisson distributed random variable is
$f(t) = e^{i m t + \lambda(e^{it} -1)}$.
It appears to me that Raikov's theorem states that the characteristic function factorises into the product of the characteristic functions for $X_1$ and $X_2$, which according to the above result should be:
$f_1(t) = e^{i(m - n)t + \mu (e^{it} -1)}$,
$f_2(t) = e^{i(m+n)t + \nu (e^{it} - 1)}$.
However we see that the product of these is $f(t) = f_1(t) f_2(t) = e^{2imt + (\mu + \nu)(e^{it} -1)} = e^{2imt + \lambda(e^{it} -1)}$. There is an extra factor of 2 appearing in front of the original shifting, which leads to the result not quite being true. Where am I going wrong or what am I misunderstanding?
As a related aside, the main theorem of this paper (https://www.jstor.org/stable/pdf/1995129.pdf) seems to have the shifts included in the characteristic function but doesn't place any constraints on them other than they have to be real. Is it to be inferred?
Thanks in advance.