How to solve the following system? I need to find the function c(k), knowing that
$$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$
$$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$
$$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k+1)}{(2k+1)!}=-1$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k)}{(2k)!}=0$$
Is it possible?

You are looking for a function $\displaystyle f(z) = \sum_{k \ge 0} \frac{c(k)}{k!} z^k$ satisfying
$$f(1) = 1$$
$$f(-1) = -1$$
$$f(i) = -i.$$
Infinitely many functions have this property.  There is a unique quadratic polynomial $p(z)$ with this property (for example by Lagrange interpolation), and for any entire function $q(z)$ the function $p(z) + (x - 1)(x + 1)(x - i) q(z)$ has this property.  In fact these are all entire functions with this property.
More generally I think the theory of interpolation by entire functions is fairly well-understood, but I don't know of a good reference.  If the set of $z$ at which you fix the value of $f$ has a limit point, then $f$ is unique by the identity theorem.  If the set of $z$ at which you fix the value of $f$ is countable and does not have a limit point, then $f$ is non-unique by Weierstrass factorization.
