Primeness in divisor sense for certain entire functions (Q798799)

An entire function F(z), with zeros, is said to be prime in divisor sense, if, for every identical relation such as \(F(z)=f(g(z))\exp (A(z)),\) where f,g (\(\not\equiv const.)\) and A are entire functions, we can deduce the following assertion: f has just one simple zero or g is a linear polynomial. A divisor D, a discrete countable set in \({\mathbb{C}}\), is said to be prime, if an entire function f(z) whose zeros-set is identical with D is prime in divisor sense, as in the definition above. The author proves: Theorem 1. Let P(z) be a polynomial. Then the function \(F(z)=z+P(\exp z)\) is prime in divisor sense. - Theorem 2. Let P(z) and Q(z) be two nonconstant polynomials. Assume that, for any natural number k and C, the function \(\exp (-kz)[Q(\exp z)+C]\) is nonconstant. Then \(F(z)=P(z)+Q(\exp z)\) is prime in divisor sense.