I am stuck on the following problem. This came up while studying an application of Fourier Transform to provide a fast solution of a Renewal Equation .
Say $F(s) = \displaystyle\sum_{n=1}^{\infty} f_n s^n$ be a probability generating function. Show that if $\gcd \{ n : f_n > 0\} = 1$ then, $F(s) = 1$ has the only solution $s = 1$ on the unit circle $|s| = 1$.
Is it actually easy to see that the reverse conclusion is also true. However, I have been stuck on this for quite some time, and haven't been to show this.
Suppose $\sum f_n s^{n}=1$ with $|s|=1$. Write $s=e^{it}$ with $t$ real. Taking real part we get $\sum f_n \cos (nt)=1$. So $1 =\sum f_n \cos (nt) \leq \sum f_n=1$ which forces $\cos (nt)$ to be $1$ for all $n$ such that $f_n\neq 0$. Can you finish? [You will need the hypothesis $\gcd \{ n : f_n > 0\} = 1$ here].