I'm working on this problem
"Find all entire functions $f(\frac{1}{p})=\frac{1}{1+p}$ for all prime $p$."
My approach is using identity theorem. But in this case it does not seem good. We have $$f(1/p)=\frac{1/p}{1+1/p}$$ So naturally, I set $g(z)=f(z)-\frac{z}{z+1}$. Consider the plane excluded a circle around $z=-1$. Then $g$ is analytic in that region. And $g=0$ on the set of $\{1/p\}$ which has a limit point. So $g=0$ on the constructed given. But how do I extend it to $-1$? Or I can't?
There is no such entire function $f$. Take$$\begin{array}{rccc}g\colon&\Bbb C\setminus\{1\}&\longrightarrow&\Bbb C\\&z&\mapsto&\dfrac z{z+1}.\end{array}$$Then, for each prime $p$, $f\left(\frac1p\right)=g\left(\frac1p\right)$. Since $f$ and $g$ are holomorphic, since $\Bbb C\setminus\{1\}$ is connected and since $\{z\in\Bbb C\setminus\{1\}\mid f(z)=g(z)\}$ has an accumulation point ($0$), the identity theorem implies that$$(\forall z\in\Bbb C\setminus\{1\}):f(z)=g(z).$$But that's impossible, since $\lim_{z\to-1}f(z)=f(-1)$, whereas the limit $\lim_{z\to-1}g(z)$ doesn't exist in $\Bbb C$.