Is analytic function on whole complex plane taking only integer values constant? Let $f(z)$ be an analytic function on the whole complex plane such that it takes only integer values. Is $f(z)$ a constant function?
 A: Note that $f$ is continuous and $\mathbb{R}^2$ is connected. Hence $f(\mathbb{R}^2)$ is also connected but the only connected subsets of $\mathbb{Z}$ as a subspace of $\mathbb{R}^2$ are singleton sets. So $f$ is constant.
A: Let $f$ be such a function, $w \in \Bbb C$ and $n = f(w)$. Since $f$ is analytic, $f$ is continuous, so there's an neighborhood $U$ of $w$ in $\Bbb C$ such that $|f(z) - f(w)|<0.5$. Using your hypothesis, $f(w)$ is also an integer and there's no integer $m\neq n$ such that $|n-m|< 0.5$ therefore $n=m$, hence $f$ is constant in $U$ and you know this fact implies that $f$ is constant in the whole plane.
A: Since $\mathbb{C}$ is uncountable and $\mathbb{N}$ is
countable it means that there is a $n_0$ such $A=\{z\in \mathbb{C} | f(z)=n_0\}$ is uncountable.
$$\mathbb{C}= \bigcup _{n\in \mathbb{N}}f^{-1}(n)$$
But since $\mathbb{C}= \bigcup _{n \in \mathbb{N}}(B_n \cap A)$ where $B_n= \{z\in \mathbb{C} | |z|\leq n\}$ there is a $n_1$ such that $B_{n_1}\cap A$ is uncountable. Which implies that there is $z_0 \in B_{n_1}$ and a sequence $(y_n), y_n \in B_{n_1}\cap A$ such that $y_i \not =y_j$ wherever $i\not =j$ and $y_n \rightarrow z_0$. Now analytic continuation guarantees that 
$f(z)= n_0 \forall z \in \mathbb{C}$
