Theorem : Every entire function $f$ so that $$f(z)=f(z+1)=f(z+i) $$ on $\Bbb{C}$ is necessarily constant
Proof:
Consider the solid square $S:=\{x+iy : 0\leq x,y \leq 1\}$.
Since $f$ is continuous on $S$, there exists an $M$ so that $\vert f(z) \vert \leq M$ on $S$.
Now, let $z \in \Bbb{C}$ be arbitrary.
$\textbf{Clearly, we can find two integers $m$ and $n$ so that $z+m+in$ in S }$ and therefore,
$\vert f(z+m+in) \vert \leq M$ for $z \in \Bbb{C}$. Further, the hypothesis imply that $$f(z)=f(z \pm 1)=...=f(z+m+in)$$ showing that $f$ is bounded on $\Bbb{C}$ as the behaviour of $f(z)$ over $\Bbb{C}$ is completely characterized by its behaviour on the compact set $S$. Hence $f$ is constant by Liouville's theorem.
I cannot understand the boldface line. If the chosen $z$ is from $S$, then $m=n=0$ is easy. What about other $z's$?
Can anyone explain this ?
Assume that $z=x+iy$ for real numbers $x,y$. Now let $x=m.a_{1}a_{2}\cdots$, the decimal expression, then $x-m=0.a_{1}a_{2}\cdots\in[0,1]$.