Let $f$ be an analytic function in the whole complex plane. If there exists $M>0$ such that $|f(z)|>M$ for all $z \in \Bbb Z$, show that $f$ is constant in the whole complex plane.
I tried to approach with contradiction supposing that $f$ is not constant. If so then there exists $z_1$ and $z_0$ such that $f(z_1) \ne f(z_0)$. Then $f(z_1)-f(z_0)\ne 0$ and $|f(z_1)-f(z_0)| \ne 0$. I then tried to use the triange inequality trick to break $|f(z_1)-f(z_0)| \ne 0$ apart as $$|(f(z_1)-f(z'))+(f(z')-f(z_0))| \le |f(z_1)-f(z')|+|f(z')-f(z_0)|$$
but I don't think I'm going anywhere with this approach. I think this resemble the Liouville theorem a bit, but it's like the converse of it?