Let $f$ be a nonconstant meromorphic function on the complex plane such that $f(z+1)=f(z)$ and $f(z+i)=f(z)$ for every $z\in\mathbb C$.
Prove that in the square $\{z\in\mathbb C\colon 0\le\Re(z)<1, 0\le\Im(z)< 1\}$ the function $f$ must have either
two (or more) different poles, or else one (or more) single pole of order $\ge 2$.
I know that $f(z)$ must have poles since the range of $f$ over $\mathbb C$ is determined by the range of $f$ over the square. If $f$ were analytic, the modulus of $f$ would be bounded over $\mathbb C$ and by Liouville theorem $f$ is constant. Contradiction.
But how to show that if $f$ only have one pole, then the order $\ge 2$? I can hardly invoke anything to deal with the order of poles.
The sum of the residues in the unit square is $0$. (Say $S$ is that sum. Concoct a roughly square curve $\gamma_r$ with side length roughly $r$, in such a way that $|f|\le c<\infty$ on $\gamma_r$, independent of $r$. Then the residue theorem shows that $r^2S=O(r)$ as $r\to\infty$, hence $S=0$.)
So if there is only one pole the residue vanishes, so it can't be a simple pole.