If $f$ is entire function such that $f(z)=f(z+1)$ for all $z$ is not that enough to consider that $f$ is constant. Why? Because we have $f(0)=f(1)=f(2)=\cdots$ and then $f(z)=f(z+1)=f(z+2)=\cdots$ so the function $f$ is determined by the image of points in the unit desk but $f$ is bounded in the unit desk no poles since it is entire so by Liouville’s theorem $f$ is a constant? Is there something missing! Am I correct
In fact the original question give another condition $f(z)=f(z+1)=f(z+a)$ for an irrational $a$ to figure out that $f$ is a constant.
The function $f$ is constant on the set $C=\{m+na\,|\,m,n\in\mathbb Z\}$; it is equal to $f(0)$ in each element of $C$. But $\overline C=\mathbb R$ and $f$ is continuous. So, the restriction of $f$ to $\mathbb R$ is constant. It follows from the identity theorem that $f$ is constant.