Proving entire function constant $f$ is entire and $f(z)=f(z+1)=f(z+i)$
prove $f(z)=const$
I have no clue how to solve it
 A: The key point in showing boundedness is that $|f|$ is continuous (since $f$ is entire) on  a "period" or unit square in the complex plane: $\{x+iy; 0\le x,y \le 1\}$ which is compact (analogous in this case to a closed interval in $\mathbb{R}$), so $|f|$ attains its maximum $M$ on this square. Then every adjacent square has an identical image, and since $\mathbb{C}$  is equal to the union of these squares, for all $z\in\mathbb{C}$, $f(z) = f(x+iy)$ for some $x,y\in[0,1]$. 
For example, in case it is not clear enough, the unit square to the right: $\{x+iy; 1\le x\le 2, 0\le y \le 1 \}$. We have $f(x+iy) = f((x-1)+iy+1) = f((x-1)+iy)$ where the argument $(x-1)+iy$ belongs to the original unit square, and so $|f(x+iy)|\le M$.
From there you do as others said and apply Louisville's.
A: As many of the comments suggest, your function $ f $ is bounded.
Look up "doubly periodic functions."
The reason for the bounded comes from the fact that $ f $ can be thought of as a function from a compact set. The continuity of $ f $ makes the image compact, which is closed and bounded (as far as Euclidean spaces go). 
