A recent Putnam problem:

> 
> Let $f$ be a real-valued function on the plane such that for every square $ABCD$ in the plane, $f(A)+f(B)+f(C)+f(D)=0$. Does it follow that $f$ is identically zero?
> 
> 
> 

The answer is yes. Without loss of generality consider only the origin. We have

$$f(1,1)+f(-1,1)+f(1,-1)+f(-1,-1)\tag{a}$$
$$f(0,0)+f(0,1)+f(1,0)+f(1,1)\tag{b}$$
$$f(0,0)+f(0,1)+f(-1,0)+f(-1,1)\tag{c}$$
$$f(0,0)+f(0,-1)+f(1,0)+f(1,-1)\tag{d}$$
$$f(0,0)+f(0,-1)+f(-1,0)+f(-1,-1)\tag{e}$$
$$f(1,0)+f(-1,0)+f(0,-1)+f(0,1)\tag{f}$$

are all zero. Then $f(0,0)=[\rm(b)+(c)+(d)+(e)-(a)-2(f)]/4=0$. A similar argument works when we replace the condition in the problem with triangles $\triangle ABC$. I have pictured both ideas as so-called "proof without words" below:

![picture](https://i.stack.imgur.com/ZSM0P.png)

What happens when we replace the codomain of $f$ with an abelian group of exponent dividing $4$ or $6$ respectively? Do there then exist non-constant maps with these properties?