TITLE: On the problem $1$ of Putnam $2009$
QUESTION [9 upvotes]: (This is adapted from problem $1$ of Putnam $2009$)
Find all values of $n$ such that the following is true: 
There is a non-constant function $f : \Bbb{R}^2 \longrightarrow \Bbb{Z}_2$ such that for any regular $n$-gon $A_1...A_n$, $f(A_1) + \cdots + f(A_n) =0$.

REPLY [2 votes]: A few easy cases:
A) Such a function does not exist for $n=4$. Assume contrariwise that we have such a non-constant function. By translating, rotating and scaling the coordinate system we can assume that
$f(0,0)=1$ and $f(2,2)=0$. Let us write $f(1,0)=a$, $f(2,0)=b$ and $f(0,1)=c$. Then

The square with vertices $(0,0)$, $(1,0)$, $(1,1)$ and $(0,1)$ forces $f(1,1)=a+c+1$.
Then the square with vertices $(1,0)$, $(2,0)$, $(2,1)$ and $(1,1)$ forces $f(2,1)=b+c+1$.
Then the square with vertices $(1,0)$, $(2,1)$, $(1,2)$ and $(0,1)$ forces $f(1,2)=a+b+1$.
Then the square with vertices $(1,1)$, $(2,1)$, $(2,2)$ and $(2,1)$ forces $f(2,2)=(a+c+1)+(b+c+1)+(a+b+1)=1\neq0$.

This is a contradiction.
B) The case $n=3$ is also impossible. Let's use a complex coordinate this time. Let $\omega=(1+i\sqrt3)/2$ be a sixth root of unity. We can assume that
$f(0)=0$ and $f(1)=1$. Then we get (rotating the obvious 60-degree sector about the origin) that $f(\omega)=1$, $f(\omega)^2=1$, and continuing $f(\omega^k)=1$
for all exponents $k$. But $1,\omega^2$ and $\omega^4$ also form an equilateral triangle, and we have ran into a contradiction.
Something similar might work for hexagons, as then we can tile the plane with them, use different sizes of regular $n$-gons, and possibly run into a contradiction.
But a different idea is needed to handle the general case.