On the problem $1$ of Putnam $2009$ (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 $A_1...A_n$, $f(A_1) + \cdots + f(A_n) =0$. 
 A: Suppose that there is such a function for all $n$-gons. We try to prove that the function should be zero for any arbitrary point $O$. Without loss of generality we assume that $O$ is at origin. 
Consider an $n$-gon placed on unit circle with the vertex $A_i$ placed on $e^{\frac{2(i-1)\pi}{n}}$ for $i=1,...n$. Then the internal angle at each vertex of regular $n-$gon is $\frac{(n-2)\pi}{n}$. 
Consider $A_1=1$. Then one can construct an $n-$gon starting from $OA_1$ as one edge and finding the points $A^{1}_{j}$ such that  $\hat{A_1OA^1_{j}}=\frac{(j-1)\pi}{n}$ and $A^1_{j}A^1_{j-1}=OA_{1}$ for $j=2,....n-1$. Call this polygon $P_1$ with vertices $A^1_{j}$ with $A^1_{1}=A_1$ and $A^1_{n}=O$.
Similarly, for each $A_i$, one can construct a regular $n$-gon with $OA_i$ as one edge, called polygon $P_i$ with vertices $A^i_{j}$ and $A^i_{1}=A_i$ and $A^i_{n-1}=O$.  $P_i$ is the rotation of $P_1$ by ${\frac{2(i-1)\pi}{n}}$ around $O$.
Now consider the sum of the function $f$ of all vertex of these polygons which should be zero according to assumption:
$$
\displaystyle\sum_{i,j=1}^{n} f(A^{i}_j)
$$
where $A^{i}_j$ is the vertex $j$ of $i$th polygon. It is easy to verify that $A^{1}_jA^{2}_j...A^{n}_j$ form a regular $n$-gon for $j=1,2,...,n-1$ (due to the fact that $P_j$ is rotation of $P_1$ around origin ), which means:
$$
\displaystyle\sum_{i=1}^{n} f(A^{i}_j)=0
$$
Therefore we have: 
$$
\displaystyle\sum_{i,j=1}^{n} f(A^{i}_j)=nf(O)+\sum^{n-1}_{j=1}\sum^{n}_{i=1} f(A^{i}_j)=nf(O)=0.
$$
Hence the function should be constant zero for all points in the plane for all odd $n$. 
A: 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.
