On a Putnam's 2009 problem Find all even natural numbers $n$ 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 $A_1...A_n$, $f(A_1) + \cdots + f(A_n) =0$. 
NOTE: This is the unsolved part of THIS  problem. I chose to state it in a separate question first to get more attention and second since some people may want to put a bounty on this part. 
 A: Below are pictures that prove the $n=8$ and make progress on the $n=10$ case. I think this is very close to a complete argument for all $n$. For the left hand picture there are 2 black $8$-gons and $4$ green $8$-gons. Most points get hit twice. The origin gets hit 4 times. What's left over are the $4$ vertices of a square. Thus we have shown that the sum over vertices of a square is zero, so we have reduced to a known case. For the second picture, there are 2 black $10$-gons and $5$ green $10$-gons. Summing over everything gives the sum of the vertices of a $5$-gon plus its center is zero. One can probably go further with this, but my wife will get angry with me if I spend any more time on it today.

A: Hint: [Special case of 4] Suppose that $f$ is not constant and w.l.g. suppose $f(-2,0)\neq f(2,0)$. Then consider two squares $(-2,0),(-1,-1),(-1,1),(0,0)$ and also  $(2,0),(1,-1),(1,1),(0,0)$. Now the sum of the value of the function over vertices of two squares should be zero. But $(0,0)$ appears twice in the sum ($0$ in $\mathbb{Z}_2$) and also $(1,-1),(-1,-1),(-1,1),(1,1)$ form a square and the sum of function on these points is zero. Therefore $f(-2,0)= f(2,0)$ which is a contradiction.

I do not think that the current approach can be generalized for all even numbers.
A: This is to show that for $n=6$ there is no such map. Consider the following layout of points:
\begin{matrix}
&&a&&b&&c\\&d&&e&&f&&g\\h&&i&&j&&k&&l\\&m&&n&&o&&p\\&&q&&r&&s
\end{matrix}
where each small triangle is imagined as equilateral of sidelength $1.$ Take the three hexagons centered at $e,k,n$ along with the larger hexagon $dbgprm$ of sidelength $\sqrt{3}$ and add the sums. Since the hexagons are $abfjid,fglpoj,ijorqm,dbgprm$, and since all their sums are to be zero mod $2$, we obtain that $a+l+q+j=0$, since each of $a,l,q$ occurs only once in the four hexagons, while $j$ occurs three times, and the other letters occur twice each. What this means geometrically is that, from the assumption that all hexagons have vertex sums zero, we have shown:
Given an equilateral triangle, the sum of $f$ over its vertices together with its center is zero.
Now by placing two such equilateral triangles so they share a common side, one gets a set of four points on a line which sum to zero, since the vertices on the common side cancel in the sum. The remaining four points consist of the third vertices of the two adjacent triangles and the two centers of the two adjacent triangles; these points are collinear and equally spaced.
But now we can take two such lines of four and lay the second one starting at the second point of the first, and then the inner three points which these two lines of four have in common will cancel in that sum, causing the sum of the first and fifth points of the union of the two segments to have sum zero. Since given any two points $P,Q$ of the plane this arrangement may be made so as to arrive at $P+Q=0$ (by appropriately scaling/rotating the figure), we get all points of the plane in the same coset mod $2$.
