mathematics - Black and White

OK guys, I think this is my best puzzle yet. Hope you enjoy it, the solution is neat and simple.

A boy draws 2015 unit squares on a piece of paper, all oriented the same way. The squares can overlap. Then he colors the resulting picture in black and white chess-wise, such that any area belonging to an even number of squares is painted white and any area belonging to an odd number of squares is painted black.

Prove that there is at least one square unit of black on the paper when he is finished.

Answer

Let's think of the problem as taking XOR's (symmetric differences) of the square areas, with black as 1 and white as 0. The black area is the XOR of all the squares.

Imagine doing this all on graph paper whose cells are sized and oriented like the squares. Take the resulting figure, cut up the occupied cells, and stack them on top of each other without rotating. Now, take the XOR of that stack.

Note that any square on the graph paper gets cut up and rearranged to make exactly the full unit cell. So, the XOR of the stack is the XOR of 2015 fully black squares, which is all black, since 2015 is odd. Then, since each black point in the XOR requires at least one black point the contribute, the total area of black in the stack must be at least 1.

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Needlessly formally:

We are interested in the group $G$ of (well-behaved) finite-area subsets of the plane under symmetric difference (XOR). We quotient out the plane by integer translations to get $(\mathbb R \times \mathbb R) / (\mathbb Z \times \mathbb Z)$, which induces a homomorphism $\phi$ from $G$ to the corresponding group $G'$ on the unit square.

We have $2015$ single-square elements $x_1, ..., x_{2015}$ of $G$ and are interested in their sum $x$. As single-square elements, they all map to the same full-square element $\phi(x_i) = s$ in $G'$, so $\phi(x) = s + \cdots + s$ summed $2015$ times. Since $G'$ has order $2$, this is simply $s$. But the homomorphism $\phi$ cannot increase the size of the subset, so

$$\mathrm{Area}(x) \geq \mathrm{Area}(\phi(x)) = \mathrm{Area}(s) =1 $$