A special chess board

Here is the answer for a):

Consider the $2\times 2$ squares from the bottom left corner to the top right corner. Since $n$ is odd, the top right $1\times 1$ square is not in either of the $2\times 2$ squares. Write an A into each square of the main diagonal (from the bottom left to the top right corner), and a B into each square of the adjacent diagonals.

The bottom left $1\times 1$ square’s adjacent white square is in the bottom left $2\times 2$ square in a B square. Hence the other A now has an adjacent white square. Therefore in the next $2\times 2$ square the bottom left A’s adjacent white square must be one of the Bs in that $2\times 2$ square. The othet A now has an adjacent white square. And so on... Untill we arrive to the top right $1\times 1$ square. We get that it can’t have an adjacent white square, because in that case the A under it would have two white adjacent squares.

I've thought this problem for a few days, but I can't solve part (a). To enable others to edit my answer (and to avoid losing reputation), I'm making this community wiki.

A story-like introduction to the problem (optional)

If you want your kid to think about this problem, you may tell them this story.

Once upon a time, there's a kingdom with $n^2$ sites. To protect the land, the King 🤴 has decided to build castles 🏰 (white squares □), so that each site □/■ is protected by a neighbouring castle 🏰. (Each castle has to pair up with another one. 🏰 ↔ 🏰)

Being too lazy to draw something like this, \begin{array}{|c|c|c|c|c|c|} \hline & & \unicode{x1f3f0} & \unicode{x1f3f0} & & \\ \hline \unicode{x1f3f0} & & & & & \unicode{x1f3f0} \\ \hline \unicode{x1f3f0} & & & & & \unicode{x1f3f0} \\ \hline & & \unicode{x1f3f0} & \unicode{x1f3f0} & & \\ \hline & & & & & \\ \hline \unicode{x1f3f0} & \unicode{x1f3f0} & & & \unicode{x1f3f0} & \unicode{x1f3f0} \\ \hline \end{array} I'll use matrices in the rest of this post, but you may imagine that they are castles 🏰 in your kingdom ♔.

Construction of beautiful squares

I use $1$ to denote white squares, and $0$ to denote black squares. The proof for part (b) will be constructive and inductive.

The base cases are either trivial or given. \begin{array}{cc} \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} & \begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{bmatrix} \\ n=2 & n=4 \end{array}

The inductive hypothesis are

• the possibility of such construction for even $n$
• on the outermost layer, the digits repeat in the pattern $1,1,0,0$
• the existence of $1$ at exactly two corners.
• We can add one more observation here: symmetry about the vertical axis. I'm not going to prove this point since I don't know how.

The inductive step is to construct a square matrix with size $(n+4)^2$. In other words, we want to expand our kingdom by $4$ rows and columns. To do this, we focus on the outermost layer. \begin{array}{cc} \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} & \begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & & & 0 \\ 1 & & & 1 \\ 1 & 0 & 0 & 1 \end{bmatrix} \\ n=2 & n=4 \end{array} The second outermost layer consists of $0$ only, while each digit in the outermost layer is opposite to the digit that it faces. \begin{array}{cc} \begin{bmatrix} 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 \end{bmatrix} & \begin{bmatrix} 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & & & & & & & 0 \\ 1 & & 0 & 1 & 1 & 0 & & 1 \\ 1 & & 0 & & & 0 & & 1 \\ 0 & & 1 & & & 1 & & 0 \\ 0 & & 1 & 0 & 0 & 1 & & 0 \\ 1 & & & & & & & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \end{bmatrix} \\ n=2+4 & n=4+4 \end{array}

My proof that such construction satisfies the question is to divide the scenario into two types: side and corner. Up to any rotational and reflectional symmetry, we have these three situations.

Side

$$ \left[ \begin{array}{ccccccc|c} 1 & 1 & 0 & 0 & 1 & \dots & 0 & \text{outermost} \\ 0 & 0 & 0 & 0 & 0 & \dots & 0 & \text{second outermost} \\ \hline 0 & 0 & 1 & 1 & 0 & \dots & 1 & \text{apply induction} \\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \text{hypothesis here} \end{array} \right] $$

Corner

$$ \begin{array}{cc} \begin{array}{|cc} \hline \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} & \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \\ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} & \begin{array}{|cc} \hline 1 & 1 \\ 0 & 0 \end{array} \\ \end{array} & \begin{array}{|cc} \hline \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} & \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \\ \begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array} & \begin{array}{|cc} \hline 0 & 1 \\ 0 & 0 \end{array} \\ \end{array} \\ \text{corner is 1} & \text{corner is 0} \end{array} $$ Observations: - From corner $1$'s, we get corner $1$'s. This shows that we've exactly two corner $1$'s and two corner $0$'s in each square.
- the pattern $1,1,0,0$ is preserved on the sides and at the corners on the outermost layer.

Ideas for part (c)

In my construction, half of each even layer is a white square. We have the recurrence sequence $w_{n+4}=w_n+2(n+3)$, where $w_n$ denotes and number of white squares in a $n \times n$ chess board. The initial conditions are $w_0=0$ and $w_2=2$.

\begin{alignat}{2} w_{4k}-w_0 & = \sum_{i=0}^{k-1}{2(4i+3)} \quad && \text{(sum of arithmetic progression)} \\ & = \frac{2[4(k-1)+3]+2[4(0)+3]}{2} \cdot k \quad && \left(\frac{\text{first term + last term}}{2} \times \text{no. of terms}\right) \\ & = (4k+2)k \\ & = 2k(2k+1) \end{alignat}

The calculations for $w_{4k+2}$ are similar, and we arrive at $$\bbox[1px, border: 1px solid black]{w_{2n}=n(n+1)} \quad\forall\,n\in\Bbb{N}.$$