, 2007

Solution:

Every square with even side length will have an equal number of black and white $1 \times 1$ squares, so it isn't a $B$-square. In a square with odd side length, there is one more $1 \times 1$ black square than white squares, if it has black corner squares. So, a square with odd side length is a $B$-square either if it is a $1 \times 1$ black square or it has black corners.

Let the given $(2n+1) \times (2n+1)$ chessboard be a $B$-square and denote by $b_{i}$ ($i=1,2, \ldots, n+1$) the lines of the chessboard which have $n+1$ black $1 \times 1$ squares, by $w_{i}$ ($i=1,2, \ldots, n$) the lines of the chessboard which have $n$ black $1 \times 1$ squares, and by $T_{m}$ ($m=1,3,5, \ldots, 2n-1,2n+1$) the total number of $B$-squares of dimension $m \times m$ of the given chessboard.

For $T_{1}$ we obtain $T_{1} = (n+1)(n+1) + n \cdot n = (n+1)^{2} + n^{2}$.

For computing $T_{3}$ we observe that there are $n$ $3 \times 3$ $B$-squares, which have the black corners on each pair of lines $(b_{i}, b_{i+1})$ for $i=1,2, \ldots, n$ and there are $n-1$ $3 \times 3$ $B$-squares, which have the black corners on each pair of lines $(w_{i}, w_{i+1})$ for $i=1,2, \ldots, n-1$. So, we have
$$T_{3} = n \cdot n + (n-1)(n-1) = n^{2} + (n-1)^{2}.$$

By using similar arguments for each pair of lines $(b_{i}, b_{i+2})$ for $i=1,2, \ldots, n-1$ and for each pair of lines $(w_{i}, w_{i+2})$ for $i=1,2, \ldots, n-2$ we compute
$$T_{5} = (n-1)(n-1) + (n-2)(n-2) = (n-1)^{2} + (n-2)^{2}.$$

Step by step, we obtain
$$\begin{gathered} T_{7} = (n-2)(n-2) + (n-3)(n-3) = (n-2)^{2} + (n-3)^{2} \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ T_{2n-1} = 2 \cdot 2 + 1 \cdot 1 = 2^{2} + 1^{2} \\ T_{2n+1} = 1 \cdot 1 = 1^{2} \end{gathered}$$

The total number of $B$-squares of the given chessboard equals to
$$\begin{gathered} T_{1} + T_{3} + T_{5} + \ldots + T_{2n+1} = 2\left(1^{2} + 2^{2} + \ldots + n^{2}\right) + (n+1)^{2} = \\ \frac{n(n+1)(2n+1)}{3} + (n+1)^{2} = \frac{(n+1)\left(2n^{2} + 4n + 3\right)}{3} \end{gathered}$$