How Many Squares are there in a Chessboard?

Every square on the chessboard can be uniquely identified by its sidelength and its bottom-left cell

For every $k\times k$ square, we count how many cells can be the bottom-left:

• $8 \times 8$ squares: Then the only possible cell on the bottom-left is the $1^2$ in row 1 and column a.
• $7 \times 7$ squares: Then the only cells that can be the bottom-left are the $2^2$ squares in rows 1-2, and columns a-b
• $6 \times 6$ squares: Then the only cells that can be the bottom-left are the $3^2$ squares in rows 1-3 and columns a-c
• $\vdots$
• $k \times k$ squares: Then the only cells that can be the bottom-left are the $(9-k)^2$ squares in the first $9-k$ rows and the first $9-k$ columns
• $\vdots$
• $2 \times 2$ squares: The only cells that can be the bottom-left are the $7^2$ squares in rows 1-7 and columns a-g
• $1 \times 1$ squares: Any of the $8^2$ cells in rows 1-8 and columns a-h can be the bottom-left square (since the "bottom-left" cell of a $1\times 1$ square is the square itself)

It follows that the number of squares is $1^2 + 2^2 + 3^2 + \cdots + 7^2 + 8^2 = \frac{8\cdot (8+1) \cdot (2(8)+1)}{6} = \boxed{204}$

---

Bonus Answer : The same pattern persists for $n \times n$ boards, giving $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ squares.

There are : $\begin{array}{c}~1 : \quad 8 \times 8~square \\ 4 : \quad 7 \times 7~squares \\ 9 : \quad 6 \times 6~squares \\ 16 : \quad 5 \times 5~squares \\ 25 : \quad 4 \times 4~squares \\ 36 : \quad 3 \times 3~squares \\ 49 : \quad 2 \times 2~squares \\ 64 : \quad 1 \times 1~squares \\ \end{array}$ So, the total number of squares = $64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204~squares$

---

For a $n \times n$ chessboard the total number of squares can be obtained by adding all the squares of integers till $n$ starting from $1$ . $\begin{array}{c}~\implies 1^2 + 2^2 + 3^2 + 4^2 + ... + (n - 1)^2 + n^2 \\ \implies \dfrac{n(n + 1)(2n + 1)}{6} \\ \end{array}$