Squirly Squares

To count the squares, let's first divide them into groups so that within each of the groups all the squares can be generated from one another by sliding them up and down and/or left and right. In the listing below, each group is identified only by the square closest to the lower left corner of the grid, point (1, 1), and each square is described only by the coordinates the two of its vertices closest to the lower left corner. For each group so identified, the number of the squares in that group is then listed.

(2, 1), (1, 1), 49 ... (2, 1), (1, 2), 36 ... (2, 1),(1, 3), 25 ... (2, 1), (1, 4), 16 ... (2, 1), (1, 5), 9... (2, 1), (1, 6), 4 ... (2, 1), (1, 7), 1

(3, 1), (1, 1), 36 ... (3, 1), (1, 2), 25 ... (3, 1), (1, 3), 16 ...(3, 1), (1, 4), 9 ... (3, 1), (1, 5), 4 ... (3, 1), (1, 5), 1

(4, 1), (1, 1), 25 ... (4, 1), (1, 2), 16 ... (4, 1), (1, 3),9 ... (4, 1), (1, 4), 4 ... (4, 1), (1, 5), 1

(5, 1), (1, 1), 16 ... (5, 1), (1, 2), 9 ... (5, 1), (1, 3), 4 ... (5, 1), (1, 4), 1

(6, 1), (1, 1), 9 ... (6, 1), (1, 2), 4 ... (6, 1), (1, 3), 1

(7, 1), (1, 1), 4 ... (7, 1), (1, 2), 1

(8, 1), (1, 1), 1

Adding them all together can be done by using formulas for the sum of squares, but it's not a real saving in this case.

To paraphrase Marta Reese's solution, consider squares of side lengths $1, 2, 3, 4, 5, 6, 7$ . Then for these size squares, parallel to the axes, there are ${7}^{2}, {6}^{2}, {5}^{2}, {4}^{2}, {3}^{2}, {2}^{2}, {1}^{2}$ such squares having vertices in the $8 \times 8$ dot array. Now, but for a square of size $1$ , there is only $1$ possible square that can be formed by the $4$ points along its perimeter. For a square of side $2$ , there's $2$ possible using the $8$ points along its perimeter, one of them at an angle like a diamond. For a square of side $3$ , there are $3$ possible using the $12$ points along its perimeter. Etc. So, for the total number of squares, we have

${7}^{2}\cdot 1+{6}^{2} \cdot 2+{5}^{2} \cdot 3 + {4}^{2} \cdot 4+ {3}^{2} \cdot 5+ {2}^{2} \cdot 6+ {1}^{2} \cdot 7 = 336$