The number of squares with side 8x8 = 1
Let us consider a mxm square at the upper left corner.
Each time we move the square towards right we get a new square.
Each time we move the square downwards we get a new square.
Since there are 8-m options while going towards left as well as down.
No. of mxm squares = (8-m)*(8-m) = (8-m)2
\dpi{200} \sum_{i=0}^{8} (8-m)^{2} = \sum_{i=0}^{8}m^{2} = \frac{8(8+1)(2\cdot 8+1)}{6}
Answer-204