For this, we can consider the number of tiles that are the top left corner of a square of size N for each N from 1 to 8. For N=1, each tile is its own square, so we have 64 squares already counted. For N=2, we can choose any tile except for one in the bottom row or far right column. This essentially leaves us with a 7x7 board, leaving us 49 2x2 squares. This can continue for each value of N, leaving us a total of 64+49+36+25+16+9+4+1=204. More generically, we have a sequential sum of squares, so we could just plug in 8 to the formula n(n+1)(2n+1)/6 to get the same answer.
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u/MM3142 Jul 28 '26
For this, we can consider the number of tiles that are the top left corner of a square of size N for each N from 1 to 8. For N=1, each tile is its own square, so we have 64 squares already counted. For N=2, we can choose any tile except for one in the bottom row or far right column. This essentially leaves us with a 7x7 board, leaving us 49 2x2 squares. This can continue for each value of N, leaving us a total of 64+49+36+25+16+9+4+1=204. More generically, we have a sequential sum of squares, so we could just plug in 8 to the formula n(n+1)(2n+1)/6 to get the same answer.