Squares in a Square

I found 105 squares in this grid.

First off - I am not certain I am correct. However, I know the problem is solvable. So a better solution is forthcoming from the community (wink). There are "level" as well as "tilted" squares. I partition the set of all squares S into subsets based on the slope of their edges.

$S=S_{0}+S_{1/1}+S_{1/2}+S_{1/3}+S_{1/4}+S_{2/3}$

Squares have slopes a/b where $a+b <= n$ and n is the number of squares along a side. In this case n = 5, which constrains the possible slopes that can fit on the grid.

I further partition each subset based on the scaling of the square. The smallest squares of a given type that fit on the grid are called 1x, the next scaling of that square are 2x, and so on...

$\textbf{S}_{0}:$ Slope 0 and undefined There are squares parallel/perpendicular with the edges of the grid. The edges of these squares have slope 0 and undefined.

$S_{0}^{1x}= \text{one-by-one level squares}; \text{for n = 5, } S_{0}^{1x} = 25 = 5^2$

$S_{0}^{2x}= \text{two-by-two level squares}; \text{for n = 5, } S_{0}^{2x} = 16 = 4^2$

$S_{0}^{3x}= \text{three-by-three level squares}; \text{for n = 5, } S_{0}^{3x} = 9 = 3^2$

$S_{0}^{4x}= \text{four-by-four level squares}; \text{for n = 5, } S_{0}^{4x} = 4 = 2^2$

$S_{0}^{5x}= \text{five-by-five level squares}; \text{for n = 5, } S_{0}^{5x} = 1 = 1^2$

These squares follow a sum of squares pattern up to n .

$\sum_{n=1}^{n} n^{2} = \frac{n(n+1)(2n+1)}{6}$

For n = 5 this sum is the total number of level squares and the first distractor, 55 .

$\textbf{S}_{1/1}:$ Slope 1/1 (a+b = 2) There are diamond-looking tilted squares with edges 45 degrees with the edges of the grid, or slopes 1/1 with scaling 1x and 2x fitting within this grid.

$S_{1/1}^{1x}= \text{slope 1/1 squares}; \text{for n = 5, } S_{1/1}^{1x} = 16 = 4^2$

$S_{1/1}^{2x}= \text{slope 2/2 squares}; \text{for n = 5, } S_{1/1}^{2x} = 4 = 2^2$

For n = 5 this sum is the total number of slope 1/1 tilted squares is 20 .

$\textbf{S}_{1/2}:$ Slope 1/2 (a+b = 3) There are tilted squares with slopes 1/2 and 2/1 with only scaling 1x fitting within this grid.

$S_{1/2}^{1x}= \text{slope 1/2 squares}; \text{for n = 5, } S_{1/2}^{1x} = 9 = 3^2$

$S_{2/1}^{1x}= \text{slope 2/1 squares}; \text{for n = 5, } S_{2/1}^{1x} = 9 = 3^2$

For n = 5 this sum is the total number of slope 1/2 tilted squares is 18 .

$\textbf{S}_{1/3}:$ Slope 1/3 (a+b = 4) There are tilted squares with slopes 1/3 and 3/1 with only scaling 1x fitting within this grid.

$S_{1/3}^{1x}= \text{slope 1/3 squares}; \text{for n = 5, } S_{1/3}^{1x} = 4 = 2^2$

$S_{3/1}^{1x}= \text{slope 3/1 squares}; \text{for n = 5, } S_{3/1}^{1x} = 4 = 2^2$

For n = 5 this sum is the total number of slope 1/3 tilted squares is 8 .

$\textbf{S}_{1/4}:$ Slope 1/4 (a+b = 5) There are tilted squares with slopes 1/4 and 4/1 with only scaling 1x fitting within this grid.

$S_{1/4}^{1x}= \text{slope 1/4 squares}; \text{for n = 5, } S_{1/4}^{1x} = 1 = 1^2$

$S_{4/1}^{1x}= \text{slope 4/1 squares}; \text{for n = 5, } S_{2/1}^{1x} = 1 = 1^2$

For n = 5 this sum is the total number of slope 1/4 tilted squares is 2 .

$\textbf{S}_{2/3}:$ Slope 2/3 (a+b = 5) There are tilted squares with slopes 2/3 and 3/2 with only scaling 1x fitting within this grid.

$S_{2/3}^{1x}= \text{slope 2/3 squares}; \text{for n = 5, } S_{2/3}^{1x} = 1 = 1^2$

$S_{3/2}^{1x}= \text{slope 3/2 squares}; \text{for n = 5, } S_{3/2}^{1x} = 1 = 1^2$

For n = 5 this sum is the total number of slope 1/2 tilted squares is 2 .

In general, it seems that the difference between the total of the slope a+b and n relates to the number of squares that fit in a given grid-size n. An explicit general rule eludes me.

kop _crush

$\frac{n^4-n^2}{12}$ ,put it in n=6 and get 105.