Okay, so I need to figure out the value of n in a 3×n grid where there are a total of 70 squares of three sizes. Let me start by understanding the problem. The grid has 3 rows and n columns of points. Each square is defined by four points that form the vertices. The squares can be of different sizes, so I need to count all possible squares of different sizes and types (axis-aligned and maybe others?) and set the total equal to 70. Then solve for n. 

First, I should recall how to count squares in a grid. In a standard m×n grid of points, the number of axis-aligned squares is given by the sum over k from 1 to min(m-1, n-1) of (m - k)(n - k). But here, the grid is 3×n. So for axis-aligned squares, the formula would be sum_{k=1}^{2} (3 - k)(n - k). Since 3 is the number of rows, so the maximum size of a square is 2x2. Wait, because the grid has 3 rows, so the height of the square can be 1 or 2 units. Similarly, the width can be from 1 to n-1. So for each square size k x k (where k is 1 or 2 in this case), the number of such squares is (3 - k)(n - k). So total axis-aligned squares would be (3 - 1)(n - 1) + (3 - 2)(n - 2) = 2(n - 1) + 1(n - 2) = 2n - 2 + n - 2 = 3n - 4.

Wait, but the problem mentions three sizes of squares. So maybe there are also squares that are not axis-aligned? Because in a 3×n grid, you can have squares that are rotated, like diamonds, but in a grid, those might not be possible unless the grid is larger. Let me think. In a 3-row grid, the maximum horizontal distance between two points is n-1 units. The vertical distance is up to 2 units. So, for a square to be rotated, its side length must be such that the diagonal can fit within the grid. But squares have equal sides and right angles. Hmm. Let me consider if there are any squares that are not axis-aligned in a 3×n grid.

For example, a square with side length sqrt(2) would have vertices at points that are diagonally adjacent. But in a 3×n grid, the vertical distance between rows is 1 unit, so the diagonal would be sqrt(2) units. But then, for such a square, the horizontal distance between points would also need to be 1 unit. So, the square would have vertices like (i,j), (i+1,j+1), (i,j+2), (i-1,j+1). Wait, but in a 3-row grid, j can be 0,1,2. So for a square like that, starting at (0,0), the next point would be (1,1), then (0,2), then (-1,1). But (-1,1) is outside the grid. So maybe such squares can't exist in a 3-row grid. Alternatively, perhaps starting at (1,0), the square would go to (2,1), (1,2), (0,1). But (0,1) is still within the grid if n is at least 2. Wait, maybe in a 3×n grid, there can be such squares of size sqrt(2). Let me check.

Suppose n is at least 2. Then the points (1,0), (2,1), (1,2), (0,1) form a square. The side length is sqrt(2), and the area is 2. So that's a different size. Similarly, if n is 3, there might be more such squares. So if the problem mentions three sizes, it's possible that there are axis-aligned squares (size 1 and 2), and these rotated squares (size sqrt(2)).

So let's confirm. The problem states there are three sizes of squares. So likely, we have:

1. Axis-aligned squares of size 1x1
2. Axis-aligned squares of size 2x2
3. Rotated squares of size sqrt(2)x sqrt(2)

So total squares would be the sum of these three. Let me verify if other sizes are possible. For example, in a 3x5 grid, could there be squares of size sqrt(5) or something? But that would require points that are 2 units apart horizontally and 1 unit vertically, but in a 3-row grid, the vertical distance is limited. Wait, sqrt(5) would be the diagonal of a 2x1 rectangle. But in a 3-row grid, the vertical distance is 2 units. So a square of size sqrt(5) would require a vertical distance of 2 units and horizontal distance of 1 unit, but in a 3-row grid, the vertical distance is 2 units, but the horizontal distance would need to be 2 units as well for a square. Wait, no. For a square to have sides of sqrt(5), the points would need to be spaced 1 unit horizontally and 2 units vertically apart. But that would form a rectangle, not a square. Wait, no, a square requires all sides equal and right angles. So, the distance between consecutive points should be the same. So, if the side is sqrt(5), then the horizontal and vertical components must satisfy sqrt(a^2 + b^2) = sqrt(5). So possible combinations are a=1, b=2. But in a 3-row grid, the vertical distance between rows is 1 unit. So if the vertical component is 2 units, that would span two rows. But then the horizontal component would have to be 1 unit. So the square would have vertices like (i,j), (i+1,j+2), but wait, in a 3-row grid, j can be 0,1,2. So if j is 0, then j+2 would be 2, which is allowed. Then the next point would be (i-1, j+1), but if i is 0, that would be (-1,1), which is outside. So maybe such squares can't exist in a 3-row grid unless there are enough columns. Wait, maybe in a 3x5 grid, starting at (1,0), the square would go to (2,2), (0,1), (-1,0). But again, (-1,0) is outside. So maybe such squares can't exist. Therefore, perhaps the only rotated squares possible are those of size sqrt(2). So maybe the problem refers to three sizes: 1x1, 2x2, and sqrt(2)x sqrt(2). Let me check.

So, for axis-aligned squares:

- 1x1 squares: In a 3×n grid, the number is (3 - 1)(n - 1) = 2(n - 1)
- 2x2 squares: (3 - 2)(n - 2) = 1(n - 2) = n - 2

For rotated squares (sqrt(2)x sqrt(2)):

These squares have vertices forming a square where each side is the diagonal of a 1x1 square. So, each such square is formed by points (i,j), (i+1,j+1), (i,j+2), (i-1,j+1). But wait, in a 3-row grid, the vertical distance is 2 units (from row 0 to row 2). So, the square would span two rows. For example, starting at (i,j), moving to (i+1,j+1) (next column, next row), then to (i,j+2) (same column, next row), then to (i-1,j+1) (previous column, same row). But this would only be possible if i-1 >= 0. So the leftmost such square would start at column 1. Similarly, the rightmost would end at column n-1. So the number of such rotated squares would be (n - 1 - 0) but considering the starting column. Wait, maybe the number is (n - 2)(3 - 2). Wait, perhaps similar to axis-aligned squares. Let me think again.

Each rotated square requires a 2x2 block of points. Because to form a square with side sqrt(2), you need two adjacent columns and two adjacent rows. So, in a 3×n grid, the number of such squares would be (n - 1) * (3 - 2). Wait, but 3 - 2 is 1, so (n - 1)*1 = n - 1. But wait, in a 3×n grid, how many 2x2 blocks are there? For columns, from 0 to n-2, so n - 1 columns. For rows, 0 to 1, so 2 rows. But since the grid has 3 rows, the vertical span for a 2x2 block is two rows. So the number of 2x2 blocks is (n - 1)(3 - 2) = (n - 1)(1) = n - 1. But each 2x2 block can form one rotated square. Wait, but actually, in a 2x2 block of points, you can form two squares: one axis-aligned 1x1 and one rotated sqrt(2)x sqrt(2). But in this case, we are only considering the rotated ones. So maybe the number of rotated squares is (n - 1)(3 - 2) = n - 1. But let me verify with an example.

Take n=3. Then the grid is 3x3 points. How many rotated squares are there? Starting at (0,0): (0,0), (1,1), (0,2), ( -1,1) – but that's outside. So invalid. Starting at (1,0): (1,0), (2,1), (1,2), (0,1). That's a valid square. Similarly, starting at (0,1): (0,1), (1,2), (0,3) – but there is no row 3, so invalid. So only one rotated square in a 3x3 grid. So n=3: rotated squares = 1, which is n - 2. Wait, 3 - 2 =1. Hmm. Wait, n=4: how many? Starting at (1,0): (1,0), (2,1), (1,2), (0,1). Starting at (2,0): (2,0), (3,1), (2,2), (1,1). Starting at (1,1): (1,1), (2,2), (1,3) – invalid. Starting at (0,1): (0,1), (1,2), (0,3) – invalid. So two rotated squares. n=4: 2 rotated squares, which is n - 2 = 4 - 2 = 2. So yes, the number of rotated squares is n - 2. So that seems to hold. Therefore, the total number of rotated squares is n - 2.

Wait, but in the 3x3 grid, n=3: n - 2 =1, which matches. For n=2: n - 2 =0, which is correct since you can't have a rotated square in a 3x2 grid. So yes, the number of rotated squares is indeed n - 2.

Therefore, total squares would be:

Axis-aligned 1x1: 2(n - 1)

Axis-aligned 2x2: n - 2

Rotated sqrt(2)x sqrt(2): n - 2

Total squares: 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6.

Wait, but according to the problem, there are 70 squares. So 4n - 6 =70. Then 4n=76, so n=19. But wait, that seems straightforward. But let me check if my reasoning is correct.

Wait, axis-aligned 1x1: 2(n -1). Axis-aligned 2x2: (3 - 2)(n -2)=1*(n -2). Rotated squares: (n -2). So total: 2(n -1) + (n -2) + (n -2) = 2n - 2 + n - 2 + n -2 = 4n -6. So 4n -6=70. 4n=76. n=19. That seems correct. But let me check with n=19: 4*19=76, 76-6=70. So yes.

But wait, let me verify this formula with a smaller n. For example, n=3. Then total squares should be 4*3 -6=6. Let's compute:

Axis-aligned 1x1: 2*(3-1)=4

Axis-aligned 2x2: 3-2=1

Rotated squares:3-2=1

Total:4+1+1=6. Which matches. So formula works for n=3. Similarly, n=4: 4*4 -6=10. Compute:

Axis-aligned 1x1:2*(4-1)=6

Axis-aligned 2x2:4-2=2

Rotated squares:4-2=2

Total:6+2+2=10. Correct. So formula is correct.

Therefore, solving 4n -6=70 gives n=19. So the answer is 19.

But let me think again. Wait, in the Asymptote code provided, the grid seems to be 3x5? Because the points go up to (5,2). Wait, in the Asymptote code, the grid is drawn with points (0,0) to (5,2). So columns from 0 to 5 (6 points), rows from 0 to 2 (3 points). So n=5? Wait, but the code draws points up to (5,2). So n=5. But the problem says "rectangular 3×n grid of points". So n is the number of columns. So in the Asymptote code, the grid is 3x6? Because columns are 0 to 5, which is 6 columns, but maybe the grid is 3×n, so n=5? Wait, the Asymptote code shows a grid that's 3 rows (y from 0 to 2) and 6 columns (x from 0 to 5). So the grid is 3×6. But in the problem, it's a 3×n grid. So perhaps the Asymptote code is just an illustration, not necessarily to scale with the problem's n. So maybe the Asymptote code is just an example, not the actual grid in the problem. So the answer is 19, regardless of the Asymptote code. So I think the calculation is correct. Therefore, the value of n is 19.

**Final Answer**
\boxed{19}
Therefore, the value of $n$ is $\boxed{19}$.
</think>