<Question>
There is a total of 70 squares of three sizes whose vertices are points on this rectangular $3\times n$ grid of points. What is the value of $n$?

[asy]

unitsize(0.4 inch);

dot((0,0),linewidth(9bp));
dot((1,0),linewidth(9bp));
dot((2,0),linewidth(9bp));
dot((0,1),linewidth(9bp));
dot((0,2),linewidth(9bp));
dot((1,1),linewidth(9bp));
dot((2,1),linewidth(9bp));
dot((1,2),linewidth(9bp));
dot((2,2),linewidth(9bp));

filldraw((2.95,-0.05)--(3.05,-0.05)--(3.05,0.05)--(2.95,0.05)--cycle,black);
filldraw((2.45,-0.05)--(2.55,-0.05)--(2.55,0.05)--(2.45,0.05)--cycle,black);
filldraw((3.45,-0.05)--(3.55,-0.05)--(3.55,0.05)--(3.45,0.05)--cycle,black);

filldraw((2.95,0.95)--(3.05,0.95)--(3.05,1.05)--(2.95,1.05)--cycle,black);
filldraw((2.45,0.95)--(2.55,0.95)--(2.55,1.05)--(2.45,1.05)--cycle,black);
filldraw((3.45,0.95)--(3.55,0.95)--(3.55,1.05)--(3.45,1.05)--cycle,black);

filldraw((2.95,1.95)--(3.05,1.95)--(3.05,2.05)--(2.95,2.05)--cycle,black);
filldraw((2.45,1.95)--(2.55,1.95)--(2.55,2.05)--(2.45,2.05)--cycle,black);
filldraw((3.45,1.95)--(3.55,1.95)--(3.55,2.05)--(3.45,2.05)--cycle,black);

dot((4,0),linewidth(9bp));
dot((5,0),linewidth(9bp));
dot((4,1),linewidth(9bp));
dot((5,1),linewidth(9bp));
dot((4,2),linewidth(9bp));
dot((5,2),linewidth(9bp));

[/asy]
</Question>

<think>
<Plan_1>
Okay, so I have this problem here: There's a 3×n grid of points, and the total number of squares formed by these points is 70. The squares can be of three different sizes. I need to find the value of n. Hmm, let me think about how to approach this.

First, I should visualize the grid.
It's a 3-row by n-column grid. So, the points are arranged in 3 horizontal lines and n vertical lines, creating a series of unit squares, but also larger squares if possible.

Since the grid is 3 units high (in terms of rows of points), the maximum size of a square that can fit vertically is 3×3. But horizontally, it can extend up to n units. However, since the number of columns is n, the number of squares of different sizes will depend on how many positions they can occupy in both rows and columns.

Wait, actually, the grid has 3 rows and n columns of points. So, the number of unit squares (1×1) would be (3-1)×(n-1) because between 3 rows, there are 2 gaps, and between n columns, there are (n-1) gaps.

<prompt_1.1> Calculate the number of unit squares (1x1) in the 3×n grid using the above plan. </prompt_1.1>

Next, 2×2 squares. For a 2×2 square, you need 3 consecutive points in both rows and columns. Since there are 3 rows, vertically, a 2×2 square can fit. But horizontally, it needs 3 columns. So, the number of positions horizontally is (n - 2), and vertically, since there are 3 rows, the number of vertical positions for a 2×2 square is (3 - 1) = 2. Wait, no.

<prompt_1.2> Calculate the number of 2x2 squares in the 3×n grid using the above plan. </prompt_1.2>

Now, 3×3 squares. These would require 4 consecutive points in both rows and columns. Vertically, since there are 3 rows, a 3×3 square needs 4 rows, but we only have 3.
Therefore, vertically, a 3×3 square can't fit. Wait, that can't be right. Wait, maybe I need to think in terms of units. If it's a 3×n grid, the vertical distance between rows is 1 unit, so a square of side length k must have k units in both height and width. So, for a 3×3 square, the height would be 3 units, which requires 4 horizontal lines. But since the grid only has 3 horizontal lines, we can't form a 3×3 square. So, the maximum square size is 2×2.

<prompt_1.3> Confirm the maximum square size and the need for an alternative approach using the above plan. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, no, actually, in a grid with h rows and w columns of points, the number of unit squares is (h - 1) * (w - 1). So here, h = 3, w = n. Therefore, unit squares would be (3 - 1) * (n - 1) = 2(n - 1). But wait, that's if it's a grid of points. Let me confirm.

For example, if it's a 3×3 grid of points (so 3 rows and 3 columns), the number of unit squares would be 2*2 = 4. That's correct: a 2x2 grid of squares. So yes, the formula is (rows - 1)*(columns - 1). So here, rows = 3, columns = n, so unit squares are 2*(n - 1).
</execution_1.1>

<execution_1.2>
Wait, no. For a 2×2 square, the vertical dimension is 2 units, so in a 3-row grid, the number of vertical positions is (3 - 2) = 1. Wait, maybe I need a better approach.

Let me recall that in a grid with h rows and w columns of points, the number of k×k squares (both in height and width) is (h - k) × (w - k). But here, h = 3, so for k=1, 2, 3. But since h=3, for k=3, (3 - 3) = 0, so no 3×3 squares. Wait, but maybe in the problem, they consider squares of different orientations? Wait, no, the problem just says squares, so axis-aligned squares.
</execution_1.2>

<execution_1.3>
Wait, but the problem states there are three sizes. Hmm, maybe I'm missing something here.

Wait, hold on. Maybe the problem is referring to squares of different orientations? Like not just axis-aligned squares, but also squares that are rotated? Because if that's the case, then even on a grid, you can have squares that are diamond-shaped, which might fit in different positions. The Asymptote code shows some black squares filled, which might be these rotated ones. Let me check the problem statement again.

The problem says "squares whose vertices are points on this rectangular 3×n grid of points". So, the squares don't have to be axis-aligned. That's an important point. So, there can be squares rotated at 45 degrees or other angles, whose vertices still lie on the grid points. That complicates things a bit, but also means there are more squares possible.

So, maybe the three sizes refer to both axis-aligned and rotated squares. Let me try to figure this out.

First, let's clarify: in a 3×n grid of points, how many axis-aligned squares are there, and how many rotated squares are there? The total would be the sum of both, and that total is given as 70. So, we need to compute the number of squares for each possible size and orientation, then set up an equation where the sum equals 70, and solve for n.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
First, axis-aligned squares:

<prompt_2.1> Calculate the total number of axis-aligned squares using the above plan. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
- 1×1: As we said, 2*(n - 1)
- 2×2: For vertical dimension, since each square is 2 units tall, and we have 3 rows, the number of vertical positions is 3 - 2 = 1. For horizontal, it's n - 2. So total 2×2 squares: 1*(n - 2) = n - 2
- 3×3: Not possible, since vertical space is 3 units, needing 4 rows, which we don't have. So 0.

So total axis-aligned squares: 2(n - 1) + (n - 2) = 2n - 2 + n - 2 = 3n - 4
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_1.3:
Now, rotated squares ( diamonds ). In a grid, a square can be rotated such that its sides are not aligned with the grid axes. For such squares, the side length in terms of grid units would be sqrt(k), where k is the number of units moved horizontally and vertically. But to have vertices on the grid points, the coordinates must be integers. So, a rotated square would require that the distance between points is the same in both x and y directions, forming a square.

<prompt_3.1> Identify possible sizes of rotated squares in the grid using the above plan. </prompt_3.1>

Next, 2x2 rotated squares.

<prompt_3.2> Calculate the number of 2x2 rotated squares using the above plan. </prompt_3.2>

Wait, let me check.

For a square with side length sqrt(2), which is a 1x1 square rotated 45 degrees. Its bounding box would be 2x2. So, to fit such a square in the grid, we need a 2x2 area.

<prompt_3.3> Confirm the number of 2x2 rotated squares using the above plan. </prompt_3.3>

Now, 3x3 rotated squares. Wait, can we have a square with side length sqrt(5)?

<prompt_3.4> Determine if 3x3 rotated squares can exist and their count using the above plan. </prompt_3.4>
</Plan_3>

<Execution_3>
<execution_3.1>
For example, a square with vertices at (0,0), (1,1), (0,2), (-1,1) would be a diamond shape with side length sqrt(2). But in our grid, points are at integer coordinates, so such squares must have vertices with integer coordinates. Therefore, the possible rotated squares would have sides that are diagonals of squares in the grid, i.e., their side lengths would be sqrt(k^2 + l^2) where k and l are integers. However, for a square, the side must be equal in all directions, so the possible squares would have sides that correspond to vectors (a, b) such that the perpendicular vector ( -b, a ) has the same length, so sqrt(a^2 + b^2) = sqrt(b^2 + a^2), which is always true. Therefore, any such square can be defined by a vector (a, b), and the square would have sides of length sqrt(a^2 + b^2). But in our case, the grid is 3 rows high and n columns wide. So, the possible vectors (a, b) must be such that when you move a units right and b units up/down, you stay within the grid.

Given the grid has 3 rows, the vertical movement (b) can only be 0, 1, or 2 units, because there are only 3 rows. Similarly, horizontal movement (a) can be up to n - 1 units. But since we need to form a square, the vectors (a, b) and (-b, a) must both fit within the grid. Therefore, for each possible vector (a, b), we can compute the number of such squares.

This seems complicated, but maybe there are only a few possible types of rotated squares in a 3×n grid.
</execution_3.1>

<execution_3.2>
The smallest rotated square would have vertices forming a diamond shape. For example, in a 2x2 grid, you can have such a square. But in our case, the grid is 3 rows high. Let's see.

A rotated square of side length sqrt(2) (which is a 1x1 square rotated 45 degrees) would require a 2x2 grid of points. So, how many such squares can fit in a 3×n grid?

Vertically, to have a 2x2 area, we need 2 consecutive rows. Since there are 3 rows, there are 2 such vertical positions (rows 1-2 and rows 2-3). Horizontally, it's similar to the 2×2 axis-aligned squares: (n - 2) horizontal positions. So total 2x2 rotated squares: 2*(n - 2)

Wait, no. Wait, each 2x2 block in the grid can have one rotated square. So, the number of 2x2 blocks in the grid is (3 - 1) * (n - 1) = 2*(n - 1). Wait, no, that's for axis-aligned squares. For rotated squares, each 2x2 block can contain one rotated square (the diamond). So, the number of 2x2 rotated squares would be (number of 2x2 blocks) * 1 = (3 - 1)*(n - 1) = 2*(n - 1). Wait, but when n is the number of columns, the number of 2x2 blocks horizontally is (n - 1). And vertically, since there are 3 rows, the number of vertical positions for 2x2 blocks is 2. So total 2*(n - 1). Therefore, 2x2 rotated squares: 2(n - 1)
</execution_3.2>

<execution_3.3>
Because the diagonal of the 1x1 square would be sqrt(2), but the axis-aligned bounding box is 2x2. So, in a 3×n grid, how many such positions are there?

Vertically, to have a 2x2 area, we need 2 consecutive rows and 2 consecutive columns. Since there are 3 rows, the number of vertical positions is 3 - 1 = 2. Horizontally, it's n - 1. Wait, no. Wait, the 2x2 area in terms of points. Wait, if the grid has n columns, the number of 2x2 areas horizontally is (n - 1). Wait, maybe I need to think in terms of positions. Let's take an example.

If n = 4, then the number of 2x2 axis-aligned squares is 3 (since 4 - 1 = 3). But for rotated squares, how many positions? For a 2x2 area (in terms of points), the number of positions would be (number of ways to choose 2 consecutive columns) * (number of ways to choose 2 consecutive rows). Since there are 3 rows, the number of vertical 2-consecutive rows is 2 (rows 1-2 and 2-3). Horizontally, it's (n - 1). So, total rotated squares of size 1x1 (rotated) would be 2*(n - 1). Wait, but in a 3x4 grid, how many?

Wait, let's take n=4. Then, horizontally, the number of 2-column spans is 3 (columns 1-2, 2-3, 3-4). Vertically, there are 2 possible vertical spans (rows 1-2 and 2-3). So, total rotated squares would be 3*2=6. Each of these spans a 2x2 area, allowing for a square rotated 45 degrees. So, each such 2x2 area contains exactly one such rotated square. Therefore, the number of 1x1 rotated squares is 2*(n - 1). Wait, but in n=4, that would be 2*3=6, which matches. So, general formula: for each possible 2x2 area in the grid, there is one rotated square. The number of 2x2 areas in the grid is (number of horizontal 2-column spans) * (number of vertical 2-row spans). Since there are 3 rows, vertical 2-row spans are 2. Horizontal 2-column spans are (n - 1). Therefore, rotated squares of the first type (1x1 rotated) are 2*(n - 1). Wait, but in the Asymptote code, they have a 3x3 grid with some black squares. Let me check if that's consistent.

Wait, in the Asymptote code, there's a 3x3 grid with some filled squares. The coordinates given are (2.95,-0.05) etc. , which seems like they are drawing a 3x3 grid with some extra points. But maybe the Asymptote code is just a small example. Anyway, back to the problem.

So, for rotated squares, the first type is the 1x1 rotated (which is actually a square of side sqrt(2)), and they are counted as 2*(n - 1). Then, there might be larger rotated squares. Let's see.

Next, 2x2 rotated squares. Wait, what's the side length here? If we have a square that's rotated 45 degrees and has side length sqrt(5). Wait, how does that work?
</execution_3.3>

<execution_3.4>
Wait, maybe not. Let me think.

Alternatively, maybe there are squares with different side lengths. Let me approach this systematically.

For axis-aligned squares, we have side lengths of 1, 2 (since 3 is too big vertically). For rotated squares, the possible side lengths would correspond to squares that fit within the grid. The minimal rotated square is a 1x1 square rotated 45 degrees, which would have vertices at (0,0), (1,1), (0,2), (-1,1), but wait, that goes out of bounds if we're only considering positive coordinates. Wait, actually, in our grid, points are from (0,0) to (n,2), right? Because it's a 3×n grid, so 3 rows (y from 0 to 2) and n columns (x from 0 to n-1). Wait, actually, the Asymptote code shows points up to (4,2) and (5,2), so maybe the coordinates are (x, y) where x ranges from 0 to n-1 and y from 0 to 2. So, the grid is n points wide and 3 points tall.

So, the coordinates are (x, y) where x is from 0 to n-1 and y is 0, 1, 2. So, for example, in the Asymptote code, they have points like (0,0), (1,0), (2,0), (0,1), etc. , up to (5,2). So, n is the number of columns, each column having 3 points (rows).

So, to form a rotated square, let's consider squares that are diamond-shaped. The smallest such square would have vertices like (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). But for this to be a valid square, all these points must lie within the grid. So, the x-coordinate must be such that x-1 >= 0 and x+1 < n (assuming columns are indexed from 0 to n-1). Similarly, y+2 must be <= 2, since there are only 3 rows (y=0,1,2). Therefore, y+2 <= 2 implies y <= 0. So, such a square can only exist in the first row (y=0). Wait, but if y starts at 0, then y+2 would be 2, which is allowed. So, for example, a square with vertices (x,0), (x+1,1), (x,2), (x-1,1). This is a diamond-shaped square with vertical diagonal of length 2 and horizontal diagonal of length 2, but rotated 45 degrees. Wait, actually, the side length in terms of grid steps would be sqrt(2). So, such a square would have side length sqrt(2).

But in terms of the grid, how many such squares can fit? Let's see. The top vertex is at y=2, so the bottom vertex is at y=0. The horizontal positions must be such that x-1 >=0 and x+1 <n. So, x must be at least 1 and at most n-2. Therefore, the number of such squares in the vertical direction is 1 (since y is fixed at 0 for the bottom vertex, and the top vertex is at y=2). Horizontally, the number of positions is (n - 2). So, total number of 1x1 rotated squares (side length sqrt(2)) is (n - 2).
</execution_3.4>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_3.2, execution_3.3, execution_3.4:
Wait, but in the example above, if n=3, then x can be 1, so only one position. So, for n=3, number of such squares is 1. Let's verify.

<prompt_4.1> Verify the count of rotated squares with side length sqrt(2) using the above plan. </prompt_4.1>

Next, side length sqrt(5). How?

Wait, sqrt(5) would correspond to a vector (2,1), since 2^2 + 1^2 = 5. So, moving 2 units right and 1 unit up. Then, such a square would have vertices at (x,y), (x+2, y+1), (x+1, y+2), (x-1, y+1). Wait, but need to check if such a square can fit in the grid.

<prompt_4.2> Calculate the number of rotated squares with side length sqrt(5) using the above plan. </prompt_4.2>

But wait, let's check if that's correct. For example, starting at (x,y), moving (2,1), then ( -1,2 ), etc. , but maybe the vectors need to be consistent.

Alternatively, perhaps the side length sqrt(5) squares would have their vertices at (x,y), (x+1, y+2), (x, y+4), (x-1, y+2). But y+4 would exceed the grid's 3 rows (y=0,1,2). Therefore, y+4 <= 2 implies y <= -2, which is impossible. Therefore, such squares can't exist.

Therefore, the only possible rotated squares are those with side length sqrt(2) (the 1x1 rotated squares) and maybe another size.

<prompt_4.3> Verify the existence of other rotated squares and finalize the count using the above plan. </prompt_4.3>
</Plan_4>

<Execution_4>
<execution_4.1>
If n=3, columns are x=0,1,2. So, x-1=0 is allowed (x=1), and x+1=2 is the last column (x=2). So, the square with vertices (1,0), (2,1), (1,2), (0,1) is valid. So, yes, one such square. So, formula (n - 2) gives 1 when n=3, which is correct.

Similarly, for n=4, x can be 1 and 2. So, two such squares: one starting at x=1 and x=2. So, (4 - 2) = 2, correct.

Therefore, rotated squares of side length sqrt(2) have count (n - 2).

But wait, in the earlier example, n=4 would have (4 - 2)=2 such squares. Let's check:

For n=4, columns x=0,1,2,3.

First square: x=1: (1,0), (2,1), (1,2), (0,1)

Second square: x=2: (2,0), (3,1), (2,2), (1,1)

So, two squares, which matches (4 - 2)=2. Correct.

So, rotated squares of this type (side sqrt(2)) count is (n - 2).
</execution_4.1>

<execution_4.2>
Let's take n=4. Then x can be 1 or 2. So, two positions. So, the squares would have vertices like (1,0), (3,1), (2,2), (0,1). Wait, but (0,1) is a valid point if x starts at 0. So, for n=4, x can be 1 and 2. So, two squares. Therefore, the number is (n - 2). So, for general n, the number of rotated squares with side length sqrt(2) is (n - 2).

But wait, in the vertical direction, the square spans from y=0 to y=2, so it's fixed. So, the only variation is horizontal. So, the number of such squares is (n - 2).
</execution_4.2>

<execution_4.3>
Wait, perhaps there's another type of rotated square with different dimensions. Let's think.

Alternatively, maybe there are squares with different orientations. For example, squares that are not centered at the same points. Wait, but in a 3-row grid, the maximum vertical distance between two points is 2 units (from y=0 to y=2). So, a square that spans vertically must have sides that are either aligned or at 45 degrees. Wait, but if a square is rotated, its vertical diagonal can be up to 2 units. For example, a square with vertical diagonal of 2 units would have side length sqrt(2), which we already considered. Alternatively, a square with horizontal diagonal of 2 units (but that would be a different orientation). Wait, perhaps I need to clarify.

Wait, perhaps the squares can have different side lengths based on the vectors. For example, in a 3×n grid, the possible vectors (a, b) where a and b are integers, and the square can be formed. The side length would be sqrt(a^2 + b^2), and the square must fit within the grid.

Given the grid has 3 rows (y from 0 to 2) and n columns (x from 0 to n-1), the possible vectors (a, b) must satisfy:

- For vertical movement: b can be -2, -1, 0, 1, 2 (but since we need to stay within 3 rows, the actual possible b values are limited). Wait, actually, the vectors must be such that starting from a point (x, y), moving (a, b) keeps the point within the grid. So, for example, if we have a vector (1, 1), then from (x, y), we go to (x+1, y+1). To form a square, we need to check if all four vertices are within the grid.

Alternatively, perhaps it's easier to consider all possible square sizes and count them.

But maybe there's a better way. Let's think of all possible squares:

1. Axis-aligned:

- 1x1: as before, 2(n - 1)

- 2x2: n - 2

Total axis-aligned: 3n - 4

2. Rotated squares:

a. Squares with side length sqrt(2): These are the diamonds we discussed earlier. Each such square is defined by a 2x2 block of points. The number of such squares is (number of 2x2 blocks) * 1 (since each 2x2 block can contain one such square). The number of 2x2 blocks in the grid is (n - 1) horizontally and 2 vertically (since 3 rows allow for 2 vertical positions). Therefore, total number is 2*(n - 1). So, count: 2(n - 1)

b. Squares with side length sqrt(5): These would be larger rotated squares. Let's see if they can fit. For example, a square with side vector (2,1). Starting at (x, y), the next point is (x+2, y+1), then (x+3, y+2), but y+2 must be <= 2, so y <= 0. Then, x+3 must be < n. So, x can be from 0 to n - 4. So, the number of such squares would be (n - 3) * 1 (since y must be 0). Therefore, count: (n - 3)

Wait, but let's verify. For example, if n=4, then x can be 0 (since 0 + 3 = 3 < 4). So, one such square. If n=5, x can be 0 and 1, so two squares. So, yes, (n - 3).

But does such a square actually form a valid square? Let's check.

Take points (x, y), (x+2, y+1), (x+3, y+2), (x+1, y+1). Wait, connecting these points: from (x, y) to (x+2, y+1) is a vector of (2,1). Then from (x+2, y+1) to (x+3, y+2) is (1,1). Wait, that's not a 90-degree rotation. Hmm, maybe my approach is wrong.

Alternatively, perhaps the square is defined by vectors that are perpendicular. For a square, the sides must be perpendicular and of equal length. So, if one side is vector (a, b), the next side should be vector (-b, a) or (b, -a) to form a 90-degree rotation.

So, for example, if we take a vector (1, 2), then the next vector should be (-2, 1) to form a square. Let's see if such a square can fit in the grid.

Starting at (x, y), moving (1,2) to (x+1, y+2). Then, from there, moving (-2,1) to (x+1 -2, y+2 +1) = (x -1, y+3). But y+3 would exceed the grid's 3 rows (max y=2). Therefore, y+3 > 2 implies y <= -1, which is impossible. Therefore, such a square can't exist.

Alternatively, maybe a different vector. Let's take vector (2,1). Then, the next vector should be (-1,2). Starting at (x, y), moving (2,1) to (x+2, y+1). Then, moving (-1,2) to (x+2 -1, y+1 +2) = (x+1, y+3). Again, y+3 exceeds the grid. So, that's not possible.

Alternatively, vector (1,1). Then next vector (-1,1). Starting at (x, y), moving (1,1) to (x+1, y+1). Then moving (-1,1) to (x+1 -1, y+1 +1) = (x, y+2). Then, moving (-1, -1) to (x -1, y+2 -1) = (x -1, y+1). Then back to (x, y). Wait, but does this form a square?

Wait, points would be (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). Let's check distances. Distance between (x, y) and (x+1, y+1) is sqrt(1 + 1) = sqrt(2). Between (x+1, y+1) and (x, y+2): sqrt(1 + 1) = sqrt(2). Between (x, y+2) and (x-1, y+1): sqrt(1 + 1) = sqrt(2). And between (x-1, y+1) and (x, y): sqrt(1 + 1) = sqrt(2). So, all sides equal. Now, check the angles. The vectors from (x, y) to (x+1, y+1) is (1,1), and from (x+1, y+1) to (x, y+2) is (-1,1). The dot product is (1)(-1) + (1)(1) = -1 + 1 = 0. So, they are perpendicular. Therefore, this is a square. Therefore, such squares exist.

To fit this square in the grid, the points must satisfy:

- y+2 <= 2 => y <= 0

- x-1 >= 0 => x >= 1

- x+1 < n => x < n -1

Therefore, x must be between 1 and n -2 (inclusive), so the number of x positions is (n - 2) - 1 + 1 = n - 2. Wait, x starts at 1 and goes up to n - 2, so the number of positions is (n - 2) - 1 + 1 = n - 2. Wait, no. The number of integers from a to b inclusive is b - a + 1. So, from 1 to n - 2, it's (n - 2) - 1 + 1 = n - 2. Therefore, for each y = 0 (since y must be <=0 and y >=0, so y=0), the number of such squares is (n - 2). Therefore, total count is (n - 2).

But wait, let's check with n=3. Then, x must be 1, so n - 2 =1. So, 1 square. For n=4, x can be 1 and 2, so 2 squares. That seems correct. Therefore, the number of such squares is (n - 2). But these squares are of side length sqrt(2), same as the previous rotated squares. Wait, but this seems conflicting with the previous count.

Wait, no. Wait, the squares with side length sqrt(2) can be in two orientations: one where the square is diamond-shaped with vectors (1,1) and (-1,1), and another with vectors (2,1) and (-1,2), but the latter might not fit. Wait, no, the squares we just considered with side length sqrt(2) are the ones with vectors (1,1) and (-1,1). The ones with vectors (2,1) and (-1,2) would have side length sqrt(5), but as we saw, they can only exist if y+3 <=2, which is impossible. Therefore, perhaps there's a confusion here.

Wait, actually, the square we just considered with points (x, y), (x+1, y+1), (x, y+2), (x-1, y+1) has side length sqrt(2), same as the diamond squares. Therefore, maybe they are the same as the 1x1 rotated squares. Therefore, perhaps my earlier count was correct, and there's only one type of rotated square with side length sqrt(2). Then, where does the problem mention three sizes? Hmm.

Wait, the problem states: "there is a total of 70 squares of three sizes". So, there must be three distinct sizes. If axis-aligned have sizes 1x1 and 2x2, and rotated squares have size sqrt(2), but sqrt(2) is a different size. However, perhaps the problem considers squares of different dimensions, regardless of orientation. So, the three sizes are 1x1 (axis-aligned), 2x2 (axis-aligned), and 1x1 rotated (sqrt(2) side length). But the problem mentions three sizes, so maybe there's another size.
</execution_4.3>
</Execution_4>

<Plan_5>
Based on execution_4.3:
Hmm, perhaps there's another orientation of squares. Let me think again.

Alternatively, maybe there are squares of size sqrt(5). Let's see.

<prompt_5.1> Re-evaluate the possibility of larger rotated squares and confirm the count of rotated squares using the above plan. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
If we consider a square with side length sqrt(5), which would correspond to a vector (2,1). Let's see if such a square can fit. Starting at (x, y), moving (2,1) to (x+2, y+1), then moving (-1,2) to (x+2 -1, y+1 +2) = (x+1, y+3). But y+3 exceeds the grid's 3 rows, so y+3 <=2 implies y <= -1, which is impossible. Therefore, such squares can't exist. Therefore, there are no squares with side length sqrt(5) in a 3×n grid.

Alternatively, maybe a different vector. For example, (1,2). Then, starting at (x, y), moving (1,2) to (x+1, y+2), then moving (-2,1) to (x+1 -2, y+2 +1) = (x-1, y+3). Again, y+3 is out of bounds. So, no.

Therefore, the only rotated squares are those with side length sqrt(2), count (n - 2). Therefore, total squares would be axis-aligned (3n -4) plus rotated (n - 2), total 3n -4 +n -2=4n -6. But the problem states total is 70. So, 4n -6=70 →4n=76→n=19. But the problem mentions three sizes. If rotated squares are considered a separate size, then axis-aligned are two sizes (1x1 and 2x2), and rotated is another size. Therefore, three sizes. So, maybe the answer is n=19. But let me check if this is correct.

Wait, but let me verify again. The problem states "three sizes". If axis-aligned are two sizes (1x1 and 2x2), and rotated squares are considered a third size (sqrt(2)), then the formula would be 2(n-1) + (n -2) + (n -2) = 2n -2 +2n -4=4n -6. Wait, no. Wait, axis-aligned 1x1:2(n -1), 2x2: (n -2), rotated: (n -2). So total 2(n -1) + (n -2) + (n -2) =2n -2 +2n -4=4n -6. So, 4n -6=70 →4n=76→n=19. But the problem mentions three sizes, which this would satisfy. However, the Asymptote code shows a 3x3 grid with some filled squares, but the problem states it's a 3×n grid. So, perhaps n=19 is the answer.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But let me check if there are other sizes. Wait, maybe there's another orientation of squares. For example, squares that are 2 units tall and 1 unit wide, but rotated. Wait, but that's the same as the 2x2 axis-aligned. Hmm.

Alternatively, perhaps the problem considers squares of different dimensions, not just by side length but also by the number of squares. Wait, but in the problem statement, it's "three sizes" of squares. So, likely, the three different side lengths. But in our count, we have three types: 1x1 axis-aligned, 2x2 axis-aligned, and 1x1 rotated. Therefore, three sizes. So, the formula would be 2(n -1) + (n -2) + (n -2) =4n -6. Therefore, 4n -6=70→n=19.

But let me check with n=5. If n=5, total squares would be 4*5 -6=14. Let's verify:

<prompt_6.1> Verify the formula with specific examples and conclude the solution using the above plan. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Axis-aligned 1x1:2*(5 -1)=8
Axis-aligned 2x2:5 -2=3
Rotated:5 -2=3
Total:8+3+3=14. Correct. So, for n=5, total 14 squares. So, formula works. Therefore, for 70 squares, n=19.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
But wait, let me check the Asymptote code again. The Asymptote code shows a 3×3 grid (points from (0,0) to (4,2)), but the problem mentions a 3×n grid. So, perhaps the Asymptote code is just an example, and the actual problem is general. So, the answer is n=19.

But wait, let me check another example. If n=4:

<prompt_7.1> Re-confirm the formula and solve for n using the above plan. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Axis-aligned 1x1:2*(4-1)=6
Axis-aligned 2x2:4-2=2
Rotated:4-2=2
Total:6+2+2=10. 4*4 -6=10. Correct.

Therefore, the formula seems consistent. Therefore, solving 4n -6=70 gives n=19.
**Final Answer**
The value of \( n \) is \boxed{19}.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
**Final Answer**
\boxed{19}
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

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