Explanation:
Okay, so I need to figure out the value of n for a 3×n grid of points where there are a total of 70 squares of three sizes. The Asymptote code shows a 3×3 grid with some squares drawn, but maybe the actual problem is for a larger grid? Wait, the problem says "three sizes" of squares. Let me think.

First, let me recall how to count squares in a grid. In a grid of points, the number of squares depends on the size of the grid. For example, in a grid with m rows and n columns of points, the number of squares can be calculated by considering all possible square sizes.

But here, the grid is 3×n, meaning there are 3 rows and n columns of points. Wait, actually, the grid is 3×n, so the number of squares would depend on the number of squares of different sizes. The problem mentions three sizes of squares. Let me think about what sizes are possible in a 3×n grid.

First, the smallest squares would be 1×1 squares. Then, maybe 2×2 squares? But in a 3×n grid, can we have 2×2 squares? Let me visualize. If there are 3 rows of points, then vertically, the distance between rows is 1 unit. So, a square of size 1×1 would have sides of length 1. A square of size 2×2 would have sides of length 2. But to have a square of size 2×2, we need at least 3 points in each direction? Wait, no. Wait, the grid is 3×n points. So, the number of horizontal lines is 3, and vertical lines is n. Wait, actually, the grid is 3 rows and n columns of points. So, the number of horizontal lines is 3, and vertical lines is n. Wait, but squares are formed by the intersections of these lines.

Wait, maybe I need to think in terms of the number of squares of different sizes. Let me recall the formula for counting squares in a grid. In a grid with m rows and n columns of points, the number of squares is the sum over k from 1 to min(m-1, n-1) of (m - k) * (n - k). Wait, but that's for squares of size k×k. Wait, actually, if you have a grid with m rows and n columns of points, then the number of squares of size 1×1 is (m-1)*(n-1). For size 2×2, it's (m-2)*(n-2), and so on. So, in general, for each square size k×k, the number is (m - k) * (n - k). But here, the grid is 3×n points. So m = 3, n = n. Therefore, the number of squares of size 1×1 is (3 - 1)*(n - 1) = 2*(n - 1). The number of squares of size 2×2 is (3 - 2)*(n - 2) = 1*(n - 2). For squares of size 3×3, it would be (3 - 3)*(n - 3) = 0. So, only squares of size 1×1 and 2×2. But the problem says there are three sizes. Hmm, that's conflicting. Maybe I'm misunderstanding the grid?

Wait, the problem says "three sizes whose vertices are points on this rectangular 3×n grid of points." Maybe there are squares of different orientations? Like squares that are not aligned with the grid? For example, squares that are rotated? But the Asymptote code shows some squares, maybe?

Wait, looking at the Asymptote code, it seems like there's a 3×3 grid of points (since there are points from (0,0) to (2,2)), and then some filled squares. But the problem says it's a 3×n grid. Maybe the Asymptote code is just an example? Wait, the Asymptote code draws a 3×3 grid of points, but then adds some squares. However, the problem says there are three sizes of squares. Maybe in a 3×n grid, there are squares of different sizes, including squares that are not axis-aligned?

Wait, let me think again. If the grid is 3×n points, then the number of squares can be calculated considering both axis-aligned squares and squares at other angles. However, the problem mentions "three sizes," so maybe there are three different sizes of squares, regardless of orientation? Or maybe three different sizes of axis-aligned squares?

But earlier, in a 3×n grid, axis-aligned squares would be 1×1 and 2×2. So only two sizes. Therefore, maybe the problem includes squares of different orientations? For example, squares that are tilted?

For example, in a grid, you can have squares that are rotated 45 degrees. For example, in a 3×3 grid, there is a square with vertices at (0,1), (1,2), (2,1), (1,0). That's a square of side length √2. So that's a different size. So maybe the problem counts squares of different orientations as different sizes?

So, if that's the case, then in a 3×n grid, there could be multiple sizes of squares. Let me try to figure out how many different sizes there are.

First, axis-aligned squares:

- 1×1 squares: side length 1.

- 2×2 squares: side length 2.

Then, squares that are tilted:

- For example, squares with side length √2 (like the one I mentioned above). These would be squares formed by connecting points that are diagonally adjacent.

- Also, maybe larger squares? For example, squares with side length √5? Wait, how?

Wait, in a grid, the possible squares can be determined by the distances between points. For example, a square can have sides that are vectors (a,b) and (-b,a), so that they are perpendicular and of equal length. So, for example, a square with side vector (1,1) would have side length √2, and another square with side vector (2,1) would have side length √(2² + 1²) = √5. So, depending on the grid size, these squares can exist.

But in a 3×n grid, how many different sizes of squares can we have?

First, axis-aligned squares:

- 1×1: side length 1.

- 2×2: side length 2.

Then, tilted squares:

- Squares with side length √2: these are squares formed by moving 1 unit right and 1 unit up, then turning left, etc. For example, in a 3×n grid, such squares can exist if there are enough points. For example, in a 3×n grid, to have a square of side length √2, you need at least 2 units in both directions. Wait, actually, the square with side length √2 would require that there are points at positions (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). Wait, but in a 3×n grid, the vertical direction has only 3 points, so y can be 0,1,2. So, for example, a square with vertices at (0,0), (1,1), (0,2), (-1,1) would require a point at (-1,1), which is outside the grid. Therefore, in a 3×n grid, such squares can only exist if there are enough columns. For example, if n is at least 2, then you can have squares that go from column 0 to 1, but to have the square, you need to have points in columns 0,1, and also maybe column 2? Wait, maybe I need to think more carefully.

Let me think of specific examples. Let's take a square with side length √2. For example, in a 3×3 grid, the square with vertices at (0,1), (1,2), (2,1), (1,0). This square is centered in the grid. So, in order to have such a square, you need that the grid has at least 3 columns (so that you can go from column 0 to 2) and 3 rows (so that you can go from row 0 to 2). Therefore, in a 3×n grid, if n is at least 3, then such squares can exist. Similarly, there might be other squares of different sizes.

Another square could be of side length √5. For example, a square with side vector (2,1). So, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and up 2 to (1,3), but that would require a row 3, which is outside the grid. Alternatively, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and down 2 to (1,-1), which is also outside. So maybe such squares can't exist in a 3×n grid? Wait, unless there's another configuration.

Alternatively, maybe squares with side length √(1² + 2²) = √5. For example, starting at (0,0), moving 1 right and 2 up to (1,2), then moving left 2 and up 1 to (-1,3), which is outside. Not helpful. Alternatively, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and up 2 to (1,3), which is outside. So maybe in a 3×n grid, such squares can't exist? Therefore, maybe only squares of side length 1, 2, and √2?

Wait, but the problem says there are three sizes. So maybe the three sizes are 1×1, 2×2, and the tilted squares with side length √2? If that's the case, then we need to calculate the number of squares of each size and set the total to 70.

But let me check if there are more sizes. For example, squares with side length √(2² + 2²) = √8? But that would require moving 2 right and 2 up, but in a 3×n grid, moving 2 right and 2 up would require that there are at least 3 columns (since starting from column 0, moving 2 right gets to column 2) and 3 rows (starting from row 0, moving 2 up gets to row 2). So, such a square would have vertices at (0,0), (2,0), (2,2), (0,2). Wait, that's actually a 2×2 square, which is axis-aligned. So that's already counted in the 2×2 squares. So maybe that's not a different size.

Alternatively, squares with side length √(1² + 3²) = √10? But that would require moving 1 right and 3 up, but since there are only 3 rows, moving 3 up would go beyond. So no.

Alternatively, squares with side length √(2² + 1²) = √5, but as before, seems not possible in a 3×n grid. So maybe only three sizes: 1×1, 2×2, and the tilted squares with side length √2. So total squares would be:

Number of 1×1 squares: ?

Number of 2×2 squares: ?

Number of tilted squares with side length √2: ?

Let me calculate each.

First, axis-aligned squares:

Number of 1×1 squares: In a grid with 3 rows and n columns of points, the number of 1×1 squares is (number of horizontal positions) × (number of vertical positions). Wait, the formula for axis-aligned squares is (rows - 1) × (columns - 1) for 1×1 squares. Wait, rows are 3, columns are n. So number of 1×1 squares is (3 - 1) × (n - 1) = 2(n - 1). Similarly, number of 2×2 squares is (3 - 2) × (n - 2) = 1 × (n - 2) = n - 2.

Now, for the tilted squares with side length √2. How many of those are there?

A square with side length √2 is formed by connecting points that are diagonally adjacent. For example, in a grid, such squares can be formed by selecting a 2×2 block of points and then connecting the diagonals. However, each such square is counted once per 2×2 block. Wait, but in a 2×2 block of points, there are two squares: one with vertices (0,0), (1,0), (1,1), (0,1) and the other with vertices (0,1), (1,1), (1,0), (0,0). Wait, no, actually, the squares with side length √2 would be the ones that are rotated, like the square with vertices (0,0), (1,1), (0,2), (-1,1). But again, that requires points outside the grid. Wait, maybe I need to think differently.

Wait, in a grid, a square with side length √2 can be formed by four points: (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). But for this square to be entirely within the grid, we need that x-1 ≥ 0 and y+2 ≤ 2 (since there are 3 rows). So y+2 ≤ 2 implies y ≤ 0. Therefore, y can only be 0. Similarly, x-1 ≥ 0 implies x ≥ 1. Therefore, such squares can only exist when y=0 and x ≥ 1. Similarly, there might be squares in other orientations?

Alternatively, maybe squares that are formed by points (x, y), (x+1, y), (x+1, y+1), (x, y+1) – but that's a 1×1 square. Wait, no. Wait, the square with side length √2 would have sides that are diagonals of the 1×1 squares. So, for example, in a 2×2 block of points, there is one square of side length √2. Wait, actually, in a 2×2 block of points, you can form two squares: one axis-aligned (1×1) and one rotated (side length √2). Wait, but the rotated square would have vertices at (0,0), (1,1), (0,2), (-1,1). But again, that requires points outside the grid. Wait, maybe I need to think of squares that are entirely within the grid.

Wait, in a 3×n grid, how can we have a square with side length √2? Let me think of coordinates. Suppose we have points (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). For this square to be entirely within the grid, we need:

- x-1 ≥ 0 ⇒ x ≥ 1

- y+2 ≤ 2 ⇒ y ≤ 0

So y must be 0, and x ≥ 1. Therefore, for each x from 1 to n-1 (since x+1 must be ≤ n), we can have such a square. Similarly, there might be squares in other orientations. For example, squares that go in the other direction: (x, y), (x+1, y-1), (x, y-2), (x-1, y-1). But y-2 ≥ 0 ⇒ y ≥ 2. Since there are only 3 rows (y=0,1,2), y=2. Then x ≥ 1. So for y=2, x from 1 to n-1. So that's another set of squares.

Additionally, there might be squares that are shifted horizontally. Wait, but if we consider squares that are not aligned in the vertical direction. For example, squares that go from (x, y), (x+1, y), (x+1, y+1), (x, y+1) – that's a 1×1 square. Wait, no. Wait, the square with side length √2 would have sides that are diagonals of the 1×1 squares. So, for example, in a 2×2 block of points, you can have a square that connects the midpoints? Wait, no, the points are grid points. So, for example, in a 2×2 block of points, the square with vertices at (0,0), (1,1), (0,2), (-1,1) is not entirely within the grid. However, if we have a 3×3 grid, then there is a square with vertices at (0,1), (1,2), (2,1), (1,0). This square is entirely within the grid. So, how many such squares are there?

In a 3×n grid, how many squares of this type (side length √2) are there? Let's think. The square with vertices (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). Wait, but in the example above, the square is (0,1), (1,2), (2,1), (1,0). So, this square is centered at (1,1). So, to form such a square, we need that there are points in columns x-1, x, x+1 and rows y, y+1, y+2. Wait, but in our grid, there are only 3 rows. So y+2 must be ≤ 2 ⇒ y ≤ 0. Therefore, y can only be 0. Then, the square would be (x, 0), (x+1, 1), (x, 2), (x-1, 1). So, for this square to exist, we need x-1 ≥ 0 ⇒ x ≥ 1, and x+1 ≤ n ⇒ x ≤ n - 1. Therefore, for each x from 1 to n - 1, there is such a square. Similarly, there are squares that go in the other direction, like (x, 2), (x+1, 1), (x, 0), (x-1, 1). Wait, that's the same square. So, actually, each such square is counted once for each x from 1 to n - 1. Therefore, the number of these squares is (n - 1). Similarly, there are squares that are mirrored in the other direction? Wait, no, because if we consider squares that go from (x, 1), (x+1, 2), (x, 3)... but there is no row 3. So, only the ones with y=0 and y=2?

Wait, actually, the square with vertices (x, 1), (x+1, 2), (x, 3)... but row 3 doesn't exist. So, only the squares that are centered in the middle row? Wait, the square I mentioned earlier is centered at (1,1) in a 3×3 grid. So, in general, in a 3×n grid, how many such squares are there?

If we consider squares that are "diamond" shaped, with their center at (x, 1), then they would require points in rows 0, 1, 2 and columns x-1, x, x+1. Therefore, for each column x from 1 to n - 1, there is such a square. Therefore, the number of these squares is (n - 1). Similarly, are there other squares of the same size?

Wait, for example, if we consider squares that are shifted horizontally? For example, starting at (x, 0), going to (x+1, 1), (x+2, 0), (x+1, -1)... but that would go out of bounds. So no. Similarly, starting at (x, 2), going to (x+1, 1), (x+2, 2), (x+1, 3)... which is out of bounds. Therefore, only the squares centered in the middle row (y=1) can exist. Therefore, the number of such squares is (n - 1). Therefore, total number of tilted squares with side length √2 is (n - 1).

Wait, but in the Asymptote code, there are three squares drawn. Let me check. The Asymptote code draws a 3×3 grid of points, and then fills in some squares. The first set of squares is at the bottom, then middle, then top. Each row has three squares? Wait, the Asymptote code has:

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);

So, three squares at the bottom, which are 1×1 squares? Wait, but they are drawn as small squares. Then similarly in the middle and top. So maybe in the Asymptote code, there are 3×3 grid of points, and the number of squares is more than 70? Wait, but the problem says that in a 3×n grid, there are 70 squares of three sizes. So maybe the Asymptote code is just an example, but the actual problem is for a different n.

But going back, if we have three sizes of squares: 1×1, 2×2, and tilted squares with side length √2. Then total number of squares would be:

Number of 1×1 squares: 2(n - 1)

Number of 2×2 squares: (n - 2)

Number of tilted squares: (n - 1)

Wait, but that would be total squares: 2(n - 1) + (n - 2) + (n - 1) = 2n - 2 + n - 2 + n - 1 = 4n - 5. But the problem says total is 70. So 4n - 5 = 70 ⇒ 4n = 75 ⇒ n = 75/4 = 18.75. Which is not an integer. Therefore, something is wrong here.

Therefore, my assumption about the number of tilted squares must be incorrect. Maybe there are more sizes of squares?

Alternatively, maybe there are more tilted squares. Let me think again.

Another type of square: squares with side length √5. How can that be formed? For example, a square with side vector (1,2). So, starting at (x, y), moving 1 right and 2 up to (x+1, y+2), then moving left 2 and up 1 to (x-1, y+3), which is outside. Not helpful. Alternatively, starting at (x, y), moving 2 right and 1 up to (x+2, y+1), then moving left 1 and up 2 to (x+1, y+3), which is outside. Not helpful. Alternatively, moving 2 right and 1 up, then moving left 1 and down 2 to (x+1, y-1). So, if we start at (x, y), then (x+2, y+1), (x+1, y-1), (x-1, y). Wait, this seems complicated. Maybe such squares can't exist in a 3×n grid.

Alternatively, maybe squares that are larger? For example, squares with side length 2√2? But that would require moving 2 right and 2 up, which would be a 2×2 square, which is already counted.

Alternatively, maybe squares that are not aligned with the grid and not the √2 ones. Wait, maybe squares with side length √( (n-1)^2 + ... )? Not sure.

Alternatively, maybe I made a mistake in counting the tilted squares. Let me think again.

In a 3×n grid, how many squares of side length √2 are there? Each such square is determined by its topmost point? Or maybe by its center?

Wait, in the 3×3 grid, there is one square of side length √2 centered at (1,1). But if n is larger, say n=4, then there would be more such squares. For example, in a 3×4 grid, we can have squares centered at (1,1), (2,1), (1,1), (2,1)... Wait, no. Wait, for each column x from 1 to n-1, there is a square centered at (x, 1). Therefore, the number of such squares is (n - 1). But in the 3×3 grid, n=3, so n-1=2. But in reality, in a 3×3 grid, there is only one such square. Wait, that contradicts. Wait, maybe my previous reasoning is wrong.

Wait, in a 3×3 grid, columns are from x=0 to x=2. So, for x from 1 to n-1 = 2. So x=1 and x=2. For x=1, the square would be (1,0), (2,1), (1,2), (0,1). For x=2, the square would be (2,0), (3,1), (2,2), (1,1). But wait, in a 3×3 grid, the columns go up to x=2, so x=3 is outside. Therefore, for x=2, the square would require column x+1 = 3, which is outside. Therefore, actually, only x=1 is valid. Therefore, in a 3×3 grid, there is only 1 such square. Therefore, my previous formula of (n - 1) is wrong. It should be (n - 2). Because for x from 1 to n - 2, since x+1 must be ≤ n - 1? Wait, no. Wait, in the 3×n grid, columns are from 0 to n - 1. Therefore, to have a square with vertices (x, 0), (x+1, 1), (x, 2), (x-1, 1), we need x-1 ≥ 0 ⇒ x ≥ 1, and x+1 ≤ n - 1 ⇒ x ≤ n - 2. Therefore, x ranges from 1 to n - 2. Therefore, the number of such squares is (n - 2). Similarly, there are squares in the other direction? For example, squares that go from (x, 2), (x+1, 1), (x, 0), (x-1, 1). But that's the same square as before. Therefore, total number of such squares is (n - 2). Therefore, in the 3×3 grid, n=3, so n - 2 = 1, which matches. In a 3×4 grid, n=4, so n - 2 = 2 squares. That seems correct.

Therefore, correcting my previous mistake, the number of tilted squares with side length √2 is (n - 2). Therefore, total number of squares would be:

Number of 1×1 squares: 2(n - 1)

Number of 2×2 squares: (n - 2)

Number of tilted squares: (n - 2)

Therefore, total squares: 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6.

Set this equal to 70:

4n - 6 = 70 ⇒ 4n = 76 ⇒ n = 19.

Wait, so n=19? Let me check if this makes sense.

But let me verify with a small n. Let's take n=3. Then total squares would be 4*3 - 6 = 6. Let's count manually.

In a 3×3 grid:

Number of 1×1 squares: 2*(3 - 1) = 4. Wait, but in reality, in a 3×3 grid of points, the number of 1×1 squares is (3 - 1)*(3 - 1) = 2*2 = 4. Correct.

Number of 2×2 squares: (3 - 2)*(3 - 2) = 1*1 = 1. Correct.

Number of tilted squares: (3 - 2) = 1. Which is the square I mentioned earlier. So total squares: 4 + 1 + 1 = 6. Which matches 4*3 - 6 = 6. So that works.

Another check: n=4. Total squares: 4*4 - 6 = 10.

Number of 1×1 squares: 2*(4 - 1) = 6.

Number of 2×2 squares: (4 - 2) = 2.

Number of tilted squares: (4 - 2) = 2.

Total: 6 + 2 + 2 = 10. Correct.

So, formula seems correct. Therefore, if total squares is 70, then 4n - 6 = 70 ⇒ 4n = 76 ⇒ n = 19. Therefore, the answer is 19? But let me check again.

Wait, but the problem says "three sizes of squares". In our calculation, we have three sizes: 1×1, 2×2, and √2. So that's three sizes. Therefore, the answer is n=19. But let me check if there are more sizes.

Wait, could there be another size of square? For example, squares with side length √( (1)^2 + (2)^2 ) = √5? Let me think. In a 3×n grid, can we have such squares?

For example, starting at (0,0), moving 1 right and 2 up to (1,2), then moving left 2 and up 1 to (-1,3), which is outside. Not possible. Alternatively, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and up 2 to (1,3), which is outside. Not possible. Alternatively, starting at (1,0), moving 2 right and 1 up to (3,1), which is outside if n=3. So unless n is larger, but even then, the square would require points outside. Therefore, maybe not. Therefore, only three sizes.

Alternatively, squares with side length 2√2? Which would be a square that spans from (0,0) to (2,2), which is a 2×2 square, already counted. So no.

Therefore, the total number of squares is indeed 4n - 6. Therefore, solving 4n - 6 = 70 gives n = 19. Therefore, the answer is 19. But let me check again with the Asymptote code.

The Asymptote code shows a 3×3 grid with some squares. If n=3, total squares would be 6. The Asymptote code draws three squares at the bottom, three in the middle, and three at the top? Wait, looking at the Asymptote code:

It draws three small squares at the bottom, then three in the middle, then three at the top. But each of those is probably 1×1 squares. However, in reality, in a 3×3 grid, there are 4 1×1 squares, 1 2×2 square, and 1 tilted square. So total 6 squares. The Asymptote code might be drawing some of them, but maybe not all. Anyway, the math seems to check out.

Therefore, if the total number of squares is 70, then n=19. Therefore, the answer is \boxed{19}.

Wait, but let me check once more. Maybe I missed some squares?

Wait, another type of square: squares that are larger than 2×2? For example, in a 3×n grid, can we have a square of size 2×2? Yes, which we already counted. What about squares that are not aligned and have different sizes?

Wait, for example, in a 3×n grid, can we have squares with side length √( (n - 1)^2 + ... )? Not sure. Wait, for example, if n is large enough, can we have squares that span multiple columns?

Wait, for example, a square that goes from (0,0) to (k, 1) to (k, 2) to (0,1)... but that would be a rectangle, not a square. Wait, unless k is such that the sides are equal. For example, if we have a square with side length √(k² + 1²). For that to be a square, the sides must be equal and the angles must be 90 degrees. So, for example, if we have a square with side vector (a, b) and (-b, a), then the length is √(a² + b²). So, for example, if a=1, b=2, then the square would have side length √5. But to have such a square in the grid, we need that the points exist.

For example, starting at (0,0), moving 1 right and 2 up to (1,2), then moving -2 right and 1 up to (-1,3), which is outside. Not possible. Starting at (0,0), moving 2 right and 1 up to (2,1), then moving -1 right and 2 up to (1,3), which is outside. Not possible. Starting at (1,0), moving 2 right and 1 up to (3,1), which is outside if n=3. So unless n is larger, but even then, the square would require points outside. Therefore, such squares cannot exist in a 3×n grid. Therefore, there are no other square sizes. Therefore, the total number of squares is indeed 4n - 6. Therefore, solving 4n - 6 = 70 gives n=19.

Therefore, I think the answer is 19.

**Final Answer**
\boxed{19}

To determine the value of $ n $ in a $ 3 \times n $ grid of points where there are a total of 70 squares of **three distinct sizes**, we need to analyze the types of squares that can be formed in such a grid.

---

### Step 1: Understanding the Grid

A $ 3 \times n $ grid of points means there are 3 rows and $ n $ columns of points. The total number of squares that can be formed includes:

- **Axis-aligned squares** (sides parallel to the grid axes)
- **Tilted squares** (squares rotated at 45°, with sides not aligned with the grid)

We are told that there are **three sizes** of squares. These are:

1. **1×1 squares** (axis-aligned, side length = 1)
2. **2×2 squares** (axis-aligned, side length = 2)
3. **Squares with side length $ \sqrt{2} $** (tilted, formed by diagonals of 1×1 squares)

---

### Step 2: Counting the Squares

#### 1. **1×1 Squares**

These are the smallest axis-aligned squares. In a $ 3 \times n $ grid, the number of such squares is:

$$
(3 - 1) \times (n - 1) = 2(n - 1)
$$

#### 2. **2×2 Squares**

These are larger axis-aligned squares. The number of such squares is:

$$
(3 - 2) \times (n - 2) = 1 \times (n - 2) = n - 2
$$

#### 3. **Tilted Squares with Side Length $ \sqrt{2} $**

These are squares formed by connecting diagonally adjacent points. Each such square is centered in the middle row (row 1), and spans from column $ x $ to $ x+1 $, with the square's top and bottom points in rows 0 and 2.

To form such a square, we need:

- $ x \geq 1 $ (to have a left neighbor)
- $ x+1 \leq n - 1 $ (to have a right neighbor)

So, the number of such squares is:

$$
(n - 2)
$$

---

### Step 3: Total Number of Squares

Adding up all the square types:

$$
\text{Total Squares} = 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6
$$

We are told that the total number of squares is 70:

$$
4n - 6 = 70
$$

Solving for $ n $:

$$
4n = 76 \Rightarrow n = \frac{76}{4} = 19
$$

---

### Final Answer

$$
\boxed{19}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of n for a 3×n grid of points where there are a total of 70 squares of three sizes. The Asymptote code shows a 3×3 grid with some squares drawn, but maybe the actual problem is for a larger grid? Wait, the problem says "three sizes" of squares. Let me think.

First, let me recall how to count squares in a grid. In a grid of points, the number of squares depends on the size of the grid. For example, in a grid with m rows and n columns of points, the number of squares can be calculated by considering all possible square sizes.

But here, the grid is 3×n, meaning there are 3 rows and n columns of points. Wait, actually, the grid is 3×n, so the number of squares would depend on the number of squares of different sizes. The problem mentions three sizes of squares. Let me think about what sizes are possible in a 3×n grid.

First, the smallest squares would be 1×1 squares. Then, maybe 2×2 squares? But in a 3×n grid, can we have 2×2 squares? Let me visualize. If there are 3 rows of points, then vertically, the distance between rows is 1 unit. So, a square of size 1×1 would have sides of length 1. A square of size 2×2 would have sides of length 2. But to have a square of size 2×2, we need at least 3 points in each direction? Wait, no. Wait, the grid is 3×n points. So, the number of horizontal lines is 3, and vertical lines is n. Wait, actually, the grid is 3 rows and n columns of points. So, the number of horizontal lines is 3, and vertical lines is n. Wait, but squares are formed by the intersections of these lines.

Wait, maybe I need to think in terms of the number of squares of different sizes. Let me recall the formula for counting squares in a grid. In a grid with m rows and n columns of points, the number of squares is the sum over k from 1 to min(m-1, n-1) of (m - k) * (n - k). Wait, but that's for squares of size k×k. Wait, actually, if you have a grid with m rows and n columns of points, then the number of squares of size 1×1 is (m-1)*(n-1). For size 2×2, it's (m-2)*(n-2), and so on. So, in general, for each square size k×k, the number is (m - k) * (n - k). But here, the grid is 3×n points. So m = 3, n = n. Therefore, the number of squares of size 1×1 is (3 - 1)*(n - 1) = 2*(n - 1). The number of squares of size 2×2 is (3 - 2)*(n - 2) = 1*(n - 2). For squares of size 3×3, it would be (3 - 3)*(n - 3) = 0. So, only squares of size 1×1 and 2×2. But the problem says there are three sizes. Hmm, that's conflicting. Maybe I'm misunderstanding the grid?

Wait, the problem says "three sizes whose vertices are points on this rectangular 3×n grid of points." Maybe there are squares of different orientations? Like squares that are not aligned with the grid? For example, squares that are rotated? But the Asymptote code shows some squares, maybe?

Wait, looking at the Asymptote code, it seems like there's a 3×3 grid of points (since there are points from (0,0) to (2,2)), and then some filled squares. But the problem says it's a 3×n grid. Maybe the Asymptote code is just an example? Wait, the Asymptote code draws a 3×3 grid of points, but then adds some squares. However, the problem says there are three sizes of squares. Maybe in a 3×n grid, there are squares of different sizes, including squares that are not axis-aligned?

Wait, let me think again. If the grid is 3×n points, then the number of squares can be calculated considering both axis-aligned squares and squares at other angles. However, the problem mentions "three sizes," so maybe there are three different sizes of squares, regardless of orientation? Or maybe three different sizes of axis-aligned squares?

But earlier, in a 3×n grid, axis-aligned squares would be 1×1 and 2×2. So only two sizes. Therefore, maybe the problem includes squares of different orientations? For example, squares that are tilted?

For example, in a grid, you can have squares that are rotated 45 degrees. For example, in a 3×3 grid, there is a square with vertices at (0,1), (1,2), (2,1), (1,0). That's a square of side length √2. So that's a different size. So maybe the problem counts squares of different orientations as different sizes?

So, if that's the case, then in a 3×n grid, there could be multiple sizes of squares. Let me try to figure out how many different sizes there are.

First, axis-aligned squares:

- 1×1 squares: side length 1.

- 2×2 squares: side length 2.

Then, squares that are tilted:

- For example, squares with side length √2 (like the one I mentioned above). These would be squares formed by connecting points that are diagonally adjacent.

- Also, maybe larger squares? For example, squares with side length √5? Wait, how?

Wait, in a grid, the possible squares can be determined by the distances between points. For example, a square can have sides that are vectors (a,b) and (-b,a), so that they are perpendicular and of equal length. So, for example, a square with side vector (1,1) would have side length √2, and another square with side vector (2,1) would have side length √(2² + 1²) = √5. So, depending on the grid size, these squares can exist.

But in a 3×n grid, how many different sizes of squares can we have?

First, axis-aligned squares:

- 1×1: side length 1.

- 2×2: side length 2.

Then, tilted squares:

- Squares with side length √2: these are squares formed by moving 1 unit right and 1 unit up, then turning left, etc. For example, in a 3×n grid, such squares can exist if there are enough points. For example, in a 3×n grid, to have a square of side length √2, you need at least 2 units in both directions. Wait, actually, the square with side length √2 would require that there are points at positions (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). Wait, but in a 3×n grid, the vertical direction has only 3 points, so y can be 0,1,2. So, for example, a square with vertices at (0,0), (1,1), (0,2), (-1,1) would require a point at (-1,1), which is outside the grid. Therefore, in a 3×n grid, such squares can only exist if there are enough columns. For example, if n is at least 2, then you can have squares that go from column 0 to 1, but to have the square, you need to have points in columns 0,1, and also maybe column 2? Wait, maybe I need to think more carefully.

Let me think of specific examples. Let's take a square with side length √2. For example, in a 3×3 grid, the square with vertices at (0,1), (1,2), (2,1), (1,0). This square is centered in the grid. So, in order to have such a square, you need that the grid has at least 3 columns (so that you can go from column 0 to 2) and 3 rows (so that you can go from row 0 to 2). Therefore, in a 3×n grid, if n is at least 3, then such squares can exist. Similarly, there might be other squares of different sizes.

Another square could be of side length √5. For example, a square with side vector (2,1). So, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and up 2 to (1,3), but that would require a row 3, which is outside the grid. Alternatively, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and down 2 to (1,-1), which is also outside. So maybe such squares can't exist in a 3×n grid? Wait, unless there's another configuration.

Alternatively, maybe squares with side length √(1² + 2²) = √5. For example, starting at (0,0), moving 1 right and 2 up to (1,2), then moving left 2 and up 1 to (-1,3), which is outside. Not helpful. Alternatively, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and up 2 to (1,3), which is outside. So maybe in a 3×n grid, such squares can't exist? Therefore, maybe only squares of side length 1, 2, and √2?

Wait, but the problem says there are three sizes. So maybe the three sizes are 1×1, 2×2, and the tilted squares with side length √2? If that's the case, then we need to calculate the number of squares of each size and set the total to 70.

But let me check if there are more sizes. For example, squares with side length √(2² + 2²) = √8? But that would require moving 2 right and 2 up, but in a 3×n grid, moving 2 right and 2 up would require that there are at least 3 columns (since starting from column 0, moving 2 right gets to column 2) and 3 rows (starting from row 0, moving 2 up gets to row 2). So, such a square would have vertices at (0,0), (2,0), (2,2), (0,2). Wait, that's actually a 2×2 square, which is axis-aligned. So that's already counted in the 2×2 squares. So maybe that's not a different size.

Alternatively, squares with side length √(1² + 3²) = √10? But that would require moving 1 right and 3 up, but since there are only 3 rows, moving 3 up would go beyond. So no.

Alternatively, squares with side length √(2² + 1²) = √5, but as before, seems not possible in a 3×n grid. So maybe only three sizes: 1×1, 2×2, and the tilted squares with side length √2. So total squares would be:

Number of 1×1 squares: ?

Number of 2×2 squares: ?

Number of tilted squares with side length √2: ?

Let me calculate each.

First, axis-aligned squares:

Number of 1×1 squares: In a grid with 3 rows and n columns of points, the number of 1×1 squares is (number of horizontal positions) × (number of vertical positions). Wait, the formula for axis-aligned squares is (rows - 1) × (columns - 1) for 1×1 squares. Wait, rows are 3, columns are n. So number of 1×1 squares is (3 - 1) × (n - 1) = 2(n - 1). Similarly, number of 2×2 squares is (3 - 2) × (n - 2) = 1 × (n - 2) = n - 2.

Now, for the tilted squares with side length √2. How many of those are there?

A square with side length √2 is formed by connecting points that are diagonally adjacent. For example, in a grid, such squares can be formed by selecting a 2×2 block of points and then connecting the diagonals. However, each such square is counted once per 2×2 block. Wait, but in a 2×2 block of points, there are two squares: one with vertices (0,0), (1,0), (1,1), (0,1) and the other with vertices (0,1), (1,1), (1,0), (0,0). Wait, no, actually, the squares with side length √2 would be the ones that are rotated, like the square with vertices (0,0), (1,1), (0,2), (-1,1). But again, that requires points outside the grid. Wait, maybe I need to think differently.

Wait, in a grid, a square with side length √2 can be formed by four points: (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). But for this square to be entirely within the grid, we need that x-1 ≥ 0 and y+2 ≤ 2 (since there are 3 rows). So y+2 ≤ 2 implies y ≤ 0. Therefore, y can only be 0. Similarly, x-1 ≥ 0 implies x ≥ 1. Therefore, such squares can only exist when y=0 and x ≥ 1. Similarly, there might be squares in other orientations?

Alternatively, maybe squares that are formed by points (x, y), (x+1, y), (x+1, y+1), (x, y+1) – but that's a 1×1 square. Wait, no. Wait, the square with side length √2 would have sides that are diagonals of the 1×1 squares. So, for example, in a 2×2 block of points, there is one square of side length √2. Wait, actually, in a 2×2 block of points, you can form two squares: one axis-aligned (1×1) and one rotated (side length √2). Wait, but the rotated square would have vertices at (0,0), (1,1), (0,2), (-1,1). But again, that requires points outside the grid. Wait, maybe I need to think of squares that are entirely within the grid.

Wait, in a 3×n grid, how can we have a square with side length √2? Let me think of coordinates. Suppose we have points (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). For this square to be entirely within the grid, we need:

- x-1 ≥ 0 ⇒ x ≥ 1

- y+2 ≤ 2 ⇒ y ≤ 0

So y must be 0, and x ≥ 1. Therefore, for each x from 1 to n-1 (since x+1 must be ≤ n), we can have such a square. Similarly, there might be squares in other orientations. For example, squares that go in the other direction: (x, y), (x+1, y-1), (x, y-2), (x-1, y-1). But y-2 ≥ 0 ⇒ y ≥ 2. Since there are only 3 rows (y=0,1,2), y=2. Then x ≥ 1. So for y=2, x from 1 to n-1. So that's another set of squares.

Additionally, there might be squares that are shifted horizontally. Wait, but if we consider squares that are not aligned in the vertical direction. For example, squares that go from (x, y), (x+1, y), (x+1, y+1), (x, y+1) – that's a 1×1 square. Wait, no. Wait, the square with side length √2 would have sides that are diagonals of the 1×1 squares. So, for example, in a 2×2 block of points, you can have a square that connects the midpoints? Wait, no, the points are grid points. So, for example, in a 2×2 block of points, the square with vertices at (0,0), (1,1), (0,2), (-1,1) is not entirely within the grid. However, if we have a 3×3 grid, then there is a square with vertices at (0,1), (1,2), (2,1), (1,0). This square is entirely within the grid. So, how many such squares are there?

In a 3×n grid, how many squares of this type (side length √2) are there? Let's think. The square with vertices (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). Wait, but in the example above, the square is (0,1), (1,2), (2,1), (1,0). So, this square is centered at (1,1). So, to form such a square, we need that there are points in columns x-1, x, x+1 and rows y, y+1, y+2. Wait, but in our grid, there are only 3 rows. So y+2 must be ≤ 2 ⇒ y ≤ 0. Therefore, y can only be 0. Then, the square would be (x, 0), (x+1, 1), (x, 2), (x-1, 1). So, for this square to exist, we need x-1 ≥ 0 ⇒ x ≥ 1, and x+1 ≤ n ⇒ x ≤ n - 1. Therefore, for each x from 1 to n - 1, there is such a square. Similarly, there are squares that go in the other direction, like (x, 2), (x+1, 1), (x, 0), (x-1, 1). Wait, that's the same square. So, actually, each such square is counted once for each x from 1 to n - 1. Therefore, the number of these squares is (n - 1). Similarly, there are squares that are mirrored in the other direction? Wait, no, because if we consider squares that go from (x, 1), (x+1, 2), (x, 3)... but there is no row 3. So, only the ones with y=0 and y=2?

Wait, actually, the square with vertices (x, 1), (x+1, 2), (x, 3)... but row 3 doesn't exist. So, only the squares that are centered in the middle row? Wait, the square I mentioned earlier is centered at (1,1) in a 3×3 grid. So, in general, in a 3×n grid, how many such squares are there?

If we consider squares that are "diamond" shaped, with their center at (x, 1), then they would require points in rows 0, 1, 2 and columns x-1, x, x+1. Therefore, for each column x from 1 to n - 1, there is such a square. Therefore, the number of these squares is (n - 1). Similarly, are there other squares of the same size?

Wait, for example, if we consider squares that are shifted horizontally? For example, starting at (x, 0), going to (x+1, 1), (x+2, 0), (x+1, -1)... but that would go out of bounds. So no. Similarly, starting at (x, 2), going to (x+1, 1), (x+2, 2), (x+1, 3)... which is out of bounds. Therefore, only the squares centered in the middle row (y=1) can exist. Therefore, the number of such squares is (n - 1). Therefore, total number of tilted squares with side length √2 is (n - 1).

Wait, but in the Asymptote code, there are three squares drawn. Let me check. The Asymptote code draws a 3×3 grid of points, and then fills in some squares. The first set of squares is at the bottom, then middle, then top. Each row has three squares? Wait, the Asymptote code has:

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);

So, three squares at the bottom, which are 1×1 squares? Wait, but they are drawn as small squares. Then similarly in the middle and top. So maybe in the Asymptote code, there are 3×3 grid of points, and the number of squares is more than 70? Wait, but the problem says that in a 3×n grid, there are 70 squares of three sizes. So maybe the Asymptote code is just an example, but the actual problem is for a different n.

But going back, if we have three sizes of squares: 1×1, 2×2, and tilted squares with side length √2. Then total number of squares would be:

Number of 1×1 squares: 2(n - 1)

Number of 2×2 squares: (n - 2)

Number of tilted squares: (n - 1)

Wait, but that would be total squares: 2(n - 1) + (n - 2) + (n - 1) = 2n - 2 + n - 2 + n - 1 = 4n - 5. But the problem says total is 70. So 4n - 5 = 70 ⇒ 4n = 75 ⇒ n = 75/4 = 18.75. Which is not an integer. Therefore, something is wrong here.

Therefore, my assumption about the number of tilted squares must be incorrect. Maybe there are more sizes of squares?

Alternatively, maybe there are more tilted squares. Let me think again.

Another type of square: squares with side length √5. How can that be formed? For example, a square with side vector (1,2). So, starting at (x, y), moving 1 right and 2 up to (x+1, y+2), then moving left 2 and up 1 to (x-1, y+3), which is outside. Not helpful. Alternatively, starting at (x, y), moving 2 right and 1 up to (x+2, y+1), then moving left 1 and up 2 to (x+1, y+3), which is outside. Not helpful. Alternatively, moving 2 right and 1 up, then moving left 1 and down 2 to (x+1, y-1). So, if we start at (x, y), then (x+2, y+1), (x+1, y-1), (x-1, y). Wait, this seems complicated. Maybe such squares can't exist in a 3×n grid.

Alternatively, maybe squares that are larger? For example, squares with side length 2√2? But that would require moving 2 right and 2 up, which would be a 2×2 square, which is already counted.

Alternatively, maybe squares that are not aligned with the grid and not the √2 ones. Wait, maybe squares with side length √( (n-1)^2 + ... )? Not sure.

Alternatively, maybe I made a mistake in counting the tilted squares. Let me think again.

In a 3×n grid, how many squares of side length √2 are there? Each such square is determined by its topmost point? Or maybe by its center?

Wait, in the 3×3 grid, there is one square of side length √2 centered at (1,1). But if n is larger, say n=4, then there would be more such squares. For example, in a 3×4 grid, we can have squares centered at (1,1), (2,1), (1,1), (2,1)... Wait, no. Wait, for each column x from 1 to n-1, there is a square centered at (x, 1). Therefore, the number of such squares is (n - 1). But in the 3×3 grid, n=3, so n-1=2. But in reality, in a 3×3 grid, there is only one such square. Wait, that contradicts. Wait, maybe my previous reasoning is wrong.

Wait, in a 3×3 grid, columns are from x=0 to x=2. So, for x from 1 to n-1 = 2. So x=1 and x=2. For x=1, the square would be (1,0), (2,1), (1,2), (0,1). For x=2, the square would be (2,0), (3,1), (2,2), (1,1). But wait, in a 3×3 grid, the columns go up to x=2, so x=3 is outside. Therefore, for x=2, the square would require column x+1 = 3, which is outside. Therefore, actually, only x=1 is valid. Therefore, in a 3×3 grid, there is only 1 such square. Therefore, my previous formula of (n - 1) is wrong. It should be (n - 2). Because for x from 1 to n - 2, since x+1 must be ≤ n - 1? Wait, no. Wait, in the 3×n grid, columns are from 0 to n - 1. Therefore, to have a square with vertices (x, 0), (x+1, 1), (x, 2), (x-1, 1), we need x-1 ≥ 0 ⇒ x ≥ 1, and x+1 ≤ n - 1 ⇒ x ≤ n - 2. Therefore, x ranges from 1 to n - 2. Therefore, the number of such squares is (n - 2). Similarly, there are squares in the other direction? For example, squares that go from (x, 2), (x+1, 1), (x, 0), (x-1, 1). But that's the same square as before. Therefore, total number of such squares is (n - 2). Therefore, in the 3×3 grid, n=3, so n - 2 = 1, which matches. In a 3×4 grid, n=4, so n - 2 = 2 squares. That seems correct.

Therefore, correcting my previous mistake, the number of tilted squares with side length √2 is (n - 2). Therefore, total number of squares would be:

Number of 1×1 squares: 2(n - 1)

Number of 2×2 squares: (n - 2)

Number of tilted squares: (n - 2)

Therefore, total squares: 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6.

Set this equal to 70:

4n - 6 = 70 ⇒ 4n = 76 ⇒ n = 19.

Wait, so n=19? Let me check if this makes sense.

But let me verify with a small n. Let's take n=3. Then total squares would be 4*3 - 6 = 6. Let's count manually.

In a 3×3 grid:

Number of 1×1 squares: 2*(3 - 1) = 4. Wait, but in reality, in a 3×3 grid of points, the number of 1×1 squares is (3 - 1)*(3 - 1) = 2*2 = 4. Correct.

Number of 2×2 squares: (3 - 2)*(3 - 2) = 1*1 = 1. Correct.

Number of tilted squares: (3 - 2) = 1. Which is the square I mentioned earlier. So total squares: 4 + 1 + 1 = 6. Which matches 4*3 - 6 = 6. So that works.

Another check: n=4. Total squares: 4*4 - 6 = 10.

Number of 1×1 squares: 2*(4 - 1) = 6.

Number of 2×2 squares: (4 - 2) = 2.

Number of tilted squares: (4 - 2) = 2.

Total: 6 + 2 + 2 = 10. Correct.

So, formula seems correct. Therefore, if total squares is 70, then 4n - 6 = 70 ⇒ 4n = 76 ⇒ n = 19. Therefore, the answer is 19? But let me check again.

Wait, but the problem says "three sizes of squares". In our calculation, we have three sizes: 1×1, 2×2, and √2. So that's three sizes. Therefore, the answer is n=19. But let me check if there are more sizes.

Wait, could there be another size of square? For example, squares with side length √( (1)^2 + (2)^2 ) = √5? Let me think. In a 3×n grid, can we have such squares?

For example, starting at (0,0), moving 1 right and 2 up to (1,2), then moving left 2 and up 1 to (-1,3), which is outside. Not possible. Alternatively, starting at (0,0), moving 2 right and 1 up to (2,1), then moving left 1 and up 2 to (1,3), which is outside. Not possible. Alternatively, starting at (1,0), moving 2 right and 1 up to (3,1), which is outside if n=3. So unless n is larger, but even then, the square would require points outside. Therefore, maybe not. Therefore, only three sizes.

Alternatively, squares with side length 2√2? Which would be a square that spans from (0,0) to (2,2), which is a 2×2 square, already counted. So no.

Therefore, the total number of squares is indeed 4n - 6. Therefore, solving 4n - 6 = 70 gives n = 19. Therefore, the answer is 19. But let me check again with the Asymptote code.

The Asymptote code shows a 3×3 grid with some squares. If n=3, total squares would be 6. The Asymptote code draws three squares at the bottom, three in the middle, and three at the top? Wait, looking at the Asymptote code:

It draws three small squares at the bottom, then three in the middle, then three at the top. But each of those is probably 1×1 squares. However, in reality, in a 3×3 grid, there are 4 1×1 squares, 1 2×2 square, and 1 tilted square. So total 6 squares. The Asymptote code might be drawing some of them, but maybe not all. Anyway, the math seems to check out.

Therefore, if the total number of squares is 70, then n=19. Therefore, the answer is \boxed{19}.

Wait, but let me check once more. Maybe I missed some squares?

Wait, another type of square: squares that are larger than 2×2? For example, in a 3×n grid, can we have a square of size 2×2? Yes, which we already counted. What about squares that are not aligned and have different sizes?

Wait, for example, in a 3×n grid, can we have squares with side length √( (n - 1)^2 + ... )? Not sure. Wait, for example, if n is large enough, can we have squares that span multiple columns?

Wait, for example, a square that goes from (0,0) to (k, 1) to (k, 2) to (0,1)... but that would be a rectangle, not a square. Wait, unless k is such that the sides are equal. For example, if we have a square with side length √(k² + 1²). For that to be a square, the sides must be equal and the angles must be 90 degrees. So, for example, if we have a square with side vector (a, b) and (-b, a), then the length is √(a² + b²). So, for example, if a=1, b=2, then the square would have side length √5. But to have such a square in the grid, we need that the points exist.

For example, starting at (0,0), moving 1 right and 2 up to (1,2), then moving -2 right and 1 up to (-1,3), which is outside. Not possible. Starting at (0,0), moving 2 right and 1 up to (2,1), then moving -1 right and 2 up to (1,3), which is outside. Not possible. Starting at (1,0), moving 2 right and 1 up to (3,1), which is outside if n=3. So unless n is larger, but even then, the square would require points outside. Therefore, such squares cannot exist in a 3×n grid. Therefore, there are no other square sizes. Therefore, the total number of squares is indeed 4n - 6. Therefore, solving 4n - 6 = 70 gives n=19.

Therefore, I think the answer is 19.

**Final Answer**
\boxed{19}

To determine the value of $ n $ in a $ 3 \times n $ grid of points where there are a total of 70 squares of **three distinct sizes**, we need to analyze the types of squares that can be formed in such a grid.

---

### Step 1: Understanding the Grid

A $ 3 \times n $ grid of points means there are 3 rows and $ n $ columns of points. The total number of squares that can be formed includes:

- **Axis-aligned squares** (sides parallel to the grid axes)
- **Tilted squares** (squares rotated at 45°, with sides not aligned with the grid)

We are told that there are **three sizes** of squares. These are:

1. **1×1 squares** (axis-aligned, side length = 1)
2. **2×2 squares** (axis-aligned, side length = 2)
3. **Squares with side length $ \sqrt{2} $** (tilted, formed by diagonals of 1×1 squares)

---

### Step 2: Counting the Squares

#### 1. **1×1 Squares**

These are the smallest axis-aligned squares. In a $ 3 \times n $ grid, the number of such squares is:

$$
(3 - 1) \times (n - 1) = 2(n - 1)
$$

#### 2. **2×2 Squares**

These are larger axis-aligned squares. The number of such squares is:

$$
(3 - 2) \times (n - 2) = 1 \times (n - 2) = n - 2
$$

#### 3. **Tilted Squares with Side Length $ \sqrt{2} $**

These are squares formed by connecting diagonally adjacent points. Each such square is centered in the middle row (row 1), and spans from column $ x $ to $ x+1 $, with the square's top and bottom points in rows 0 and 2.

To form such a square, we need:

- $ x \geq 1 $ (to have a left neighbor)
- $ x+1 \leq n - 1 $ (to have a right neighbor)

So, the number of such squares is:

$$
(n - 2)
$$

---

### Step 3: Total Number of Squares

Adding up all the square types:

$$
\text{Total Squares} = 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6
$$

We are told that the total number of squares is 70:

$$
4n - 6 = 70
$$

Solving for $ n $:

$$
4n = 76 \Rightarrow n = \frac{76}{4} = 19
$$

---

### Final Answer

$$
\boxed{19}
$$

Solution:
The number of $1\times1$ squares in the diagram is $2(n-1)$, the number of $2\times 2$ squares is $n-2$, and the number of $\sqrt{2} \times \sqrt{2}$ squares is also $n-2$ (see diagram).  Solving \[
2(n-1)+n-2+n-2=70
\] we find $n=\boxed{19}$.

[asy]
unitsize(5mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
dotfactor=4;

int i,j;

for(i=0;i<=10;i=i+1)

for(j=0;j<=2;j=j+1)

{

dot((i,j));

}

draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((3,0)--(5,0)--(5,2)--(3,2)--cycle);
draw((7,1)--(8,2)--(9,1)--(8,0)--cycle); [/asy]


Explanation:
Okay, so I need to find the product of the y-coordinates of all the distinct solutions (x, y) for the two equations y = x² - 8 and y² = -5x + 44. Hmm, let me think about how to approach this.

First, since both equations are given, I can probably substitute one into the other to find the solutions. The first equation gives y in terms of x, so maybe I can plug that expression for y into the second equation. That way, I can solve for x and then find the corresponding y-values. Let me try that.

So, substituting y = x² - 8 into the second equation y² = -5x + 44, we get:

(x² - 8)² = -5x + 44

Okay, expanding the left side. Let me compute (x² - 8)². That would be x⁴ - 16x² + 64, right? Wait, let me check:

(x² - 8)² = (x²)² - 2*(x²)*(8) + 8² = x⁴ - 16x² + 64. Yes, that's correct.

So the equation becomes:

x⁴ - 16x² + 64 = -5x + 44

Now, let me bring all terms to one side to set the equation to zero:

x⁴ - 16x² + 64 + 5x - 44 = 0

Simplify the constants: 64 - 44 = 20, so:

x⁴ - 16x² + 5x + 20 = 0

So now we have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Hmm, solving quartic equations can be tricky. Maybe I can factor this equation? Let me try to factor it.

First, let me check for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (20) divided by factors of the leading coefficient (1). So possible rational roots are ±1, ±2, ±4, ±5, ±10, ±20.

Let me test x = 1:

1⁴ - 16*(1)² + 5*(1) + 20 = 1 - 16 + 5 + 20 = 10 ≠ 0

x = -1:

(-1)^4 - 16*(-1)^2 + 5*(-1) + 20 = 1 - 16 -5 + 20 = 0. Oh! That works. So x = -1 is a root. Therefore, (x + 1) is a factor.

Let me perform polynomial division or use synthetic division to factor out (x + 1) from the quartic.

Using synthetic division with root -1:

Coefficients of the quartic: 1 (x⁴), 0 (x³), -16 (x²), 5 (x), 20 (constant)

Set up synthetic division:

-1 | 1   0   -16   5   20
    |     -1   1   15  -20
    ------------------------
      1  -1   -15  20   0

So the quartic factors into (x + 1)(x³ - x² - 15x + 20). Now, let's try to factor the cubic polynomial x³ - x² - 15x + 20.

Again, let's check for rational roots. Possible roots are ±1, ±2, ±4, ±5, ±10, ±20.

Testing x = 1:

1 - 1 - 15 + 20 = 5 ≠ 0

x = 2:

8 - 4 - 30 + 20 = -6 ≠ 0

x = 4:

64 - 16 - 60 + 20 = 8 ≠ 0

x = 5:

125 - 25 - 75 + 20 = 45 ≠ 0

x = -1:

-1 - 1 + 15 + 20 = 33 ≠ 0

x = -2:

-8 - 4 + 30 + 20 = 38 ≠ 0

x = 10: Probably too big, but let me check:

1000 - 100 - 150 + 20 = 770 ≠ 0

Hmm, maybe I made a mistake? Let me check x = 5 again:

x³ - x² -15x +20 at x=5: 125 - 25 -75 +20 = 125 -25 is 100, 100 -75 is 25, 25 +20 is 45. Yeah, 45. Not zero.

Wait, maybe x = something else? Let me try x = 2 again:

8 - 4 - 30 + 20 = (8 -4) + (-30 +20) = 4 -10 = -6. Not zero.

x = 4: 64 - 16 - 60 + 20 = 64 -16 is 48, 48 -60 is -12, -12 +20 is 8. Not zero.

x = -2: (-8) - 4 - (-30) +20 = -8 -4 +30 +20 = 38. Not zero.

x = 5/ something? Maybe not rational. Maybe I need to factor by grouping?

Looking at x³ - x² -15x +20. Let me try grouping terms:

Group as (x³ - x²) + (-15x +20) = x²(x - 1) -5(3x - 4). Hmm, not helpful.

Alternatively, maybe different grouping? Let me see:

x³ -15x -x² +20 = x(x² -15) - (x² -20). Not helpful.

Alternatively, maybe there's a typo in my previous steps? Let me check the synthetic division again.

Original quartic: x⁴ -16x² +5x +20. Divided by (x +1). Coefficients: 1, 0, -16, 5, 20.

Bring down the 1. Multiply by -1: -1. Add to next term: 0 + (-1) = -1. Multiply by -1: 1. Add to next term: -16 +1 = -15. Multiply by -1: 15. Add to next term: 5 +15=20. Multiply by -1: -20. Add to last term: 20 + (-20)=0. So that's correct. So the cubic is x³ -x² -15x +20. So that's correct.

Hmm. Maybe I need to use the rational root theorem again but perhaps I missed something? Let me check x=  something else. Wait, maybe x=  something like  sqrt? No, since we are looking for rational roots. Maybe there's a mistake in the previous steps?

Wait, let me check the original equations again. Maybe I made a mistake when substituting?

Original equations: y = x² -8 and y² = -5x +44. So substituting y = x² -8 into the second equation gives (x² -8)^2 = -5x +44. Then expanding gives x^4 -16x² +64 = -5x +44. Then moving all terms to left: x^4 -16x² +64 +5x -44 = x^4 -16x² +5x +20. Yes, that's correct. So the quartic is correct.

So the cubic is x³ -x² -15x +20. Maybe I need to try another method. Maybe factor by grouping? Let me try again.

x³ -x² -15x +20. Let me try to split the middle terms. Let me think. Suppose I write it as x³ -x² -15x +20. Maybe split into x³ -x² and -15x +20. Then factor:

x²(x -1) -5(3x -4). Not helpful. Alternatively, maybe different grouping. Let me try:

x³ -15x -x² +20. Then factor x(x² -15) - (x² -20). Still not helpful. Alternatively, maybe x³ -x² -15x +20 = x³ -x² -15x +20. Maybe try to factor as (x^3 - x^2) - (15x -20) = x²(x -1) -5(3x -4). Still not helpful.

Alternatively, maybe use the method of depressed cubic? But that might be complicated. Alternatively, maybe use the cubic formula? But that's too involved. Alternatively, maybe there's another rational root that I missed?

Wait, maybe I made a mistake in possible roots? Let me check again. The possible rational roots are ±1, ±2, ±4, ±5, ±10, ±20. Wait, maybe x=  something else? Wait, maybe x=  something like  5/ something? Wait, but the rational root theorem says that any rational root p/q must have p dividing 20 and q dividing 1, so p/q is integer. So only the ones I checked. So maybe there are no rational roots? Then perhaps I need to factor it another way?

Alternatively, maybe I made a mistake in the previous steps? Let me check again. Wait, maybe I can use the fact that we already have one root x=-1, so maybe there are other real roots? Maybe I can graph the cubic or use some other method?

Alternatively, maybe I can use substitution. Let me let z = x. Then the cubic is z³ - z² -15z +20. Maybe try to find approximate roots? But since this is a math competition problem, there should be rational roots. Maybe I made a mistake in calculation?

Wait, let me check x=  something else. Wait, maybe x=  something like  2. Let me check x= 2 again:

2³ - 2² -15*2 +20 = 8 -4 -30 +20 = (8 -4) + (-30 +20) = 4 -10 = -6. Not zero.

x= 4: 64 - 16 -60 +20 = 8. Not zero.

x= 5: 125 -25 -75 +20 = 45. Not zero.

x= -2: -8 -4 +30 +20 = 38. Not zero.

x= -4: -64 -16 +60 +20 = 0? Let me compute:

(-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = (-64 -16) + (60 +20) = -80 + 80 = 0! Oh! x = -4 is a root!

Wait, why didn't I think of that earlier? Let me check again:

x = -4:

(-4)^3 - (-4)^2 -15*(-4) +20 = -64 - 16 + 60 + 20 = (-64 -16) + (60 +20) = -80 + 80 = 0. Yes! So x = -4 is a root. Therefore, (x + 4) is a factor. So now, let's perform synthetic division on the cubic x³ -x² -15x +20 with root -4.

Set up synthetic division:

Coefficients: 1 (x³), -1 (x²), -15 (x), 20 (constant)

Root: -4

Bring down the 1. Multiply by -4: -4. Add to next term: -1 + (-4) = -5. Multiply by -4: 20. Add to next term: -15 +20 = 5. Multiply by -4: -20. Add to last term: 20 + (-20) = 0. Perfect.

So the cubic factors into (x + 4)(x² -5x +5). Therefore, the original quartic factors into (x +1)(x +4)(x² -5x +5).

So now, the quartic equation factors as (x +1)(x +4)(x² -5x +5) = 0. Therefore, the solutions are x = -1, x = -4, and the roots of x² -5x +5 = 0.

Let me solve x² -5x +5 = 0. Using quadratic formula:

x = [5 ± sqrt(25 - 20)] / 2 = [5 ± sqrt(5)] / 2.

So the four solutions for x are x = -1, x = -4, x = [5 + sqrt(5)] / 2, and x = [5 - sqrt(5)] / 2.

Now, for each x, we can find the corresponding y from the first equation y = x² -8. Then, we need to find the product of all distinct y-coordinates.

But first, let me check if all these solutions are distinct. Since x = -1, x = -4, and the two irrational roots are distinct from each other and from -1 and -4, so all four solutions are distinct. Therefore, there are four distinct solutions.

Now, let me compute the y-coordinates for each x:

1. For x = -1: y = (-1)^2 -8 = 1 -8 = -7.

2. For x = -4: y = (-4)^2 -8 = 16 -8 = 8.

3. For x = [5 + sqrt(5)] / 2: Let me compute y. Let me denote x = [5 + sqrt(5)] / 2. Then y = x² -8. Let me compute x²:

x² = ([5 + sqrt(5)] / 2)^2 = [25 + 10 sqrt(5) + 5] / 4 = [30 + 10 sqrt(5)] / 4 = [15 + 5 sqrt(5)] / 2. Therefore, y = [15 + 5 sqrt(5)] / 2 - 8 = [15 + 5 sqrt(5) - 16] / 2 = [-1 + 5 sqrt(5)] / 2.

Similarly, for x = [5 - sqrt(5)] / 2:

x² = ([5 - sqrt(5)] / 2)^2 = [25 - 10 sqrt(5) + 5] / 4 = [30 - 10 sqrt(5)] / 4 = [15 - 5 sqrt(5)] / 2. Therefore, y = [15 - 5 sqrt(5)] / 2 - 8 = [15 - 5 sqrt(5) - 16] / 2 = [-1 - 5 sqrt(5)] / 2.

So the four y-coordinates are:

-7, 8, [-1 + 5√5]/2, and [-1 - 5√5]/2.

Now, the problem asks for the product of the y-coordinates of all distinct solutions. So we need to compute (-7) * 8 * [(-1 + 5√5)/2] * [(-1 - 5√5)/2].

Let me compute this step by step.

First, multiply the first two terms: (-7) * 8 = -56.

Now, compute the product of the last two terms: [(-1 + 5√5)/2] * [(-1 - 5√5)/2]. This looks like a difference of squares. Let me recall that (a + b)(a - b) = a² - b². Here, a = -1, b = 5√5. Wait, but the terms are (-1 + 5√5) and (-1 - 5√5). So that's [(-1) + 5√5][(-1) - 5√5] = (-1)^2 - (5√5)^2 = 1 - 25*5 = 1 - 125 = -124. Therefore, the product is -124 divided by (2*2) = 4. So the product of the last two terms is -124 / 4 = -31.

Therefore, the total product is (-56) * (-31) = 56 * 31. Let me compute that.

First, 56 * 30 = 1680, and 56 * 1 = 56, so total is 1680 + 56 = 1736.

Wait, but let me check again:

Wait, the product of the last two terms was [(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)^2 - (5√5)^2]/(2*2) = (1 - 125)/4 = (-124)/4 = -31. Yes, that's correct. Then multiplying by -56 gives (-56)*(-31) = 56*31. Let me compute 56*31:

Well, 56*30 = 1680, 56*1 = 56, so 1680 + 56 = 1736. So the product is 1736.

But wait, let me check if there's a mistake here. Because the problem says "the product of the y-coordinates of all the distinct solutions". But let me check if all four y's are distinct. The y's are -7, 8, [-1 + 5√5]/2, and [-1 - 5√5]/2. Are these all distinct?

Well, -7 and 8 are distinct. The other two are irrational numbers. Let me check if they are equal to -7 or 8. Let me approximate sqrt(5) ≈ 2.236. Then 5√5 ≈ 11.18. So [-1 + 11.18]/2 ≈ 10.18/2 ≈ 5.09. Similarly, [-1 - 11.18]/2 ≈ -12.18/2 ≈ -6.09. So these are approximately 5.09 and -6.09, which are different from -7 and 8. So all four y's are distinct. Therefore, the product is indeed 1736.

But wait, let me check again the calculation. Because sometimes when you have multiple steps, it's easy to make an error. Let me verify:

First, the product of the four y's:

(-7) * 8 * [(-1 + 5√5)/2] * [(-1 - 5√5)/2]

First, compute (-7)*8 = -56.

Then compute [(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)^2 - (5√5)^2]/(2*2) = (1 - 125)/4 = (-124)/4 = -31. So then total product is (-56)*(-31) = 56*31.

Now, 56*31. Let me compute 56*31:

Well, 56*30 = 1680, 56*1 = 56, so 1680 + 56 = 1736. So that's correct.

But let me check if there's another way to compute this product, maybe using Vieta's formula? Because sometimes Vieta's can give the product of roots without having to compute each term.

Wait, but Vieta's formula applies to polynomials. Let me think. The original equations are y = x² -8 and y² = -5x +44. But we found the quartic equation in x, which had roots x=-1, x=-4, and the roots of x² -5x +5=0. Then, for each x, we have a y. However, the product of the y's is not directly given by Vieta's on the quartic. However, maybe we can find a relation?

Alternatively, since we have the quartic equation in x, and we have y = x² -8, perhaps we can find a relation between y and x?

Alternatively, maybe consider that for each solution (x, y), y = x² -8, so x² = y + 8. Then, from the second equation, y² = -5x +44. So maybe we can eliminate x?

But since x² = y + 8, then x = ±√(y + 8). But that might complicate things. Alternatively, express x from the second equation?

From y² = -5x +44, we can write x = (44 - y²)/5. Then substitute into x² = y + 8:

[(44 - y²)/5]^2 = y + 8.

Expanding this:

(44 - y²)^2 / 25 = y + 8

Multiply both sides by 25:

(44 - y²)^2 = 25y + 200

Expanding the left side:

(44)^2 - 2*44*y² + y^4 = 25y + 200

Compute 44²: 1936. So:

1936 - 88y² + y^4 = 25y + 200

Bring all terms to left:

y^4 -88y² -25y + 1936 - 200 = 0

Simplify constants: 1936 - 200 = 1736

So equation is y^4 -88y² -25y + 1736 = 0

Therefore, the y-coordinates of the solutions are the roots of this quartic equation. Therefore, the product of the roots of this quartic equation is given by Vieta's formula. For a quartic equation y^4 + a y^3 + b y^2 + c y + d = 0, the product of the roots is d if the leading coefficient is 1. However, in our case, the quartic is y^4 -88y² -25y + 1736 = 0. So written as y^4 + 0y^3 -88y² -25y + 1736 = 0. Therefore, the product of the roots is 1736. However, we need to check if all roots are distinct and if there are any multiplicities. But since we already found four distinct solutions for x, which correspond to four distinct y's, then the product of the y's is indeed 1736.

Therefore, this confirms our previous calculation. So the answer is 1736.

But let me check once more. Because sometimes when you use Vieta's formula, you have to consider the number of roots. However, since we have four distinct solutions, and the quartic equation in y has four roots, which are exactly the y-coordinates of the solutions. Therefore, the product is 1736.

But let me check with the previous calculation. Earlier, we calculated the product as (-7)*8*[(-1 + 5√5)/2]*[(-1 - 5√5)/2] = (-56)*(-31) = 1736. So both methods agree. Therefore, the answer is 1736.

But let me check if there's a possibility that some y-values are repeated? For example, maybe two different x-values give the same y? But in our case, we have four distinct x-values, and since y = x² -8, unless two different x's give the same y. For example, if x1 and x2 are such that x1² = x2², then they would have same y. However, in our case, the x-values are -1, -4, [5 + sqrt(5)]/2, and [5 - sqrt(5)]/2. Let me check if any of these x-values have squares equal.

First, x=-1 and x=-4: their squares are 1 and 16, different. Then, [5 + sqrt(5)]/2 and [5 - sqrt(5)]/2. Let me compute their squares:

Let me denote x1 = [5 + sqrt(5)]/2 and x2 = [5 - sqrt(5)]/2. Then x1² and x2²:

x1² = [25 + 10 sqrt(5) + 5]/4 = [30 + 10 sqrt(5)]/4 = [15 + 5 sqrt(5)]/2.

x2² = [25 - 10 sqrt(5) + 5]/4 = [30 - 10 sqrt(5)]/4 = [15 - 5 sqrt(5)]/2.

These are different from each other and from 1 and 16. Therefore, all four y-values are distinct. Therefore, the product is indeed 1736.

Therefore, the answer is 1736.

But just to be thorough, let me check if there's an alternative way to compute the product. For example, using the quartic equation in x and Vieta's formula. Wait, but Vieta's formula on the quartic in x would give the product of the roots, but we need the product of the y's. However, since y = x² -8, maybe we can relate the product of y's to the product of x's? But that seems more complicated.

Alternatively, since we have four y's, and we found their product via two methods, both leading to 1736, I think that's solid.

Therefore, I think the answer is 1736.

**Final Answer**
\boxed{1736}

To solve the problem, we are given two equations:

1. $ y = x^2 - 8 $
2. $ y^2 = -5x + 44 $

We are asked to find the **product of the $ y $-coordinates** of all **distinct solutions** $(x, y)$ that satisfy both equations.

---

### Step 1: Substitution

We substitute the expression for $ y $ from the first equation into the second:

$$
(x^2 - 8)^2 = -5x + 44
$$

Expanding the left-hand side:

$$
x^4 - 16x^2 + 64 = -5x + 44
$$

Bring all terms to one side:

$$
x^4 - 16x^2 + 5x + 20 = 0
$$

---

### Step 2: Factor the Quartic

We attempt to factor the quartic polynomial:

$$
x^4 - 16x^2 + 5x + 20
$$

Using the Rational Root Theorem, we test possible rational roots and find that $ x = -1 $ is a root. Using synthetic division, we factor out $ (x + 1) $:

$$
(x + 1)(x^3 - x^2 - 15x + 20)
$$

Now, we factor the cubic:

$$
x^3 - x^2 - 15x + 20
$$

Testing $ x = -4 $, we find it is a root. Using synthetic division again, we factor:

$$
(x + 4)(x^2 - 5x + 5)
$$

So the full factorization of the quartic is:

$$
(x + 1)(x + 4)(x^2 - 5x + 5)
$$

---

### Step 3: Find All $ x $-Values

The roots of the quartic are:

- $ x = -1 $
- $ x = -4 $
- $ x = \frac{5 \pm \sqrt{5}}{2} $

These are all **distinct** real roots.

---

### Step 4: Compute Corresponding $ y $-Values

Using the first equation $ y = x^2 - 8 $, we compute the $ y $-coordinates:

1. For $ x = -1 $:
   $$
   y = (-1)^2 - 8 = 1 - 8 = -7
   $$

2. For $ x = -4 $:
   $$
   y = (-4)^2 - 8 = 16 - 8 = 8
   $$

3. For $ x = \frac{5 + \sqrt{5}}{2} $:
   $$
   y = \left( \frac{5 + \sqrt{5}}{2} \right)^2 - 8 = \frac{30 + 10\sqrt{5}}{4} - 8 = \frac{-1 + 5\sqrt{5}}{2}
   $$

4. For $ x = \frac{5 - \sqrt{5}}{2} $:
   $$
   y = \left( \frac{5 - \sqrt{5}}{2} \right)^2 - 8 = \frac{30 - 10\sqrt{5}}{4} - 8 = \frac{-1 - 5\sqrt{5}}{2}
   $$

So the four distinct $ y $-coordinates are:

$$
-7, \quad 8, \quad \frac{-1 + 5\sqrt{5}}{2}, \quad \frac{-1 - 5\sqrt{5}}{2}
$$

---

### Step 5: Compute the Product of the $ y $-Coordinates

We compute the product:

$$
(-7) \cdot 8 \cdot \left( \frac{-1 + 5\sqrt{5}}{2} \right) \cdot \left( \frac{-1 - 5\sqrt{5}}{2} \right)
$$

Group the last two terms:

$$
\left( \frac{-1 + 5\sqrt{5}}{2} \right) \cdot \left( \frac{-1 - 5\sqrt{5}}{2} \right) = \frac{(-1)^2 - (5\sqrt{5})^2}{4} = \frac{1 - 125}{4} = \frac{-124}{4} = -31
$$

Now multiply all terms:

$$
(-7) \cdot 8 \cdot (-31) = (-56) \cdot (-31) = 1736
$$

---

### Final Answer

$$
\boxed{1736}
$$

[Assistant]
Explanation:
Okay, so I need to find the product of the y-coordinates of all the distinct solutions (x, y) for the two equations y = x² - 8 and y² = -5x + 44. Hmm, let me think about how to approach this.

First, since both equations are given, I can probably substitute one into the other to find the solutions. The first equation gives y in terms of x, so maybe I can plug that expression for y into the second equation. That way, I can solve for x and then find the corresponding y-values. Let me try that.

So, substituting y = x² - 8 into the second equation y² = -5x + 44, we get:

(x² - 8)² = -5x + 44

Okay, expanding the left side. Let me compute (x² - 8)². That would be x⁴ - 16x² + 64, right? Wait, let me check:

(x² - 8)² = (x²)² - 2*(x²)*(8) + 8² = x⁴ - 16x² + 64. Yes, that's correct.

So the equation becomes:

x⁴ - 16x² + 64 = -5x + 44

Now, let me bring all terms to one side to set the equation to zero:

x⁴ - 16x² + 64 + 5x - 44 = 0

Simplify the constants: 64 - 44 = 20, so:

x⁴ - 16x² + 5x + 20 = 0

So now we have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Hmm, solving quartic equations can be tricky. Maybe I can factor this equation? Let me try to factor it.

First, let me check for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (20) divided by factors of the leading coefficient (1). So possible rational roots are ±1, ±2, ±4, ±5, ±10, ±20.

Let me test x = 1:

1⁴ - 16*(1)² + 5*(1) + 20 = 1 - 16 + 5 + 20 = 10 ≠ 0

x = -1:

(-1)^4 - 16*(-1)^2 + 5*(-1) + 20 = 1 - 16 -5 + 20 = 0. Oh! That works. So x = -1 is a root. Therefore, (x + 1) is a factor.

Let me perform polynomial division or use synthetic division to factor out (x + 1) from the quartic.

Using synthetic division with root -1:

Coefficients of the quartic: 1 (x⁴), 0 (x³), -16 (x²), 5 (x), 20 (constant)

Set up synthetic division:

-1 | 1   0   -16   5   20
    |     -1   1   15  -20
    ------------------------
      1  -1   -15  20   0

So the quartic factors into (x + 1)(x³ - x² - 15x + 20). Now, let's try to factor the cubic polynomial x³ - x² - 15x + 20.

Again, let's check for rational roots. Possible roots are ±1, ±2, ±4, ±5, ±10, ±20.

Testing x = 1:

1 - 1 - 15 + 20 = 5 ≠ 0

x = 2:

8 - 4 - 30 + 20 = -6 ≠ 0

x = 4:

64 - 16 - 60 + 20 = 8 ≠ 0

x = 5:

125 - 25 - 75 + 20 = 45 ≠ 0

x = -1:

-1 - 1 + 15 + 20 = 33 ≠ 0

x = -2:

-8 - 4 + 30 + 20 = 38 ≠ 0

x = 10: Probably too big, but let me check:

1000 - 100 - 150 + 20 = 770 ≠ 0

Hmm, maybe I made a mistake? Let me check x = 5 again:

x³ - x² -15x +20 at x=5: 125 - 25 -75 +20 = 125 -25 is 100, 100 -75 is 25, 25 +20 is 45. Yeah, 45. Not zero.

Wait, maybe x = something else? Let me try x = 2 again:

8 - 4 - 30 + 20 = (8 -4) + (-30 +20) = 4 -10 = -6. Not zero.

x = 4: 64 - 16 - 60 + 20 = 64 -16 is 48, 48 -60 is -12, -12 +20 is 8. Not zero.

x = -2: (-8) - 4 - (-30) +20 = -8 -4 +30 +20 = 38. Not zero.

x = 5/ something? Maybe not rational. Maybe I need to factor by grouping?

Looking at x³ - x² -15x +20. Let me try grouping terms:

Group as (x³ - x²) + (-15x +20) = x²(x - 1) -5(3x - 4). Hmm, not helpful.

Alternatively, maybe different grouping? Let me see:

x³ -15x -x² +20 = x(x² -15) - (x² -20). Not helpful.

Alternatively, maybe there's a typo in my previous steps? Let me check the synthetic division again.

Original quartic: x⁴ -16x² +5x +20. Divided by (x +1). Coefficients: 1, 0, -16, 5, 20.

Bring down the 1. Multiply by -1: -1. Add to next term: 0 + (-1) = -1. Multiply by -1: 1. Add to next term: -16 +1 = -15. Multiply by -1: 15. Add to next term: 5 +15=20. Multiply by -1: -20. Add to last term: 20 + (-20)=0. So that's correct. So the cubic is x³ -x² -15x +20. So that's correct.

Hmm. Maybe I need to use the rational root theorem again but perhaps I missed something? Let me check x=  something else. Wait, maybe x=  something like  sqrt? No, since we are looking for rational roots. Maybe there's a mistake in the previous steps?

Wait, let me check the original equations again. Maybe I made a mistake when substituting?

Original equations: y = x² -8 and y² = -5x +44. So substituting y = x² -8 into the second equation gives (x² -8)^2 = -5x +44. Then expanding gives x^4 -16x² +64 = -5x +44. Then moving all terms to left: x^4 -16x² +64 +5x -44 = x^4 -16x² +5x +20. Yes, that's correct. So the quartic is correct.

So the cubic is x³ -x² -15x +20. Maybe I need to try another method. Maybe factor by grouping? Let me try again.

x³ -x² -15x +20. Let me try to split the middle terms. Let me think. Suppose I write it as x³ -x² -15x +20. Maybe split into x³ -x² and -15x +20. Then factor:

x²(x -1) -5(3x -4). Not helpful. Alternatively, maybe different grouping. Let me try:

x³ -15x -x² +20. Then factor x(x² -15) - (x² -20). Still not helpful. Alternatively, maybe x³ -x² -15x +20 = x³ -x² -15x +20. Maybe try to factor as (x^3 - x^2) - (15x -20) = x²(x -1) -5(3x -4). Still not helpful.

Alternatively, maybe use the method of depressed cubic? But that might be complicated. Alternatively, maybe use the cubic formula? But that's too involved. Alternatively, maybe there's another rational root that I missed?

Wait, maybe I made a mistake in possible roots? Let me check again. The possible rational roots are ±1, ±2, ±4, ±5, ±10, ±20. Wait, maybe x=  something else? Wait, maybe x=  something like  5/ something? Wait, but the rational root theorem says that any rational root p/q must have p dividing 20 and q dividing 1, so p/q is integer. So only the ones I checked. So maybe there are no rational roots? Then perhaps I need to factor it another way?

Alternatively, maybe I made a mistake in the previous steps? Let me check again. Wait, maybe I can use the fact that we already have one root x=-1, so maybe there are other real roots? Maybe I can graph the cubic or use some other method?

Alternatively, maybe I can use substitution. Let me let z = x. Then the cubic is z³ - z² -15z +20. Maybe try to find approximate roots? But since this is a math competition problem, there should be rational roots. Maybe I made a mistake in calculation?

Wait, let me check x=  something else. Wait, maybe x=  something like  2. Let me check x= 2 again:

2³ - 2² -15*2 +20 = 8 -4 -30 +20 = (8 -4) + (-30 +20) = 4 -10 = -6. Not zero.

x= 4: 64 - 16 -60 +20 = 8. Not zero.

x= 5: 125 -25 -75 +20 = 45. Not zero.

x= -2: -8 -4 +30 +20 = 38. Not zero.

x= -4: -64 -16 +60 +20 = 0? Let me compute:

(-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = (-64 -16) + (60 +20) = -80 + 80 = 0! Oh! x = -4 is a root!

Wait, why didn't I think of that earlier? Let me check again:

x = -4:

(-4)^3 - (-4)^2 -15*(-4) +20 = -64 - 16 + 60 + 20 = (-64 -16) + (60 +20) = -80 + 80 = 0. Yes! So x = -4 is a root. Therefore, (x + 4) is a factor. So now, let's perform synthetic division on the cubic x³ -x² -15x +20 with root -4.

Set up synthetic division:

Coefficients: 1 (x³), -1 (x²), -15 (x), 20 (constant)

Root: -4

Bring down the 1. Multiply by -4: -4. Add to next term: -1 + (-4) = -5. Multiply by -4: 20. Add to next term: -15 +20 = 5. Multiply by -4: -20. Add to last term: 20 + (-20) = 0. Perfect.

So the cubic factors into (x + 4)(x² -5x +5). Therefore, the original quartic factors into (x +1)(x +4)(x² -5x +5).

So now, the quartic equation factors as (x +1)(x +4)(x² -5x +5) = 0. Therefore, the solutions are x = -1, x = -4, and the roots of x² -5x +5 = 0.

Let me solve x² -5x +5 = 0. Using quadratic formula:

x = [5 ± sqrt(25 - 20)] / 2 = [5 ± sqrt(5)] / 2.

So the four solutions for x are x = -1, x = -4, x = [5 + sqrt(5)] / 2, and x = [5 - sqrt(5)] / 2.

Now, for each x, we can find the corresponding y from the first equation y = x² -8. Then, we need to find the product of all distinct y-coordinates.

But first, let me check if all these solutions are distinct. Since x = -1, x = -4, and the two irrational roots are distinct from each other and from -1 and -4, so all four solutions are distinct. Therefore, there are four distinct solutions.

Now, let me compute the y-coordinates for each x:

1. For x = -1: y = (-1)^2 -8 = 1 -8 = -7.

2. For x = -4: y = (-4)^2 -8 = 16 -8 = 8.

3. For x = [5 + sqrt(5)] / 2: Let me compute y. Let me denote x = [5 + sqrt(5)] / 2. Then y = x² -8. Let me compute x²:

x² = ([5 + sqrt(5)] / 2)^2 = [25 + 10 sqrt(5) + 5] / 4 = [30 + 10 sqrt(5)] / 4 = [15 + 5 sqrt(5)] / 2. Therefore, y = [15 + 5 sqrt(5)] / 2 - 8 = [15 + 5 sqrt(5) - 16] / 2 = [-1 + 5 sqrt(5)] / 2.

Similarly, for x = [5 - sqrt(5)] / 2:

x² = ([5 - sqrt(5)] / 2)^2 = [25 - 10 sqrt(5) + 5] / 4 = [30 - 10 sqrt(5)] / 4 = [15 - 5 sqrt(5)] / 2. Therefore, y = [15 - 5 sqrt(5)] / 2 - 8 = [15 - 5 sqrt(5) - 16] / 2 = [-1 - 5 sqrt(5)] / 2.

So the four y-coordinates are:

-7, 8, [-1 + 5√5]/2, and [-1 - 5√5]/2.

Now, the problem asks for the product of the y-coordinates of all distinct solutions. So we need to compute (-7) * 8 * [(-1 + 5√5)/2] * [(-1 - 5√5)/2].

Let me compute this step by step.

First, multiply the first two terms: (-7) * 8 = -56.

Now, compute the product of the last two terms: [(-1 + 5√5)/2] * [(-1 - 5√5)/2]. This looks like a difference of squares. Let me recall that (a + b)(a - b) = a² - b². Here, a = -1, b = 5√5. Wait, but the terms are (-1 + 5√5) and (-1 - 5√5). So that's [(-1) + 5√5][(-1) - 5√5] = (-1)^2 - (5√5)^2 = 1 - 25*5 = 1 - 125 = -124. Therefore, the product is -124 divided by (2*2) = 4. So the product of the last two terms is -124 / 4 = -31.

Therefore, the total product is (-56) * (-31) = 56 * 31. Let me compute that.

First, 56 * 30 = 1680, and 56 * 1 = 56, so total is 1680 + 56 = 1736.

Wait, but let me check again:

Wait, the product of the last two terms was [(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)^2 - (5√5)^2]/(2*2) = (1 - 125)/4 = (-124)/4 = -31. Yes, that's correct. Then multiplying by -56 gives (-56)*(-31) = 56*31. Let me compute 56*31:

Well, 56*30 = 1680, 56*1 = 56, so 1680 + 56 = 1736. So the product is 1736.

But wait, let me check if there's a mistake here. Because the problem says "the product of the y-coordinates of all the distinct solutions". But let me check if all four y's are distinct. The y's are -7, 8, [-1 + 5√5]/2, and [-1 - 5√5]/2. Are these all distinct?

Well, -7 and 8 are distinct. The other two are irrational numbers. Let me check if they are equal to -7 or 8. Let me approximate sqrt(5) ≈ 2.236. Then 5√5 ≈ 11.18. So [-1 + 11.18]/2 ≈ 10.18/2 ≈ 5.09. Similarly, [-1 - 11.18]/2 ≈ -12.18/2 ≈ -6.09. So these are approximately 5.09 and -6.09, which are different from -7 and 8. So all four y's are distinct. Therefore, the product is indeed 1736.

But wait, let me check again the calculation. Because sometimes when you have multiple steps, it's easy to make an error. Let me verify:

First, the product of the four y's:

(-7) * 8 * [(-1 + 5√5)/2] * [(-1 - 5√5)/2]

First, compute (-7)*8 = -56.

Then compute [(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)^2 - (5√5)^2]/(2*2) = (1 - 125)/4 = (-124)/4 = -31. So then total product is (-56)*(-31) = 56*31.

Now, 56*31. Let me compute 56*31:

Well, 56*30 = 1680, 56*1 = 56, so 1680 + 56 = 1736. So that's correct.

But let me check if there's another way to compute this product, maybe using Vieta's formula? Because sometimes Vieta's can give the product of roots without having to compute each term.

Wait, but Vieta's formula applies to polynomials. Let me think. The original equations are y = x² -8 and y² = -5x +44. But we found the quartic equation in x, which had roots x=-1, x=-4, and the roots of x² -5x +5=0. Then, for each x, we have a y. However, the product of the y's is not directly given by Vieta's on the quartic. However, maybe we can find a relation?

Alternatively, since we have the quartic equation in x, and we have y = x² -8, perhaps we can find a relation between y and x?

Alternatively, maybe consider that for each solution (x, y), y = x² -8, so x² = y + 8. Then, from the second equation, y² = -5x +44. So maybe we can eliminate x?

But since x² = y + 8, then x = ±√(y + 8). But that might complicate things. Alternatively, express x from the second equation?

From y² = -5x +44, we can write x = (44 - y²)/5. Then substitute into x² = y + 8:

[(44 - y²)/5]^2 = y + 8.

Expanding this:

(44 - y²)^2 / 25 = y + 8

Multiply both sides by 25:

(44 - y²)^2 = 25y + 200

Expanding the left side:

(44)^2 - 2*44*y² + y^4 = 25y + 200

Compute 44²: 1936. So:

1936 - 88y² + y^4 = 25y + 200

Bring all terms to left:

y^4 -88y² -25y + 1936 - 200 = 0

Simplify constants: 1936 - 200 = 1736

So equation is y^4 -88y² -25y + 1736 = 0

Therefore, the y-coordinates of the solutions are the roots of this quartic equation. Therefore, the product of the roots of this quartic equation is given by Vieta's formula. For a quartic equation y^4 + a y^3 + b y^2 + c y + d = 0, the product of the roots is d if the leading coefficient is 1. However, in our case, the quartic is y^4 -88y² -25y + 1736 = 0. So written as y^4 + 0y^3 -88y² -25y + 1736 = 0. Therefore, the product of the roots is 1736. However, we need to check if all roots are distinct and if there are any multiplicities. But since we already found four distinct solutions for x, which correspond to four distinct y's, then the product of the y's is indeed 1736.

Therefore, this confirms our previous calculation. So the answer is 1736.

But let me check once more. Because sometimes when you use Vieta's formula, you have to consider the number of roots. However, since we have four distinct solutions, and the quartic equation in y has four roots, which are exactly the y-coordinates of the solutions. Therefore, the product is 1736.

But let me check with the previous calculation. Earlier, we calculated the product as (-7)*8*[(-1 + 5√5)/2]*[(-1 - 5√5)/2] = (-56)*(-31) = 1736. So both methods agree. Therefore, the answer is 1736.

But let me check if there's a possibility that some y-values are repeated? For example, maybe two different x-values give the same y? But in our case, we have four distinct x-values, and since y = x² -8, unless two different x's give the same y. For example, if x1 and x2 are such that x1² = x2², then they would have same y. However, in our case, the x-values are -1, -4, [5 + sqrt(5)]/2, and [5 - sqrt(5)]/2. Let me check if any of these x-values have squares equal.

First, x=-1 and x=-4: their squares are 1 and 16, different. Then, [5 + sqrt(5)]/2 and [5 - sqrt(5)]/2. Let me compute their squares:

Let me denote x1 = [5 + sqrt(5)]/2 and x2 = [5 - sqrt(5)]/2. Then x1² and x2²:

x1² = [25 + 10 sqrt(5) + 5]/4 = [30 + 10 sqrt(5)]/4 = [15 + 5 sqrt(5)]/2.

x2² = [25 - 10 sqrt(5) + 5]/4 = [30 - 10 sqrt(5)]/4 = [15 - 5 sqrt(5)]/2.

These are different from each other and from 1 and 16. Therefore, all four y-values are distinct. Therefore, the product is indeed 1736.

Therefore, the answer is 1736.

But just to be thorough, let me check if there's an alternative way to compute the product. For example, using the quartic equation in x and Vieta's formula. Wait, but Vieta's formula on the quartic in x would give the product of the roots, but we need the product of the y's. However, since y = x² -8, maybe we can relate the product of y's to the product of x's? But that seems more complicated.

Alternatively, since we have four y's, and we found their product via two methods, both leading to 1736, I think that's solid.

Therefore, I think the answer is 1736.

**Final Answer**
\boxed{1736}

To solve the problem, we are given two equations:

1. $ y = x^2 - 8 $
2. $ y^2 = -5x + 44 $

We are asked to find the **product of the $ y $-coordinates** of all **distinct solutions** $(x, y)$ that satisfy both equations.

---

### Step 1: Substitution

We substitute the expression for $ y $ from the first equation into the second:

$$
(x^2 - 8)^2 = -5x + 44
$$

Expanding the left-hand side:

$$
x^4 - 16x^2 + 64 = -5x + 44
$$

Bring all terms to one side:

$$
x^4 - 16x^2 + 5x + 20 = 0
$$

---

### Step 2: Factor the Quartic

We attempt to factor the quartic polynomial:

$$
x^4 - 16x^2 + 5x + 20
$$

Using the Rational Root Theorem, we test possible rational roots and find that $ x = -1 $ is a root. Using synthetic division, we factor out $ (x + 1) $:

$$
(x + 1)(x^3 - x^2 - 15x + 20)
$$

Now, we factor the cubic:

$$
x^3 - x^2 - 15x + 20
$$

Testing $ x = -4 $, we find it is a root. Using synthetic division again, we factor:

$$
(x + 4)(x^2 - 5x + 5)
$$

So the full factorization of the quartic is:

$$
(x + 1)(x + 4)(x^2 - 5x + 5)
$$

---

### Step 3: Find All $ x $-Values

The roots of the quartic are:

- $ x = -1 $
- $ x = -4 $
- $ x = \frac{5 \pm \sqrt{5}}{2} $

These are all **distinct** real roots.

---

### Step 4: Compute Corresponding $ y $-Values

Using the first equation $ y = x^2 - 8 $, we compute the $ y $-coordinates:

1. For $ x = -1 $:
   $$
   y = (-1)^2 - 8 = 1 - 8 = -7
   $$

2. For $ x = -4 $:
   $$
   y = (-4)^2 - 8 = 16 - 8 = 8
   $$

3. For $ x = \frac{5 + \sqrt{5}}{2} $:
   $$
   y = \left( \frac{5 + \sqrt{5}}{2} \right)^2 - 8 = \frac{30 + 10\sqrt{5}}{4} - 8 = \frac{-1 + 5\sqrt{5}}{2}
   $$

4. For $ x = \frac{5 - \sqrt{5}}{2} $:
   $$
   y = \left( \frac{5 - \sqrt{5}}{2} \right)^2 - 8 = \frac{30 - 10\sqrt{5}}{4} - 8 = \frac{-1 - 5\sqrt{5}}{2}
   $$

So the four distinct $ y $-coordinates are:

$$
-7, \quad 8, \quad \frac{-1 + 5\sqrt{5}}{2}, \quad \frac{-1 - 5\sqrt{5}}{2}
$$

---

### Step 5: Compute the Product of the $ y $-Coordinates

We compute the product:

$$
(-7) \cdot 8 \cdot \left( \frac{-1 + 5\sqrt{5}}{2} \right) \cdot \left( \frac{-1 - 5\sqrt{5}}{2} \right)
$$

Group the last two terms:

$$
\left( \frac{-1 + 5\sqrt{5}}{2} \right) \cdot \left( \frac{-1 - 5\sqrt{5}}{2} \right) = \frac{(-1)^2 - (5\sqrt{5})^2}{4} = \frac{1 - 125}{4} = \frac{-124}{4} = -31
$$

Now multiply all terms:

$$
(-7) \cdot 8 \cdot (-31) = (-56) \cdot (-31) = 1736
$$

---

### Final Answer

$$
\boxed{1736}
$$

Solution:
Squaring $y=x^2-8$, we obtain $y^2=x^4-16x^2+64$. Setting the right-hand sides equal to each other, we find \begin{align*}
-5x+44&=x^4-16x^2+64\quad\Rightarrow\\
0&=x^4-16x^2+5x+20\quad\Rightarrow\\
&=x^2(x^2-16)+5(x+4)\quad\Rightarrow\\
&=x^2(x-4)(x+4)+5(x+4)\quad\Rightarrow\\
&=(x+4)(x^3-4x^2+5).
\end{align*} Therefore, one of the solutions has an $x$-value of $-4$. Then there is the polynomial $x^3-4x^2+5$. The only possible rational roots are now $\pm1$ and $\pm5$. Using synthetic or long division, it can be determined that $(x+1)$ is a factor: \[(x+1)(x^2-5x+5)=x^3-4x^2+5\] Therefore, one of the solutions has an $x$-value of $-1$. Because $x^2-5x+5$ does not factor easily, we use the quadratic formula to get \begin{align*}
x&=\frac{5\pm\sqrt{25-4\cdot1\cdot5}}{2}\quad\Rightarrow\\
&=\frac{5\pm\sqrt{5}}{2}.
\end{align*} The four values for $x$ are then $-4, -1, \frac{5\pm\sqrt{5}}{2}$. Squaring each: \[(-4)^2=16\] \[(-1)^2=1\] \[\left(\frac{5+\sqrt{5}}{2}\right)^2=\frac{25+10\sqrt{5}+5}{4}=\frac{15+5\sqrt{5}}{2}\] \[\left(\frac{5-\sqrt{5}}{2}\right)^2=\frac{25-10\sqrt{5}+5}{4}=\frac{15-5\sqrt{5}}{2}\] And subtracting $8$: \[16-8=8\] \[1-8=-7\] \[\frac{15+5\sqrt{5}}{2}-\frac{16}{2}=\frac{-1+5\sqrt{5}}{2}\] \[\frac{15-5\sqrt{5}}{2}-\frac{16}{2}=\frac{-1-5\sqrt{5}}{2}\] Therefore, the four solutions are $$(-4,8),(-1,-7),$$ $$\left(\frac{5+\sqrt{5}}{2},\frac{-1+5\sqrt{5}}{2}\right),\left(\frac{5-\sqrt{5}}{2},\frac{-1-5\sqrt{5}}{2}\right).$$

Multiplying the $y$-coordinates: \[8\cdot-7\cdot\frac{-1+5\sqrt{5}}{2}\cdot\frac{-1-5\sqrt{5}}{2}=\frac{-56(1-25\cdot5)}{4}=\boxed{1736}.\]