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 grid with some dots and some filled squares, but maybe it's just an example? Let me try to visualize it. The grid is 3 rows by n columns? Wait, the Asymptote code shows a 3x3 grid? Wait, the first part has points from (0,0) to (2,2), so that's 3x3 grid points? But then there are more points at (4,0), (5,0), etc. Maybe the Asymptote code is just an example for n=3? But the problem says it's a 3×n grid. So maybe the Asymptote code is just illustrating the grid with n=3? But the problem says there are squares of three sizes. Let me think.

First, let me recall that in a grid of points, the number of squares can be calculated by considering squares of different sizes. For a grid with m rows and n columns of points, the number of squares is the sum over all possible square sizes. For example, in a grid with m rows and n columns, the number of 1x1 squares is (m-1)(n-1), the number of 2x2 squares is (m-2)(n-2), and so on, up to the maximum square size. However, in this problem, it's mentioned that there are squares of three sizes. So maybe the grid is such that there are three different square sizes possible? Or maybe the problem is referring to squares of three different sizes, not necessarily all sizes?

Wait, the problem says "three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are three different sizes of squares that can be formed on the grid? Let me think. For example, in a grid with 3 rows and n columns, the possible square sizes would depend on the spacing between the points. Since it's a grid of points, the distance between adjacent points is 1 unit. So squares can be of different sizes based on their side lengths. For example, in a grid with 3 rows and n columns, the possible square sizes would be:

- 1x1 squares: these have sides of length 1 unit. The number of such squares would be (number of positions horizontally) × (number of positions vertically). Since there are 3 rows, vertically, the number of positions for vertical sides would be (3 - 1) = 2. Horizontally, for n columns, the number of positions would be (n - 1). So total 1x1 squares would be 2*(n - 1).

Wait, but actually, the number of squares of size k x k in a grid with m rows and n columns is (m - k) * (n - k). Wait, but that's if the grid is considered as a grid of squares. Wait, maybe I need to clarify.

Wait, if the grid is of points, then the number of squares of size 1x1 (i.e., squares with side length 1 unit) would be (number of horizontal positions) × (number of vertical positions). For a grid with 3 rows and n columns of points, the number of horizontal positions for a square of side length 1 is (n - 1), since each square spans 1 unit horizontally. Similarly, vertically, since there are 3 rows, the number of vertical positions is (3 - 1) = 2. Therefore, the number of 1x1 squares is 2*(n - 1).

Similarly, for squares of size 2x2 (side length 2 units), the number would be (number of horizontal positions) × (number of vertical positions). The horizontal positions would be (n - 2), since each square spans 2 units horizontally. The vertical positions would be (3 - 2) = 1. Therefore, the number of 2x2 squares is 1*(n - 2).

For squares of size 3x3 (side length 3 units), the horizontal positions would be (n - 3), and vertical positions would be (3 - 3) = 0. Therefore, there are 0 such squares. So in this case, there are only two sizes of squares: 1x1 and 2x2. But the problem says there are three sizes. Hmm. So maybe there are squares of different orientations? Like squares that are not aligned with the grid? For example, squares that are rotated?

Wait, the problem says "squares of three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are squares of different sizes, including both axis-aligned and rotated squares? For example, in a grid, you can have squares that are tilted, like with side length sqrt(2), which would be squares formed by diagonally adjacent points. But in a 3xN grid, how many such squares are there?

Let me think. For example, in a grid with 3 rows and n columns, the possible squares can be:

1. Axis-aligned squares: sizes 1x1, 2x2.

2. Squares with side length sqrt(2): these are squares that are rotated 45 degrees, with their sides along the diagonals of the 1x1 squares. For such squares, the side length is sqrt(2), and they require a 2x2 block of points. For example, the square with vertices at (0,0), (1,1), (0,2), (-1,1) – but wait, that would go outside the grid. So in a 3xN grid, how many such squares are there?

Alternatively, maybe squares that are formed by points spaced two units apart? Wait, maybe I need to think more systematically.

First, let me recall that in a grid, the number of squares can be calculated by considering all possible squares, both axis-aligned and rotated. However, the problem mentions "three sizes", so maybe there are three different sizes of squares, regardless of orientation. Let me check.

In a 3xN grid of points, the possible squares can be:

- Axis-aligned squares:

   - 1x1: side length 1

   - 2x2: side length 2

- Rotated squares:

   - Squares with side length sqrt(2): these are squares formed by moving 1 unit right and 1 unit up, then 1 unit left and 1 unit up, etc. For example, a square with vertices at (0,0), (1,1), (0,2), (-1,1). But since the grid is only 3 rows, the square would need to fit within the grid. So for such squares, the vertical span is 2 units (from row 0 to row 2), and horizontally, they need to have at least 2 columns. Wait, for example, in a 3xN grid, the square with side length sqrt(2) would require that there are two columns and two rows? Wait, maybe not. Let me think.

Actually, a square with side length sqrt(2) would have its vertices at points that are diagonally adjacent. For example, in a grid, if you have points (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). But this square would require that x-1 is within the grid. So if the grid has columns from 0 to n-1, then x must be at least 1. Similarly, vertically, since there are 3 rows, y can be 0 or 1 (since y+2 would be 2, which is the top row). Therefore, the number of such squares would be (n - 2) * 1. Wait, because horizontally, you need two columns to have the square, but actually, the square spans from column x to x+1, but the square itself is formed by points (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). Wait, so for each such square, you need columns x-1, x, x+1? Wait, no. Let me think again.

Wait, if you have a square with vertices at (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). So to have this square, you need that x-1 ≥ 0 (so x ≥ 1) and y+2 ≤ 2 (since there are 3 rows, y can be 0 or 1). Therefore, for y, there are 2 possibilities (y=0 and y=1). For x, since x must be at least 1, and the square spans from x-1 to x+1 horizontally, so the number of such squares horizontally would be n - 2 (since x can be from 1 to n-2). Therefore, total number of such squares is 2*(n - 2). Wait, but each square is counted once? Or is there another way?

Alternatively, maybe there are squares of side length sqrt(2) that are formed in different positions. Wait, but maybe I need to check for other sizes. For example, squares with side length sqrt(5)? Wait, that would be squares that are formed by moving 1 unit in one direction and 2 units in another. For example, a square with vertices at (0,0), (1,2), (3,1), (2,-1). But again, this would go outside the grid. So maybe in a 3xN grid, there are no such squares? Or maybe only some?

Alternatively, maybe there are squares of side length sqrt(5) if the grid is large enough. But since the grid has only 3 rows, vertically, the maximum distance between two points is 2 units. So if you have a square with side length sqrt(5), that would require moving 1 unit horizontally and 2 units vertically, but then the next side would have to be 2 units horizontally and 1 unit vertically, but that would require more rows or columns. Since there are only 3 rows, maybe such squares can't exist? Let me check.

Suppose we have a square with side length sqrt(5). Let's say starting at (x,y), moving right 1 and up 2 to (x+1, y+2). Then from there, moving left 2 and up 1? Wait, but that would be (-2, y+3), which is outside the grid. Alternatively, moving left 2 and down 1? That would be (x+1 - 2, y+2 - 1) = (x -1, y +1). Then from there, moving right 1 and down 2? That would be (x -1 +1, y +1 - 2) = (x, y -1). But that would go below the grid. So unless y -1 is within the grid, which would require y ≥ 1. So if y=1, then y -1 = 0. So starting at (x,1), moving to (x+1, 3), but there are only 3 rows, so y+2 = 3 would be outside. Wait, rows are from 0 to 2, so y+2 would be 2 + 2 = 4? Wait, no. Wait, if the grid has 3 rows, then the y-coordinates are 0, 1, 2. So moving up 2 units from y=0 would reach y=2, which is the top row. So starting at (x,0), moving right 1 and up 2 would reach (x+1, 2). Then from there, moving left 2 and down 1 would reach (x+1 - 2, 2 - 1) = (x -1, 1). Then from there, moving right 1 and down 2 would reach (x -1 +1, 1 - 2) = (x, -1), which is outside. So that doesn't work. Alternatively, maybe another direction? Maybe from (x,0), moving right 2 and up 1 to (x+2, 1). Then from there, moving left 1 and up 2 to (x+2 -1, 1 + 2) = (x+1, 3), which is outside. So again, no good. Therefore, maybe there are no squares of side length sqrt(5) in a 3xN grid? So maybe only squares of side length 1, sqrt(2), and 2?

Wait, but earlier we thought there are only two sizes of axis-aligned squares (1x1 and 2x2). Then, if we include the rotated squares of side length sqrt(2), that would be a third size. So total three sizes. Therefore, the problem says there are squares of three sizes, which would be 1x1, 2x2, and sqrt(2)x sqrt(2). Therefore, the total number of squares would be the sum of the number of each size.

So let me calculate that.

First, axis-aligned squares:

- 1x1: as before, number is (number of horizontal positions) × (number of vertical positions). For 1x1 squares, horizontally, there are (n - 1) positions (since each square spans 1 unit horizontally, and there are n columns, so n - 1 positions). Vertically, since there are 3 rows, the number of vertical positions is (3 - 1) = 2. Therefore, total 1x1 squares: 2*(n - 1).

- 2x2: horizontally, positions are (n - 2), vertically, positions are (3 - 2) = 1. So total 2x2 squares: 1*(n - 2).

Then, rotated squares of side length sqrt(2). How many are there?

As discussed earlier, these squares are formed by points (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). For these squares to exist, we need:

- x - 1 ≥ 0 ⇒ x ≥ 1

- y + 2 ≤ 2 ⇒ y ≤ 0. Since y can be 0 or 1, but y + 2 ≤ 2 ⇒ y ≤ 0. Therefore, y must be 0. So only y=0.

Therefore, for each x from 1 to n - 2 (since x+1 must be ≤ n - 1, so x ≤ n - 2), we have such squares. Therefore, the number of these squares is (n - 2) * 1 = n - 2.

Wait, but is that all? Are there other squares of side length sqrt(2)? For example, squares that are placed differently? For example, maybe squares that are placed with their sides going in different directions? Wait, for example, starting at (x, y), going right 1, down 1, etc. But since the grid has only 3 rows, going down from y=0 would go to y=-1, which is outside. Similarly, going up from y=2 would go to y=3, which is outside. So maybe only the squares that are placed with their top side at y=2? Wait, but we considered y=0. Wait, maybe there are squares that are placed with their bottom side at y=0? Wait, no, the square with vertices at (x,0), (x+1,1), (x,2), (x-1,1). That's the same as before, with y=0. So that's the only possibility. Therefore, the number of such squares is (n - 2). So total rotated squares of side length sqrt(2) is n - 2.

Wait, but maybe there are squares of side length sqrt(2) in other orientations? For example, squares that are placed horizontally? Wait, no, they are rotated. Wait, maybe squares that are placed with their sides not aligned with the grid? But we already considered those. So maybe that's all.

Therefore, total number of squares is:

- 1x1: 2(n - 1)

- 2x2: 1(n - 2)

- sqrt(2)x sqrt(2): 1(n - 2)

Wait, so total squares would be 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6.

But the problem says there are 70 squares. Therefore, 4n - 6 = 70 ⇒ 4n = 76 ⇒ n = 19. But wait, let me check if this is correct.

Wait, but hold on. Is that all the squares? Are there any other squares? For example, squares of side length 2*sqrt(2)? That would be squares formed by moving 2 units in one direction and 2 units in another. But in a 3xN grid, can such squares exist?

For example, a square with vertices at (x, y), (x+2, y+2), (x, y+4), (x-2, y+2). But since there are only 3 rows, y+4 would be outside. So no. Similarly, other orientations? Maybe squares with side length sqrt(5)? Wait, earlier we thought they might not exist. But let me check again.

Alternatively, maybe squares of side length 2? Wait, axis-aligned squares of size 2x2 are already counted. What about squares of side length 2 that are rotated? For example, a square with side length 2, but rotated. For example, starting at (x, y), moving right 1, up 1, then right 1, up 1, etc. Wait, that would be a square of side length sqrt(2), which we already considered. Wait, no. Wait, if you have a square with side length 2, but rotated, then the distance between two adjacent vertices would be 2. For example, a square with vertices at (x, y), (x + 1, y + sqrt(3)), etc. But since the grid is only points with integer coordinates, the distance between two points is sqrt((dx)^2 + (dy)^2). So for a square with side length 2, we need two points with distance 2. For example, (x, y) and (x + 2, y) is distance 2, which is axis-aligned. Or (x, y) and (x + 1, y + 1) is sqrt(2). Or (x, y) and (x + 1, y + sqrt(3))? No, that's not possible with integer coordinates. So the only squares with side length 2 are the axis-aligned ones (2x2) and maybe some rotated ones? Wait, for example, a square with vertices at (x, y), (x + 1, y + 1), (x + 2, y), (x + 1, y - 1). This is a square rotated by 45 degrees, with side length sqrt(2). Wait, but that's the same as before. So the side length is sqrt(2), not 2. Therefore, there are no squares with side length 2 except the axis-aligned ones. Therefore, our previous count seems okay.

Wait, but then according to our previous calculation, total squares would be 4n - 6. If that equals 70, then n = 19. But let me check with n=3. If n=3, then total squares would be 4*3 -6 = 6. Let me count manually.

For n=3, grid is 3x3 points. How many squares?

Axis-aligned:

- 1x1: 2*(3 - 1) = 4. Let me count: in a 3x3 grid, the number of 1x1 squares is 2*2=4. Correct.

- 2x2: 1*(3 - 2) = 1. Which is the big square in the center. Correct.

Rotated squares of sqrt(2):

n - 2 = 1. So 1 square. Which is the square with vertices at (1,0), (2,1), (1,2), (0,1). So yes, that's one square. So total squares: 4 + 1 + 1 = 6. Which matches 4*3 -6 = 6. So that seems correct.

But the problem says there are 70 squares. So 4n -6 =70 ⇒ 4n=76 ⇒ n=19. But let me check if there are other squares that I haven't considered. For example, squares of different sizes?

Wait, maybe there are squares of side length sqrt(5)? Let me think. For example, in a 3xN grid, can we have squares with side length sqrt(5)? For example, starting at (x, y), moving right 1 and up 2, then right 2 and down 1, etc. Wait, but in a 3xN grid, moving up 2 from y=0 would reach y=2, which is the top row. Then moving right 2 would require that x + 2 is within the grid. Let me try to construct such a square.

Let me take points (0,0), (1,2), (3,1), (2,-1). But (2,-1) is outside. Not valid. Alternatively, (1,0), (2,2), (4,1), (3,-1). Again, outside. Alternatively, maybe starting at (x, y) = (1, 0), moving right 2 and up 1 to (3,1), then left 1 and up 2 to (2, 3), which is outside. Not valid. Alternatively, moving left? Maybe not. It seems difficult to form such squares within the grid. Therefore, maybe there are no squares of side length sqrt(5). Therefore, our previous count of 4n -6 is correct.

But let me check another value. Let's take n=4. Then total squares would be 4*4 -6 = 10. Let me count manually.

For n=4, grid is 3x4 points.

Axis-aligned:

- 1x1: 2*(4 -1) = 6. Let me count: in each of the 2 vertical positions (rows 0-1, 1-2), there are 3 horizontal positions (columns 0-1, 1-2, 2-3). So 2*3=6. Correct.

- 2x2: 1*(4 - 2) = 2. Which are the two 2x2 squares in the grid. Correct.

Rotated squares of sqrt(2): n - 2 = 2. So two squares. Let me see: for y=0, x can be 1 and 2. So squares at positions (1,0), (2,1), (1,2), (0,1) and (2,0), (3,1), (2,2), (1,1). So two squares. Correct. So total squares: 6 + 2 + 2 = 10. Which matches 4*4 -6 = 10. So that seems correct.

Therefore, the formula seems to hold. Therefore, if total squares are 70, then 4n -6 =70 ⇒ 4n=76 ⇒ n=19. Therefore, the answer is 19? Wait, but let me check again. Wait, the problem says "three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are three sizes, but maybe my count is wrong? Let me check again.

Wait, the three sizes would be:

1. 1x1 squares (side length 1)

2. 2x2 squares (side length 2)

3. sqrt(2)x sqrt(2) squares (side length sqrt(2))

So three sizes. Therefore, the total number of squares is 2(n -1) + (n -2) + (n -2) = 4n -6. Therefore, solving 4n -6 =70 gives n=19. Therefore, the answer is 19? But let me check again with n=5. For n=5, total squares would be 4*5 -6=14. Let me count:

Axis-aligned:

- 1x1: 2*(5-1)=8

- 2x2: 1*(5-2)=3

Rotated squares: n-2=3. So total squares: 8 + 3 + 3=14. Correct. So formula holds. Therefore, seems correct.

But let me check if there are other squares. For example, in a 3xN grid, are there squares of side length sqrt(5)? For example, in n=5, can we have such squares?

Let me try. Suppose we have points (0,0), (1,2), (3,1), (2,-1). But (2,-1) is outside. How about (1,0), (2,2), (4,1), (3,-1). Again, (3,-1) is outside. How about (1,0), (3,1), (2,3), (0,2). But (2,3) is outside. Alternatively, (0,1), (1,3), (3,2), (2,0). (1,3) is outside. So no. Therefore, no such squares. Therefore, our count is correct.

Therefore, the answer is n=19. Wait, but let me check again the problem statement. It says "three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are three sizes, but maybe I missed some?

Wait, another thought: maybe there are squares of size 1x1, 2x2, and 3x3? But in a 3xN grid, the maximum square size is 2x2, because vertically there are only 3 rows, so a 3x3 square would require 3 rows and 3 columns. Wait, but the grid is 3 rows and n columns. So if n ≥ 3, then there is a 2x2 square. But a 3x3 square would require 3 rows and 3 columns. So if n ≥ 3, then there is a 3x3 square? Wait, no. Wait, a square of size 3x3 would have side length 3 units. But in a grid with 3 rows, the vertical distance between the first and last row is 2 units (from y=0 to y=2). Therefore, you can't have a square of side length 3. Therefore, there are no 3x3 squares. Therefore, only up to 2x2. Therefore, the three sizes are 1x1, 2x2, and sqrt(2)x sqrt(2). Therefore, our count is correct.

Therefore, solving 4n -6 =70 gives n=19. Therefore, the answer is 19. Wait, but let me check once more with the Asymptote code. The Asymptote code seems to show a 3x3 grid? Let me check:

The Asymptote code starts with dots from (0,0) to (2,2), which is 3x3 grid points. Then there are some filled squares. Then it adds more points at (4,0), (5,0), etc. So maybe the Asymptote code is just an example with n=3? But the problem says there are 70 squares. So for n=3, total squares are 6, which is way less than 70. Therefore, the answer is n=19. Therefore, I think that is correct.

But let me check once again the formula. The total number of squares is:

Number of 1x1 squares: 2(n - 1)

Number of 2x2 squares: 1(n - 2)

Number of sqrt(2)x sqrt(2) squares: 1(n - 2)

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

So 4n -6 =70 ⇒ 4n=76 ⇒ n=19. Therefore, the answer is 19. Therefore, I think that is correct.

**Final Answer**
\boxed{19}

To solve the problem, we are given a **rectangular grid of points** with **3 rows and $ n $ columns**, and we are told that there are **70 squares** of **three distinct sizes** whose vertices lie on this grid.

---

### Step 1: Understanding the Grid

The grid is defined as a **3×n grid of points**, meaning there are 3 rows and $ n $ columns of points. The coordinates of the points range from $ (0,0) $ to $ (n-1, 2) $, assuming the grid is aligned with the coordinate system.

---

### Step 2: Identifying the Three Sizes of Squares

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

1. **1×1 axis-aligned squares** (side length = 1 unit)
2. **2×2 axis-aligned squares** (side length = 2 units)
3. **Squares with side length $ \sqrt{2} $** (rotated 45°, formed by diagonally adjacent points)

These are the only possible square sizes that can be formed on a 3×n grid of points, given the constraints of the grid's dimensions.

---

### Step 3: Counting the Squares

#### 1. **1×1 Squares**

These are the smallest axis-aligned squares. Each such square spans 1 unit horizontally and 1 unit vertically.

- **Horizontal positions**: $ n - 1 $ (since each square spans 1 unit, and there are $ n $ columns)
- **Vertical positions**: $ 3 - 1 = 2 $ (since there are 3 rows)

$$
\text{Number of 1×1 squares} = 2(n - 1)
$$

#### 2. **2×2 Squares**

These are larger axis-aligned squares, spanning 2 units in both directions.

- **Horizontal positions**: $ n - 2 $
- **Vertical positions**: $ 3 - 2 = 1 $

$$
\text{Number of 2×2 squares} = 1(n - 2)
$$

#### 3. **Squares with side length $ \sqrt{2} $**

These are squares formed by diagonally adjacent points, such as from $ (x, y) $ to $ (x+1, y+1) $, etc. These squares are rotated 45° and have side length $ \sqrt{2} $.

- These squares require 2 rows and 3 columns (to fit the square and its diagonals).
- **Vertical positions**: Only possible in the bottom 2 rows (since the square spans 2 rows)
- **Horizontal positions**: $ n - 2 $ (since the square spans 2 columns)

$$
\text{Number of } \sqrt{2} \text{ squares} = 1(n - 2)
$$

---

### Step 4: Total Number of Squares

Adding up all the square counts:

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

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

$$
4n - 6 = 70
$$

$$
4n = 76 \Rightarrow n = 19
$$

---

### 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 grid with some dots and some filled squares, but maybe it's just an example? Let me try to visualize it. The grid is 3 rows by n columns? Wait, the Asymptote code shows a 3x3 grid? Wait, the first part has points from (0,0) to (2,2), so that's 3x3 grid points? But then there are more points at (4,0), (5,0), etc. Maybe the Asymptote code is just an example for n=3? But the problem says it's a 3×n grid. So maybe the Asymptote code is just illustrating the grid with n=3? But the problem says there are squares of three sizes. Let me think.

First, let me recall that in a grid of points, the number of squares can be calculated by considering squares of different sizes. For a grid with m rows and n columns of points, the number of squares is the sum over all possible square sizes. For example, in a grid with m rows and n columns, the number of 1x1 squares is (m-1)(n-1), the number of 2x2 squares is (m-2)(n-2), and so on, up to the maximum square size. However, in this problem, it's mentioned that there are squares of three sizes. So maybe the grid is such that there are three different square sizes possible? Or maybe the problem is referring to squares of three different sizes, not necessarily all sizes?

Wait, the problem says "three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are three different sizes of squares that can be formed on the grid? Let me think. For example, in a grid with 3 rows and n columns, the possible square sizes would depend on the spacing between the points. Since it's a grid of points, the distance between adjacent points is 1 unit. So squares can be of different sizes based on their side lengths. For example, in a grid with 3 rows and n columns, the possible square sizes would be:

- 1x1 squares: these have sides of length 1 unit. The number of such squares would be (number of positions horizontally) × (number of positions vertically). Since there are 3 rows, vertically, the number of positions for vertical sides would be (3 - 1) = 2. Horizontally, for n columns, the number of positions would be (n - 1). So total 1x1 squares would be 2*(n - 1).

Wait, but actually, the number of squares of size k x k in a grid with m rows and n columns is (m - k) * (n - k). Wait, but that's if the grid is considered as a grid of squares. Wait, maybe I need to clarify.

Wait, if the grid is of points, then the number of squares of size 1x1 (i.e., squares with side length 1 unit) would be (number of horizontal positions) × (number of vertical positions). For a grid with 3 rows and n columns of points, the number of horizontal positions for a square of side length 1 is (n - 1), since each square spans 1 unit horizontally. Similarly, vertically, since there are 3 rows, the number of vertical positions is (3 - 1) = 2. Therefore, the number of 1x1 squares is 2*(n - 1).

Similarly, for squares of size 2x2 (side length 2 units), the number would be (number of horizontal positions) × (number of vertical positions). The horizontal positions would be (n - 2), since each square spans 2 units horizontally. The vertical positions would be (3 - 2) = 1. Therefore, the number of 2x2 squares is 1*(n - 2).

For squares of size 3x3 (side length 3 units), the horizontal positions would be (n - 3), and vertical positions would be (3 - 3) = 0. Therefore, there are 0 such squares. So in this case, there are only two sizes of squares: 1x1 and 2x2. But the problem says there are three sizes. Hmm. So maybe there are squares of different orientations? Like squares that are not aligned with the grid? For example, squares that are rotated?

Wait, the problem says "squares of three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are squares of different sizes, including both axis-aligned and rotated squares? For example, in a grid, you can have squares that are tilted, like with side length sqrt(2), which would be squares formed by diagonally adjacent points. But in a 3xN grid, how many such squares are there?

Let me think. For example, in a grid with 3 rows and n columns, the possible squares can be:

1. Axis-aligned squares: sizes 1x1, 2x2.

2. Squares with side length sqrt(2): these are squares that are rotated 45 degrees, with their sides along the diagonals of the 1x1 squares. For such squares, the side length is sqrt(2), and they require a 2x2 block of points. For example, the square with vertices at (0,0), (1,1), (0,2), (-1,1) – but wait, that would go outside the grid. So in a 3xN grid, how many such squares are there?

Alternatively, maybe squares that are formed by points spaced two units apart? Wait, maybe I need to think more systematically.

First, let me recall that in a grid, the number of squares can be calculated by considering all possible squares, both axis-aligned and rotated. However, the problem mentions "three sizes", so maybe there are three different sizes of squares, regardless of orientation. Let me check.

In a 3xN grid of points, the possible squares can be:

- Axis-aligned squares:

   - 1x1: side length 1

   - 2x2: side length 2

- Rotated squares:

   - Squares with side length sqrt(2): these are squares formed by moving 1 unit right and 1 unit up, then 1 unit left and 1 unit up, etc. For example, a square with vertices at (0,0), (1,1), (0,2), (-1,1). But since the grid is only 3 rows, the square would need to fit within the grid. So for such squares, the vertical span is 2 units (from row 0 to row 2), and horizontally, they need to have at least 2 columns. Wait, for example, in a 3xN grid, the square with side length sqrt(2) would require that there are two columns and two rows? Wait, maybe not. Let me think.

Actually, a square with side length sqrt(2) would have its vertices at points that are diagonally adjacent. For example, in a grid, if you have points (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). But this square would require that x-1 is within the grid. So if the grid has columns from 0 to n-1, then x must be at least 1. Similarly, vertically, since there are 3 rows, y can be 0 or 1 (since y+2 would be 2, which is the top row). Therefore, the number of such squares would be (n - 2) * 1. Wait, because horizontally, you need two columns to have the square, but actually, the square spans from column x to x+1, but the square itself is formed by points (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). Wait, so for each such square, you need columns x-1, x, x+1? Wait, no. Let me think again.

Wait, if you have a square with vertices at (x,y), (x+1,y+1), (x,y+2), (x-1,y+1). So to have this square, you need that x-1 ≥ 0 (so x ≥ 1) and y+2 ≤ 2 (since there are 3 rows, y can be 0 or 1). Therefore, for y, there are 2 possibilities (y=0 and y=1). For x, since x must be at least 1, and the square spans from x-1 to x+1 horizontally, so the number of such squares horizontally would be n - 2 (since x can be from 1 to n-2). Therefore, total number of such squares is 2*(n - 2). Wait, but each square is counted once? Or is there another way?

Alternatively, maybe there are squares of side length sqrt(2) that are formed in different positions. Wait, but maybe I need to check for other sizes. For example, squares with side length sqrt(5)? Wait, that would be squares that are formed by moving 1 unit in one direction and 2 units in another. For example, a square with vertices at (0,0), (1,2), (3,1), (2,-1). But again, this would go outside the grid. So maybe in a 3xN grid, there are no such squares? Or maybe only some?

Alternatively, maybe there are squares of side length sqrt(5) if the grid is large enough. But since the grid has only 3 rows, vertically, the maximum distance between two points is 2 units. So if you have a square with side length sqrt(5), that would require moving 1 unit horizontally and 2 units vertically, but then the next side would have to be 2 units horizontally and 1 unit vertically, but that would require more rows or columns. Since there are only 3 rows, maybe such squares can't exist? Let me check.

Suppose we have a square with side length sqrt(5). Let's say starting at (x,y), moving right 1 and up 2 to (x+1, y+2). Then from there, moving left 2 and up 1? Wait, but that would be (-2, y+3), which is outside the grid. Alternatively, moving left 2 and down 1? That would be (x+1 - 2, y+2 - 1) = (x -1, y +1). Then from there, moving right 1 and down 2? That would be (x -1 +1, y +1 - 2) = (x, y -1). But that would go below the grid. So unless y -1 is within the grid, which would require y ≥ 1. So if y=1, then y -1 = 0. So starting at (x,1), moving to (x+1, 3), but there are only 3 rows, so y+2 = 3 would be outside. Wait, rows are from 0 to 2, so y+2 would be 2 + 2 = 4? Wait, no. Wait, if the grid has 3 rows, then the y-coordinates are 0, 1, 2. So moving up 2 units from y=0 would reach y=2, which is the top row. So starting at (x,0), moving right 1 and up 2 would reach (x+1, 2). Then from there, moving left 2 and down 1 would reach (x+1 - 2, 2 - 1) = (x -1, 1). Then from there, moving right 1 and down 2 would reach (x -1 +1, 1 - 2) = (x, -1), which is outside. So that doesn't work. Alternatively, maybe another direction? Maybe from (x,0), moving right 2 and up 1 to (x+2, 1). Then from there, moving left 1 and up 2 to (x+2 -1, 1 + 2) = (x+1, 3), which is outside. So again, no good. Therefore, maybe there are no squares of side length sqrt(5) in a 3xN grid? So maybe only squares of side length 1, sqrt(2), and 2?

Wait, but earlier we thought there are only two sizes of axis-aligned squares (1x1 and 2x2). Then, if we include the rotated squares of side length sqrt(2), that would be a third size. So total three sizes. Therefore, the problem says there are squares of three sizes, which would be 1x1, 2x2, and sqrt(2)x sqrt(2). Therefore, the total number of squares would be the sum of the number of each size.

So let me calculate that.

First, axis-aligned squares:

- 1x1: as before, number is (number of horizontal positions) × (number of vertical positions). For 1x1 squares, horizontally, there are (n - 1) positions (since each square spans 1 unit horizontally, and there are n columns, so n - 1 positions). Vertically, since there are 3 rows, the number of vertical positions is (3 - 1) = 2. Therefore, total 1x1 squares: 2*(n - 1).

- 2x2: horizontally, positions are (n - 2), vertically, positions are (3 - 2) = 1. So total 2x2 squares: 1*(n - 2).

Then, rotated squares of side length sqrt(2). How many are there?

As discussed earlier, these squares are formed by points (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). For these squares to exist, we need:

- x - 1 ≥ 0 ⇒ x ≥ 1

- y + 2 ≤ 2 ⇒ y ≤ 0. Since y can be 0 or 1, but y + 2 ≤ 2 ⇒ y ≤ 0. Therefore, y must be 0. So only y=0.

Therefore, for each x from 1 to n - 2 (since x+1 must be ≤ n - 1, so x ≤ n - 2), we have such squares. Therefore, the number of these squares is (n - 2) * 1 = n - 2.

Wait, but is that all? Are there other squares of side length sqrt(2)? For example, squares that are placed differently? For example, maybe squares that are placed with their sides going in different directions? Wait, for example, starting at (x, y), going right 1, down 1, etc. But since the grid has only 3 rows, going down from y=0 would go to y=-1, which is outside. Similarly, going up from y=2 would go to y=3, which is outside. So maybe only the squares that are placed with their top side at y=2? Wait, but we considered y=0. Wait, maybe there are squares that are placed with their bottom side at y=0? Wait, no, the square with vertices at (x,0), (x+1,1), (x,2), (x-1,1). That's the same as before, with y=0. So that's the only possibility. Therefore, the number of such squares is (n - 2). So total rotated squares of side length sqrt(2) is n - 2.

Wait, but maybe there are squares of side length sqrt(2) in other orientations? For example, squares that are placed horizontally? Wait, no, they are rotated. Wait, maybe squares that are placed with their sides not aligned with the grid? But we already considered those. So maybe that's all.

Therefore, total number of squares is:

- 1x1: 2(n - 1)

- 2x2: 1(n - 2)

- sqrt(2)x sqrt(2): 1(n - 2)

Wait, so total squares would be 2(n - 1) + (n - 2) + (n - 2) = 2n - 2 + n - 2 + n - 2 = 4n - 6.

But the problem says there are 70 squares. Therefore, 4n - 6 = 70 ⇒ 4n = 76 ⇒ n = 19. But wait, let me check if this is correct.

Wait, but hold on. Is that all the squares? Are there any other squares? For example, squares of side length 2*sqrt(2)? That would be squares formed by moving 2 units in one direction and 2 units in another. But in a 3xN grid, can such squares exist?

For example, a square with vertices at (x, y), (x+2, y+2), (x, y+4), (x-2, y+2). But since there are only 3 rows, y+4 would be outside. So no. Similarly, other orientations? Maybe squares with side length sqrt(5)? Wait, earlier we thought they might not exist. But let me check again.

Alternatively, maybe squares of side length 2? Wait, axis-aligned squares of size 2x2 are already counted. What about squares of side length 2 that are rotated? For example, a square with side length 2, but rotated. For example, starting at (x, y), moving right 1, up 1, then right 1, up 1, etc. Wait, that would be a square of side length sqrt(2), which we already considered. Wait, no. Wait, if you have a square with side length 2, but rotated, then the distance between two adjacent vertices would be 2. For example, a square with vertices at (x, y), (x + 1, y + sqrt(3)), etc. But since the grid is only points with integer coordinates, the distance between two points is sqrt((dx)^2 + (dy)^2). So for a square with side length 2, we need two points with distance 2. For example, (x, y) and (x + 2, y) is distance 2, which is axis-aligned. Or (x, y) and (x + 1, y + 1) is sqrt(2). Or (x, y) and (x + 1, y + sqrt(3))? No, that's not possible with integer coordinates. So the only squares with side length 2 are the axis-aligned ones (2x2) and maybe some rotated ones? Wait, for example, a square with vertices at (x, y), (x + 1, y + 1), (x + 2, y), (x + 1, y - 1). This is a square rotated by 45 degrees, with side length sqrt(2). Wait, but that's the same as before. So the side length is sqrt(2), not 2. Therefore, there are no squares with side length 2 except the axis-aligned ones. Therefore, our previous count seems okay.

Wait, but then according to our previous calculation, total squares would be 4n - 6. If that equals 70, then n = 19. But let me check with n=3. If n=3, then total squares would be 4*3 -6 = 6. Let me count manually.

For n=3, grid is 3x3 points. How many squares?

Axis-aligned:

- 1x1: 2*(3 - 1) = 4. Let me count: in a 3x3 grid, the number of 1x1 squares is 2*2=4. Correct.

- 2x2: 1*(3 - 2) = 1. Which is the big square in the center. Correct.

Rotated squares of sqrt(2):

n - 2 = 1. So 1 square. Which is the square with vertices at (1,0), (2,1), (1,2), (0,1). So yes, that's one square. So total squares: 4 + 1 + 1 = 6. Which matches 4*3 -6 = 6. So that seems correct.

But the problem says there are 70 squares. So 4n -6 =70 ⇒ 4n=76 ⇒ n=19. But let me check if there are other squares that I haven't considered. For example, squares of different sizes?

Wait, maybe there are squares of side length sqrt(5)? Let me think. For example, in a 3xN grid, can we have squares with side length sqrt(5)? For example, starting at (x, y), moving right 1 and up 2, then right 2 and down 1, etc. Wait, but in a 3xN grid, moving up 2 from y=0 would reach y=2, which is the top row. Then moving right 2 would require that x + 2 is within the grid. Let me try to construct such a square.

Let me take points (0,0), (1,2), (3,1), (2,-1). But (2,-1) is outside. Not valid. Alternatively, (1,0), (2,2), (4,1), (3,-1). Again, outside. Alternatively, maybe starting at (x, y) = (1, 0), moving right 2 and up 1 to (3,1), then left 1 and up 2 to (2, 3), which is outside. Not valid. Alternatively, moving left? Maybe not. It seems difficult to form such squares within the grid. Therefore, maybe there are no squares of side length sqrt(5). Therefore, our previous count of 4n -6 is correct.

But let me check another value. Let's take n=4. Then total squares would be 4*4 -6 = 10. Let me count manually.

For n=4, grid is 3x4 points.

Axis-aligned:

- 1x1: 2*(4 -1) = 6. Let me count: in each of the 2 vertical positions (rows 0-1, 1-2), there are 3 horizontal positions (columns 0-1, 1-2, 2-3). So 2*3=6. Correct.

- 2x2: 1*(4 - 2) = 2. Which are the two 2x2 squares in the grid. Correct.

Rotated squares of sqrt(2): n - 2 = 2. So two squares. Let me see: for y=0, x can be 1 and 2. So squares at positions (1,0), (2,1), (1,2), (0,1) and (2,0), (3,1), (2,2), (1,1). So two squares. Correct. So total squares: 6 + 2 + 2 = 10. Which matches 4*4 -6 = 10. So that seems correct.

Therefore, the formula seems to hold. Therefore, if total squares are 70, then 4n -6 =70 ⇒ 4n=76 ⇒ n=19. Therefore, the answer is 19? Wait, but let me check again. Wait, the problem says "three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are three sizes, but maybe my count is wrong? Let me check again.

Wait, the three sizes would be:

1. 1x1 squares (side length 1)

2. 2x2 squares (side length 2)

3. sqrt(2)x sqrt(2) squares (side length sqrt(2))

So three sizes. Therefore, the total number of squares is 2(n -1) + (n -2) + (n -2) = 4n -6. Therefore, solving 4n -6 =70 gives n=19. Therefore, the answer is 19? But let me check again with n=5. For n=5, total squares would be 4*5 -6=14. Let me count:

Axis-aligned:

- 1x1: 2*(5-1)=8

- 2x2: 1*(5-2)=3

Rotated squares: n-2=3. So total squares: 8 + 3 + 3=14. Correct. So formula holds. Therefore, seems correct.

But let me check if there are other squares. For example, in a 3xN grid, are there squares of side length sqrt(5)? For example, in n=5, can we have such squares?

Let me try. Suppose we have points (0,0), (1,2), (3,1), (2,-1). But (2,-1) is outside. How about (1,0), (2,2), (4,1), (3,-1). Again, (3,-1) is outside. How about (1,0), (3,1), (2,3), (0,2). But (2,3) is outside. Alternatively, (0,1), (1,3), (3,2), (2,0). (1,3) is outside. So no. Therefore, no such squares. Therefore, our count is correct.

Therefore, the answer is n=19. Wait, but let me check again the problem statement. It says "three sizes whose vertices are points on this rectangular 3×n grid of points." So maybe there are three sizes, but maybe I missed some?

Wait, another thought: maybe there are squares of size 1x1, 2x2, and 3x3? But in a 3xN grid, the maximum square size is 2x2, because vertically there are only 3 rows, so a 3x3 square would require 3 rows and 3 columns. Wait, but the grid is 3 rows and n columns. So if n ≥ 3, then there is a 2x2 square. But a 3x3 square would require 3 rows and 3 columns. So if n ≥ 3, then there is a 3x3 square? Wait, no. Wait, a square of size 3x3 would have side length 3 units. But in a grid with 3 rows, the vertical distance between the first and last row is 2 units (from y=0 to y=2). Therefore, you can't have a square of side length 3. Therefore, there are no 3x3 squares. Therefore, only up to 2x2. Therefore, the three sizes are 1x1, 2x2, and sqrt(2)x sqrt(2). Therefore, our count is correct.

Therefore, solving 4n -6 =70 gives n=19. Therefore, the answer is 19. Wait, but let me check once more with the Asymptote code. The Asymptote code seems to show a 3x3 grid? Let me check:

The Asymptote code starts with dots from (0,0) to (2,2), which is 3x3 grid points. Then there are some filled squares. Then it adds more points at (4,0), (5,0), etc. So maybe the Asymptote code is just an example with n=3? But the problem says there are 70 squares. So for n=3, total squares are 6, which is way less than 70. Therefore, the answer is n=19. Therefore, I think that is correct.

But let me check once again the formula. The total number of squares is:

Number of 1x1 squares: 2(n - 1)

Number of 2x2 squares: 1(n - 2)

Number of sqrt(2)x sqrt(2) squares: 1(n - 2)

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

So 4n -6 =70 ⇒ 4n=76 ⇒ n=19. Therefore, the answer is 19. Therefore, I think that is correct.

**Final Answer**
\boxed{19}

To solve the problem, we are given a **rectangular grid of points** with **3 rows and $ n $ columns**, and we are told that there are **70 squares** of **three distinct sizes** whose vertices lie on this grid.

---

### Step 1: Understanding the Grid

The grid is defined as a **3×n grid of points**, meaning there are 3 rows and $ n $ columns of points. The coordinates of the points range from $ (0,0) $ to $ (n-1, 2) $, assuming the grid is aligned with the coordinate system.

---

### Step 2: Identifying the Three Sizes of Squares

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

1. **1×1 axis-aligned squares** (side length = 1 unit)
2. **2×2 axis-aligned squares** (side length = 2 units)
3. **Squares with side length $ \sqrt{2} $** (rotated 45°, formed by diagonally adjacent points)

These are the only possible square sizes that can be formed on a 3×n grid of points, given the constraints of the grid's dimensions.

---

### Step 3: Counting the Squares

#### 1. **1×1 Squares**

These are the smallest axis-aligned squares. Each such square spans 1 unit horizontally and 1 unit vertically.

- **Horizontal positions**: $ n - 1 $ (since each square spans 1 unit, and there are $ n $ columns)
- **Vertical positions**: $ 3 - 1 = 2 $ (since there are 3 rows)

$$
\text{Number of 1×1 squares} = 2(n - 1)
$$

#### 2. **2×2 Squares**

These are larger axis-aligned squares, spanning 2 units in both directions.

- **Horizontal positions**: $ n - 2 $
- **Vertical positions**: $ 3 - 2 = 1 $

$$
\text{Number of 2×2 squares} = 1(n - 2)
$$

#### 3. **Squares with side length $ \sqrt{2} $**

These are squares formed by diagonally adjacent points, such as from $ (x, y) $ to $ (x+1, y+1) $, etc. These squares are rotated 45° and have side length $ \sqrt{2} $.

- These squares require 2 rows and 3 columns (to fit the square and its diagonals).
- **Vertical positions**: Only possible in the bottom 2 rows (since the square spans 2 rows)
- **Horizontal positions**: $ n - 2 $ (since the square spans 2 columns)

$$
\text{Number of } \sqrt{2} \text{ squares} = 1(n - 2)
$$

---

### Step 4: Total Number of Squares

Adding up all the square counts:

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

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

$$
4n - 6 = 70
$$

$$
4n = 76 \Rightarrow n = 19
$$

---

### 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, let me expand the left side. (x² - 8)² is x⁴ - 16x² + 64, right? Wait, let me check that again. (a - b)² is a² - 2ab + b². So here, a is x² and b is 8. Therefore, (x²)² - 2*(x²)*(8) + (8)² = x⁴ - 16x² + 64. Yeah, 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 is 20. So:

x⁴ - 16x² + 5x + 20 = 0

Hmm, so now I have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Solving quartic equations can be tricky. Maybe I can factor this? 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 for root x = -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 try 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, that's right. How about 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 = -6. x= -2: -8 -4 +30 +20= 38. x= -4: -64 -16 +60 +20= 0? Let me check:

x=-4: (-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = (-64 -16) + (60 +20) = -80 + 80 = 0! Oh! So x = -4 is a root. Therefore, (x + 4) is a factor.

So let's perform synthetic division on the cubic polynomial x³ - x² -15x +20 with root x = -4.

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

Using synthetic division with -4:

-4 | 1   -1   -15   20
    |     -4   20   -20
    ---------------------
      1   -5   5   0

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

So the quartic equation factors as (x + 1)(x + 4)(x² - 5x + 5) = 0. Therefore, the roots 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 ± √(25 - 20)] / 2 = [5 ± √5]/2.

So the roots are x = -1, x = -4, x = [5 + √5]/2, and x = [5 - √5]/2.

Therefore, there are four x-values. Now, for each x, we can find the corresponding y from the first equation y = x² - 8.

But before that, let me check if all these x-values are valid. Since we substituted y = x² - 8 into the second equation, we need to make sure that these x-values don't lead to contradictions. However, since we derived them through substitution, they should all be valid. But let me check for possible extraneous solutions? Wait, no, because we didn't square both sides or anything like that. So all solutions should be valid.

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

First, x = -1:

y = (-1)^2 - 8 = 1 - 8 = -7

Second, x = -4:

y = (-4)^2 - 8 = 16 - 8 = 8

Third, x = [5 + √5]/2:

Let me compute y:

y = ([5 + √5]/2)^2 - 8

First, square [5 + √5]/2:

= (25 + 10√5 + 5)/4 = (30 + 10√5)/4 = (15 + 5√5)/2

Then subtract 8:

= (15 + 5√5)/2 - 8 = (15 + 5√5 - 16)/2 = (-1 + 5√5)/2

Fourth, x = [5 - √5]/2:

Similarly, compute y:

y = ([5 - √5]/2)^2 - 8

Square [5 - √5]/2:

= (25 - 10√5 + 5)/4 = (30 - 10√5)/4 = (15 - 5√5)/2

Subtract 8:

= (15 - 5√5)/2 - 8 = (15 - 5√5 - 16)/2 = (-1 - 5√5)/2

So the four y-coordinates are: -7, 8, (-1 + 5√5)/2, and (-1 - 5√5)/2.

But the question says "the product of the y-coordinates of all the distinct solutions". Wait, are all these solutions distinct? Let me check if any of the y-values are repeated.

First, -7 and 8 are distinct. The other two y-values are (-1 + 5√5)/2 and (-1 - 5√5)/2. These are two distinct values because √5 is irrational, so they can't be equal. Therefore, all four solutions are distinct.

Therefore, the product of the y-coordinates is (-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. However, here it's (-1 + 5√5)/2 times (-1 - 5√5)/2. Let me write it as [( -1 + 5√5 ) * ( -1 - 5√5 ) ] / (2 * 2) = [(-1)^2 - (5√5)^2]/4.

Compute numerator:

(-1)^2 = 1

(5√5)^2 = 25 * 5 = 125

Therefore, numerator is 1 - 125 = -124

So the product is -124 / 4 = -31.

Therefore, the product of the last two y-coordinates is -31.

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

56 * 31. Let me compute 56*30 = 1680, plus 56*1 = 56, so total 1680 + 56 = 1736. Since both numbers are negative, the product is positive 1736.

Wait, but let me check again:

Original product: (-7) * 8 * [(-1 + 5√5)/2] * [(-1 - 5√5)/2] = (-56) * (-31) = 56 * 31 = 1736.

But let me verify the calculation of the product of the two terms again.

[(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)(-1) + (-1)(-5√5) + (5√5)(-1) + (5√5)(-5√5)] / 4? Wait, no, that's expanding the numerator. Wait, but actually, it's easier to use the formula (a + b)(a - b) = a² - b². Let me check if that applies here.

Let me set a = -1, b = 5√5. Then (a + b)(a - b) = a² - b². Wait, but in our case, the terms are (-1 + 5√5) and (-1 - 5√5). So that is (a + b)(a - b) where a = -1 and b = 5√5? Wait, no. Wait, (-1 + 5√5) is (a + b) where a = -1 and b = 5√5? Wait, no. Wait, if I think of it as ( (-1) + 5√5 ) * ( (-1) - 5√5 ) = (-1)^2 - (5√5)^2 = 1 - 125 = -124. Yes, that's correct. So divided by 4, gives -31. So that part is correct.

So then, total product is (-7)*8*(-31) = (-56)*(-31) = 1736.

But wait, let me check if there's another way to compute this product, maybe using Vieta's formula? Because sometimes when you have equations, you can find the product of roots without computing each one. Let me think.

But in this case, the original equations are y = x² - 8 and y² = -5x + 44. However, we substituted and found the quartic equation in x. But maybe there's another way to think about the product of y-coordinates?

Alternatively, since we have four solutions (x, y), each with their own y. The product of the y's is what we need.

Alternatively, perhaps we can consider the equations again. Let me think.

From the first equation, y = x² - 8. So if we can find the product of all y's, which is product of (x² - 8) for each solution x.

But since we have four x-values: x1 = -1, x2 = -4, x3 = [5 + √5]/2, x4 = [5 - √5]/2.

Therefore, the product of y's is (x1² - 8)(x2² - 8)(x3² - 8)(x4² - 8).

But since y = x² - 8, so each term is y_i.

Alternatively, since we already found the product as 1736, but maybe there's a smarter way? Let me check.

Alternatively, since we have the quartic equation in x: x⁴ - 16x² + 5x + 20 = 0. The roots are x1, x2, x3, x4.

But if we want the product of (x_i² - 8) for each root x_i.

Note that (x_i² - 8) is equal to y_i. So the product we need is product_{i=1}^4 (x_i² - 8).

But how can we compute this? Maybe express it in terms of the quartic equation?

Let me think. Let me denote P(x) = x⁴ - 16x² + 5x + 20. The roots are x1, x2, x3, x4.

We need to compute product_{i=1}^4 (x_i² - 8).

Note that x_i² - 8 = (x_i - sqrt(8))(x_i + sqrt(8)). Therefore, product_{i=1}^4 (x_i² - 8) = product_{i=1}^4 (x_i - 2√2)(x_i + 2√2) = P(2√2) * P(-2√2).

Wait, because if you have a polynomial P(x) with roots x1, x2, x3, x4, then P(a) = (a - x1)(a - x2)(a - x3)(a - x4). Therefore, product_{i=1}^4 (a - x_i) = P(a). Therefore, product_{i=1}^4 (x_i - a) = (-1)^4 P(a) = P(a).

But here, we have product_{i=1}^4 (x_i² - 8) = product_{i=1}^4 (x_i - 2√2)(x_i + 2√2) = P(2√2) * P(-2√2).

Therefore, if I compute P(2√2) and P(-2√2), then multiply them, that would give me the product of the y-coordinates.

Let me try this method. Maybe it's faster?

First, compute P(2√2):

P(x) = x⁴ - 16x² + 5x + 20

So P(2√2) = (2√2)^4 - 16*(2√2)^2 + 5*(2√2) + 20

Compute each term:

(2√2)^4: First, (2√2)^2 = 4*2 = 8. Then squared again: 8^2 = 64.

-16*(2√2)^2: -16*(8) = -128

5*(2√2) = 10√2

+20

So total P(2√2) = 64 - 128 + 10√2 + 20 = (64 - 128 + 20) + 10√2 = (-44) + 10√2

Similarly, compute P(-2√2):

P(-2√2) = (-2√2)^4 - 16*(-2√2)^2 + 5*(-2√2) + 20

Compute each term:

(-2√2)^4 = same as (2√2)^4 = 64

-16*(-2√2)^2 = -16*(8) = -128

5*(-2√2) = -10√2

+20

So total P(-2√2) = 64 - 128 -10√2 + 20 = (64 - 128 + 20) -10√2 = (-44) -10√2

Therefore, the product P(2√2)*P(-2√2) = [(-44) + 10√2] * [(-44) -10√2] = (-44)^2 - (10√2)^2 = 1936 - 100*2 = 1936 - 200 = 1736.

Which matches the previous result! So that's reassuring. Therefore, the product of the y-coordinates is 1736.

But let me check again with the previous method. Earlier, I calculated (-7)*8*[(-1 + 5√5)/2]*[(-1 - 5√5)/2] = (-56)*(-31) = 1736. So same answer.

Therefore, the answer is 1736.

But let me check once more if there are any mistakes in the calculation.

First, when I calculated the product of the two y-values from the quadratic roots:

[(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)^2 - (5√5)^2]/4 = (1 - 125)/4 = (-124)/4 = -31. Correct.

Then, multiplying by (-7)*8 = -56. So -56 * -31 = 1736. Correct.

Alternatively, using Vieta's formula on the quartic equation. Wait, but Vieta's formula relates to the sum and products of roots. However, since we have the product of (x_i² - 8), which is different. But we already confirmed with two methods.

Alternatively, maybe I can check with actual numbers. Let me compute the approximate values of the y-coordinates and multiply them to see if it's approximately 1736.

First, compute the y-values:

For x = -1: y = -7

For x = -4: y = 8

For x = [5 + √5]/2 ≈ [5 + 2.236]/2 ≈ 7.236/2 ≈ 3.618. Then y = x² - 8. Let me compute x²:

x ≈ 3.618, so x² ≈ 13.09, so y ≈ 13.09 - 8 ≈ 5.09. Wait, but earlier I had (-1 + 5√5)/2. Let me compute that:

(-1 + 5√5)/2 ≈ (-1 + 5*2.236)/2 ≈ (-1 + 11.18)/2 ≈ 10.18/2 ≈ 5.09. Correct.

Similarly, the other y is (-1 - 5√5)/2 ≈ (-1 - 11.18)/2 ≈ -12.18/2 ≈ -6.09.

So the four y-values are approximately -7, 8, 5.09, -6.09.

Multiply them: (-7) * 8 = -56; 5.09 * (-6.09) ≈ -5.09*6.09 ≈ Let me compute 5*6.09 = 30.45, 0.09*6.09 ≈ 0.548, so total ≈ 30.45 + 0.548 ≈ 30.998. But since it's negative, it's approximately -31. Then total product is (-56)*(-31) ≈ 1736. Which matches.

So the approximate calculation also gives around 1736. Therefore, I think the answer is correct.

But let me check once more if there are any mistakes in the factoring.

Original quartic equation: x⁴ - 16x² + 5x + 20 = 0.

Factored as (x + 1)(x + 4)(x² - 5x + 5) = 0. Let me expand this to check:

First, multiply (x + 1)(x + 4) = x² + 5x + 4. Then multiply by (x² - 5x + 5):

(x² + 5x + 4)(x² - 5x + 5). Let me compute this:

Multiply term by term:

First, x²*(x² - 5x + 5) = x⁴ -5x³ +5x²

Then 5x*(x² -5x +5) = 5x³ -25x² +25x

Then 4*(x² -5x +5) = 4x² -20x +20

Now add them all together:

x⁴ -5x³ +5x² +5x³ -25x² +25x +4x² -20x +20

Combine like terms:

x⁴ + (-5x³ +5x³) + (5x² -25x² +4x²) + (25x -20x) +20

Simplify:

x⁴ + 0x³ + (-16x²) + 5x +20.

Which is x⁴ -16x² +5x +20. Which matches the original quartic. So factoring is correct.

Therefore, the roots are correct, and the y-values are correct.

Therefore, the product of the y-coordinates is indeed 1736.

But just to make sure, let me check if there are any other solutions or if I missed something.

The original equations are y = x² - 8 and y² = -5x + 44.

We substituted y from the first equation into the second, leading to quartic equation. We found four real roots? Wait, the quartic equation had four real roots? Let me check.

The roots are x = -1, x = -4, and two roots from x² -5x +5 = 0. The discriminant of x² -5x +5 is 25 - 20 = 5, which is positive, so two real roots. Therefore, total four real roots. Therefore, four solutions.

Therefore, all solutions are accounted for, and their y-coordinates are as calculated.

Therefore, the product is 1736.

But let me check once more with another approach.

Suppose I consider the system:

y = x² - 8

y² = -5x + 44

Let me solve for x from the second equation.

From the second equation: y² + 5x = 44 => 5x = 44 - y² => x = (44 - y²)/5

Then substitute into the first equation:

y = [(44 - y²)/5]^2 - 8

So:

y = ( (44 - y²)^2 ) / 25 - 8

Multiply both sides by 25:

25y = (44 - y²)^2 - 200

Expand (44 - y²)^2:

= 44² - 2*44*y² + y⁴ = 1936 - 88y² + y⁴

Therefore:

25y = 1936 - 88y² + y⁴ - 200

Simplify constants: 1936 - 200 = 1736

So:

25y = y⁴ - 88y² + 1736

Bring all terms to one side:

y⁴ - 88y² -25y + 1736 = 0

So we have a quartic equation in y: y⁴ -88y² -25y +1736 = 0

Now, if we can find the product of the roots of this equation, which would be the product of all y-coordinates.

But for a quartic equation y⁴ + a y³ + b y² + c y + d = 0, the product of the roots is d if the leading coefficient is 1. However, in our case, the equation is y⁴ -88y² -25y +1736 = 0. So written as y⁴ + 0y³ -88y² -25y +1736 = 0.

Therefore, the product of the roots is the constant term divided by the leading coefficient, which is 1736 / 1 = 1736.

But wait, the product of all roots of the quartic equation in y is 1736. However, the question says "the product of the y-coordinates of all the distinct solutions".

But are all roots of this quartic equation distinct? Let me check.

Earlier, we found four distinct solutions for x, which correspond to four distinct y-values. Therefore, the quartic equation in y should have four distinct roots, which are the y-coordinates. Therefore, the product of the roots is 1736.

Therefore, this confirms again that the product is 1736.

Therefore, regardless of the method, the answer is 1736.

Therefore, I can be confident that 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:

$$
(x^2 - 8)^2 = x^4 - 16x^2 + 64
$$

So the equation becomes:

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

Bring all terms to one side:

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

---

### Step 2: Factor the Quartic Equation

We attempt to factor the quartic:

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

Using the Rational Root Theorem, we test possible rational roots and find that $ x = -1 $ and $ x = -4 $ are roots. Using synthetic or polynomial division, we factor the quartic as:

$$
(x + 1)(x + 4)(x^2 - 5x + 5) = 0
$$

The quadratic $ x^2 - 5x + 5 $ has roots:

$$
x = \frac{5 \pm \sqrt{5}}{2}
$$

So the four real solutions for $ x $ are:

$$
x = -1, \quad x = -4, \quad x = \frac{5 + \sqrt{5}}{2}, \quad x = \frac{5 - \sqrt{5}}{2}
$$

---

### Step 3: Find Corresponding $ y $-Coordinates

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

- For $ x = -1 $: $ y = (-1)^2 - 8 = -7 $
- For $ x = -4 $: $ y = (-4)^2 - 8 = 8 $
- For $ x = \frac{5 + \sqrt{5}}{2} $:
  $$
  y = \left( \frac{5 + \sqrt{5}}{2} \right)^2 - 8 = \frac{30 + 10\sqrt{5}}{4} - 8 = \frac{-1 + 5\sqrt{5}}{2}
  $$
- 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 $ y $-coordinates are:

$$
y_1 = -7, \quad y_2 = 8, \quad y_3 = \frac{-1 + 5\sqrt{5}}{2}, \quad y_4 = \frac{-1 - 5\sqrt{5}}{2}
$$

---

### Step 4: Compute the Product of the $ y $-Coordinates

We compute:

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

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:

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

---

### Step 5: Verification via Vieta's Formula

Alternatively, we can consider the quartic in $ y $, derived by substituting $ x = \frac{44 - y^2}{5} $ into the first equation. This leads to:

$$
y^4 - 88y^2 - 25y + 1736 = 0
$$

This is a quartic in $ y $, and the product of all its roots (by Vieta's formula) is the constant term divided by the leading coefficient:

$$
\text{Product of roots} = \frac{1736}{1} = 1736
$$

This confirms our earlier result.

---

### 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, let me expand the left side. (x² - 8)² is x⁴ - 16x² + 64, right? Wait, let me check that again. (a - b)² is a² - 2ab + b². So here, a is x² and b is 8. Therefore, (x²)² - 2*(x²)*(8) + (8)² = x⁴ - 16x² + 64. Yeah, 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 is 20. So:

x⁴ - 16x² + 5x + 20 = 0

Hmm, so now I have a quartic equation: x⁴ - 16x² + 5x + 20 = 0. Solving quartic equations can be tricky. Maybe I can factor this? 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 for root x = -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 try 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, that's right. How about 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 = -6. x= -2: -8 -4 +30 +20= 38. x= -4: -64 -16 +60 +20= 0? Let me check:

x=-4: (-4)^3 - (-4)^2 -15*(-4) +20 = -64 -16 +60 +20 = (-64 -16) + (60 +20) = -80 + 80 = 0! Oh! So x = -4 is a root. Therefore, (x + 4) is a factor.

So let's perform synthetic division on the cubic polynomial x³ - x² -15x +20 with root x = -4.

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

Using synthetic division with -4:

-4 | 1   -1   -15   20
    |     -4   20   -20
    ---------------------
      1   -5   5   0

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

So the quartic equation factors as (x + 1)(x + 4)(x² - 5x + 5) = 0. Therefore, the roots 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 ± √(25 - 20)] / 2 = [5 ± √5]/2.

So the roots are x = -1, x = -4, x = [5 + √5]/2, and x = [5 - √5]/2.

Therefore, there are four x-values. Now, for each x, we can find the corresponding y from the first equation y = x² - 8.

But before that, let me check if all these x-values are valid. Since we substituted y = x² - 8 into the second equation, we need to make sure that these x-values don't lead to contradictions. However, since we derived them through substitution, they should all be valid. But let me check for possible extraneous solutions? Wait, no, because we didn't square both sides or anything like that. So all solutions should be valid.

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

First, x = -1:

y = (-1)^2 - 8 = 1 - 8 = -7

Second, x = -4:

y = (-4)^2 - 8 = 16 - 8 = 8

Third, x = [5 + √5]/2:

Let me compute y:

y = ([5 + √5]/2)^2 - 8

First, square [5 + √5]/2:

= (25 + 10√5 + 5)/4 = (30 + 10√5)/4 = (15 + 5√5)/2

Then subtract 8:

= (15 + 5√5)/2 - 8 = (15 + 5√5 - 16)/2 = (-1 + 5√5)/2

Fourth, x = [5 - √5]/2:

Similarly, compute y:

y = ([5 - √5]/2)^2 - 8

Square [5 - √5]/2:

= (25 - 10√5 + 5)/4 = (30 - 10√5)/4 = (15 - 5√5)/2

Subtract 8:

= (15 - 5√5)/2 - 8 = (15 - 5√5 - 16)/2 = (-1 - 5√5)/2

So the four y-coordinates are: -7, 8, (-1 + 5√5)/2, and (-1 - 5√5)/2.

But the question says "the product of the y-coordinates of all the distinct solutions". Wait, are all these solutions distinct? Let me check if any of the y-values are repeated.

First, -7 and 8 are distinct. The other two y-values are (-1 + 5√5)/2 and (-1 - 5√5)/2. These are two distinct values because √5 is irrational, so they can't be equal. Therefore, all four solutions are distinct.

Therefore, the product of the y-coordinates is (-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. However, here it's (-1 + 5√5)/2 times (-1 - 5√5)/2. Let me write it as [( -1 + 5√5 ) * ( -1 - 5√5 ) ] / (2 * 2) = [(-1)^2 - (5√5)^2]/4.

Compute numerator:

(-1)^2 = 1

(5√5)^2 = 25 * 5 = 125

Therefore, numerator is 1 - 125 = -124

So the product is -124 / 4 = -31.

Therefore, the product of the last two y-coordinates is -31.

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

56 * 31. Let me compute 56*30 = 1680, plus 56*1 = 56, so total 1680 + 56 = 1736. Since both numbers are negative, the product is positive 1736.

Wait, but let me check again:

Original product: (-7) * 8 * [(-1 + 5√5)/2] * [(-1 - 5√5)/2] = (-56) * (-31) = 56 * 31 = 1736.

But let me verify the calculation of the product of the two terms again.

[(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)(-1) + (-1)(-5√5) + (5√5)(-1) + (5√5)(-5√5)] / 4? Wait, no, that's expanding the numerator. Wait, but actually, it's easier to use the formula (a + b)(a - b) = a² - b². Let me check if that applies here.

Let me set a = -1, b = 5√5. Then (a + b)(a - b) = a² - b². Wait, but in our case, the terms are (-1 + 5√5) and (-1 - 5√5). So that is (a + b)(a - b) where a = -1 and b = 5√5? Wait, no. Wait, (-1 + 5√5) is (a + b) where a = -1 and b = 5√5? Wait, no. Wait, if I think of it as ( (-1) + 5√5 ) * ( (-1) - 5√5 ) = (-1)^2 - (5√5)^2 = 1 - 125 = -124. Yes, that's correct. So divided by 4, gives -31. So that part is correct.

So then, total product is (-7)*8*(-31) = (-56)*(-31) = 1736.

But wait, let me check if there's another way to compute this product, maybe using Vieta's formula? Because sometimes when you have equations, you can find the product of roots without computing each one. Let me think.

But in this case, the original equations are y = x² - 8 and y² = -5x + 44. However, we substituted and found the quartic equation in x. But maybe there's another way to think about the product of y-coordinates?

Alternatively, since we have four solutions (x, y), each with their own y. The product of the y's is what we need.

Alternatively, perhaps we can consider the equations again. Let me think.

From the first equation, y = x² - 8. So if we can find the product of all y's, which is product of (x² - 8) for each solution x.

But since we have four x-values: x1 = -1, x2 = -4, x3 = [5 + √5]/2, x4 = [5 - √5]/2.

Therefore, the product of y's is (x1² - 8)(x2² - 8)(x3² - 8)(x4² - 8).

But since y = x² - 8, so each term is y_i.

Alternatively, since we already found the product as 1736, but maybe there's a smarter way? Let me check.

Alternatively, since we have the quartic equation in x: x⁴ - 16x² + 5x + 20 = 0. The roots are x1, x2, x3, x4.

But if we want the product of (x_i² - 8) for each root x_i.

Note that (x_i² - 8) is equal to y_i. So the product we need is product_{i=1}^4 (x_i² - 8).

But how can we compute this? Maybe express it in terms of the quartic equation?

Let me think. Let me denote P(x) = x⁴ - 16x² + 5x + 20. The roots are x1, x2, x3, x4.

We need to compute product_{i=1}^4 (x_i² - 8).

Note that x_i² - 8 = (x_i - sqrt(8))(x_i + sqrt(8)). Therefore, product_{i=1}^4 (x_i² - 8) = product_{i=1}^4 (x_i - 2√2)(x_i + 2√2) = P(2√2) * P(-2√2).

Wait, because if you have a polynomial P(x) with roots x1, x2, x3, x4, then P(a) = (a - x1)(a - x2)(a - x3)(a - x4). Therefore, product_{i=1}^4 (a - x_i) = P(a). Therefore, product_{i=1}^4 (x_i - a) = (-1)^4 P(a) = P(a).

But here, we have product_{i=1}^4 (x_i² - 8) = product_{i=1}^4 (x_i - 2√2)(x_i + 2√2) = P(2√2) * P(-2√2).

Therefore, if I compute P(2√2) and P(-2√2), then multiply them, that would give me the product of the y-coordinates.

Let me try this method. Maybe it's faster?

First, compute P(2√2):

P(x) = x⁴ - 16x² + 5x + 20

So P(2√2) = (2√2)^4 - 16*(2√2)^2 + 5*(2√2) + 20

Compute each term:

(2√2)^4: First, (2√2)^2 = 4*2 = 8. Then squared again: 8^2 = 64.

-16*(2√2)^2: -16*(8) = -128

5*(2√2) = 10√2

+20

So total P(2√2) = 64 - 128 + 10√2 + 20 = (64 - 128 + 20) + 10√2 = (-44) + 10√2

Similarly, compute P(-2√2):

P(-2√2) = (-2√2)^4 - 16*(-2√2)^2 + 5*(-2√2) + 20

Compute each term:

(-2√2)^4 = same as (2√2)^4 = 64

-16*(-2√2)^2 = -16*(8) = -128

5*(-2√2) = -10√2

+20

So total P(-2√2) = 64 - 128 -10√2 + 20 = (64 - 128 + 20) -10√2 = (-44) -10√2

Therefore, the product P(2√2)*P(-2√2) = [(-44) + 10√2] * [(-44) -10√2] = (-44)^2 - (10√2)^2 = 1936 - 100*2 = 1936 - 200 = 1736.

Which matches the previous result! So that's reassuring. Therefore, the product of the y-coordinates is 1736.

But let me check again with the previous method. Earlier, I calculated (-7)*8*[(-1 + 5√5)/2]*[(-1 - 5√5)/2] = (-56)*(-31) = 1736. So same answer.

Therefore, the answer is 1736.

But let me check once more if there are any mistakes in the calculation.

First, when I calculated the product of the two y-values from the quadratic roots:

[(-1 + 5√5)/2] * [(-1 - 5√5)/2] = [(-1)^2 - (5√5)^2]/4 = (1 - 125)/4 = (-124)/4 = -31. Correct.

Then, multiplying by (-7)*8 = -56. So -56 * -31 = 1736. Correct.

Alternatively, using Vieta's formula on the quartic equation. Wait, but Vieta's formula relates to the sum and products of roots. However, since we have the product of (x_i² - 8), which is different. But we already confirmed with two methods.

Alternatively, maybe I can check with actual numbers. Let me compute the approximate values of the y-coordinates and multiply them to see if it's approximately 1736.

First, compute the y-values:

For x = -1: y = -7

For x = -4: y = 8

For x = [5 + √5]/2 ≈ [5 + 2.236]/2 ≈ 7.236/2 ≈ 3.618. Then y = x² - 8. Let me compute x²:

x ≈ 3.618, so x² ≈ 13.09, so y ≈ 13.09 - 8 ≈ 5.09. Wait, but earlier I had (-1 + 5√5)/2. Let me compute that:

(-1 + 5√5)/2 ≈ (-1 + 5*2.236)/2 ≈ (-1 + 11.18)/2 ≈ 10.18/2 ≈ 5.09. Correct.

Similarly, the other y is (-1 - 5√5)/2 ≈ (-1 - 11.18)/2 ≈ -12.18/2 ≈ -6.09.

So the four y-values are approximately -7, 8, 5.09, -6.09.

Multiply them: (-7) * 8 = -56; 5.09 * (-6.09) ≈ -5.09*6.09 ≈ Let me compute 5*6.09 = 30.45, 0.09*6.09 ≈ 0.548, so total ≈ 30.45 + 0.548 ≈ 30.998. But since it's negative, it's approximately -31. Then total product is (-56)*(-31) ≈ 1736. Which matches.

So the approximate calculation also gives around 1736. Therefore, I think the answer is correct.

But let me check once more if there are any mistakes in the factoring.

Original quartic equation: x⁴ - 16x² + 5x + 20 = 0.

Factored as (x + 1)(x + 4)(x² - 5x + 5) = 0. Let me expand this to check:

First, multiply (x + 1)(x + 4) = x² + 5x + 4. Then multiply by (x² - 5x + 5):

(x² + 5x + 4)(x² - 5x + 5). Let me compute this:

Multiply term by term:

First, x²*(x² - 5x + 5) = x⁴ -5x³ +5x²

Then 5x*(x² -5x +5) = 5x³ -25x² +25x

Then 4*(x² -5x +5) = 4x² -20x +20

Now add them all together:

x⁴ -5x³ +5x² +5x³ -25x² +25x +4x² -20x +20

Combine like terms:

x⁴ + (-5x³ +5x³) + (5x² -25x² +4x²) + (25x -20x) +20

Simplify:

x⁴ + 0x³ + (-16x²) + 5x +20.

Which is x⁴ -16x² +5x +20. Which matches the original quartic. So factoring is correct.

Therefore, the roots are correct, and the y-values are correct.

Therefore, the product of the y-coordinates is indeed 1736.

But just to make sure, let me check if there are any other solutions or if I missed something.

The original equations are y = x² - 8 and y² = -5x + 44.

We substituted y from the first equation into the second, leading to quartic equation. We found four real roots? Wait, the quartic equation had four real roots? Let me check.

The roots are x = -1, x = -4, and two roots from x² -5x +5 = 0. The discriminant of x² -5x +5 is 25 - 20 = 5, which is positive, so two real roots. Therefore, total four real roots. Therefore, four solutions.

Therefore, all solutions are accounted for, and their y-coordinates are as calculated.

Therefore, the product is 1736.

But let me check once more with another approach.

Suppose I consider the system:

y = x² - 8

y² = -5x + 44

Let me solve for x from the second equation.

From the second equation: y² + 5x = 44 => 5x = 44 - y² => x = (44 - y²)/5

Then substitute into the first equation:

y = [(44 - y²)/5]^2 - 8

So:

y = ( (44 - y²)^2 ) / 25 - 8

Multiply both sides by 25:

25y = (44 - y²)^2 - 200

Expand (44 - y²)^2:

= 44² - 2*44*y² + y⁴ = 1936 - 88y² + y⁴

Therefore:

25y = 1936 - 88y² + y⁴ - 200

Simplify constants: 1936 - 200 = 1736

So:

25y = y⁴ - 88y² + 1736

Bring all terms to one side:

y⁴ - 88y² -25y + 1736 = 0

So we have a quartic equation in y: y⁴ -88y² -25y +1736 = 0

Now, if we can find the product of the roots of this equation, which would be the product of all y-coordinates.

But for a quartic equation y⁴ + a y³ + b y² + c y + d = 0, the product of the roots is d if the leading coefficient is 1. However, in our case, the equation is y⁴ -88y² -25y +1736 = 0. So written as y⁴ + 0y³ -88y² -25y +1736 = 0.

Therefore, the product of the roots is the constant term divided by the leading coefficient, which is 1736 / 1 = 1736.

But wait, the product of all roots of the quartic equation in y is 1736. However, the question says "the product of the y-coordinates of all the distinct solutions".

But are all roots of this quartic equation distinct? Let me check.

Earlier, we found four distinct solutions for x, which correspond to four distinct y-values. Therefore, the quartic equation in y should have four distinct roots, which are the y-coordinates. Therefore, the product of the roots is 1736.

Therefore, this confirms again that the product is 1736.

Therefore, regardless of the method, the answer is 1736.

Therefore, I can be confident that 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:

$$
(x^2 - 8)^2 = x^4 - 16x^2 + 64
$$

So the equation becomes:

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

Bring all terms to one side:

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

---

### Step 2: Factor the Quartic Equation

We attempt to factor the quartic:

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

Using the Rational Root Theorem, we test possible rational roots and find that $ x = -1 $ and $ x = -4 $ are roots. Using synthetic or polynomial division, we factor the quartic as:

$$
(x + 1)(x + 4)(x^2 - 5x + 5) = 0
$$

The quadratic $ x^2 - 5x + 5 $ has roots:

$$
x = \frac{5 \pm \sqrt{5}}{2}
$$

So the four real solutions for $ x $ are:

$$
x = -1, \quad x = -4, \quad x = \frac{5 + \sqrt{5}}{2}, \quad x = \frac{5 - \sqrt{5}}{2}
$$

---

### Step 3: Find Corresponding $ y $-Coordinates

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

- For $ x = -1 $: $ y = (-1)^2 - 8 = -7 $
- For $ x = -4 $: $ y = (-4)^2 - 8 = 8 $
- For $ x = \frac{5 + \sqrt{5}}{2} $:
  $$
  y = \left( \frac{5 + \sqrt{5}}{2} \right)^2 - 8 = \frac{30 + 10\sqrt{5}}{4} - 8 = \frac{-1 + 5\sqrt{5}}{2}
  $$
- 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 $ y $-coordinates are:

$$
y_1 = -7, \quad y_2 = 8, \quad y_3 = \frac{-1 + 5\sqrt{5}}{2}, \quad y_4 = \frac{-1 - 5\sqrt{5}}{2}
$$

---

### Step 4: Compute the Product of the $ y $-Coordinates

We compute:

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

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:

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

---

### Step 5: Verification via Vieta's Formula

Alternatively, we can consider the quartic in $ y $, derived by substituting $ x = \frac{44 - y^2}{5} $ into the first equation. This leads to:

$$
y^4 - 88y^2 - 25y + 1736 = 0
$$

This is a quartic in $ y $, and the product of all its roots (by Vieta's formula) is the constant term divided by the leading coefficient:

$$
\text{Product of roots} = \frac{1736}{1} = 1736
$$

This confirms our earlier result.

---

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