Problem 1412

1. Subtracting Equations:
We start by subtracting the second equation from the third equation:
$$(2a+4)x + (2a+2)y + (a^2+3)z - \left[(2a+4)x + (a^2+3)y + (2a+2)z\right] = (a+1) - (3a-1)$$
Simplifying, we get:
$$(2a+2-a^2-3)(y-z) = 2a-2$$
$$(a^2-2a+1)(y-z) = 2(a-1)$$
$$(a-1)^2(y-z) = 2(a-1)$$
Thus, either $a = 1$ or $y - z = \frac{2}{a-1}$.

2. **Finding Possible Values of $a$**:
For $y - z = \frac{2}{a-1}$ to be an integer, $\frac{2}{a-1}$ must be an integer. Therefore, $a-1$ must be a divisor of 2. The possible values of $a-1$ are $\pm 1, \pm 2$, leading to:
$$a = 0, 2, 3, -1$$
Including $a = 1$, we have $a \in \{-1, 0, 1, 2, 3\}$.

3. **Solving for Each $a$:
- Case $a = -1$**:
$$5x + y + z = -1$$
$$0x + 4y + 0z = -4$$
$$0x + 0y + 4z = 0$$
Solving, we get:
$$y = -1, z = 0, x = 0$$
$$(x, y, z) = (0, -1, 0)$$

- **Case $a = 0$**:
$$5x + 2y + 2z = 0$$
$$4x + 3y + 2z = -1$$
$$4x + 2y + 3z = 1$$
Solving, we get:
$$y = -1, z = 1, x = 0$$
$$(x, y, z) = (0, -1, 1)$$

- **Case $a = 1$**:
$$5x + 3y + 3z = 1$$
$$6x + 4y + 4z = 2$$
$$6x + 4y + 4z = 2$$
Solving, we get:
$$x = -1, y = 2 - t, z = t \quad \text{for all } t \in \mathbb{Z}$$
$$(x, y, z) = (-1, 2 - t, t)$$

- **Case $a = 2$**:
$$5x + 4y + 4z = 2$$
$$8x + 7y + 6z = 5$$
$$8x + 6y + 7z = 3$$
Solving, we get:
$$x = -6, y = 5, z = 3$$
$$(x, y, z) = (-6, 5, 3)$$

- **Case $a = 3$**:
$$5x + 5y + 5z = 3$$
$$10x + 12y + 8z = 8$$
$$10x + 8y + 12z = 4$$
Solving, we get:
$$x = -\frac{2}{5} - 2t, y = t + 1, z = t \quad \text{for all } t \in \mathbb{R}$$
This does not yield integer solutions.

Therefore, the integer solutions are:
- $a = -1$: $(x, y, z) = (0, -1, 0)$
- $a = 0$: $(x, y, z) = (0, -1, 1)$
- $a = 1$: $(x, y, z) = (-1, 2 - t, t)$ for all $t \in \mathbb{Z}$
- $a = 2$: $(x, y, z) = (-6, 5, 3)$

The final answer is: