Prove that the system of equations
\[2yz+x-y-z=a,\\ 2xz-x+y-z=a,\\ 2xy-x-y+z=a, \]
$a$ being a parameter, cannot have five distinct solutions. For what values of $a$ does this system have four distinct integer solutions?
To prove that the system of equations
\[2yz + x - y - z = a,\]
\[2xz - x + y - z = a,\]
\[2xy - x - y + z = a,\]
cannot have five distinct solutions, we will analyze the system step by step.

1. **Rewrite the equations:**
   \[
   \begin{cases}
   2yz + x - y - z = a \quad \text{(1)} \\
   2xz - x + y - z = a \quad \text{(2)} \\
   2xy - x - y + z = a \quad \text{(3)}
   \end{cases}
   \]

2. **Subtract equation (1) from equation (2):**
   \[
   (2xz - x + y - z) - (2yz + x - y - z) = 0
   \]
   Simplifying, we get:
   \[
   2xz - 2yz - 2x + 2y = 0
   \]
   \[
   2(xz - yz - x + y) = 0
   \]
   \[
   xz - yz - x + y = 0
   \]
   \[
   (x - y)(z - 1) = 0
   \]
   This implies either \(x = y\) or \(z = 1\).

3. **Subtract equation (2) from equation (3):**
   \[
   (2xy - x - y + z) - (2xz - x + y - z) = 0
   \]
   Simplifying, we get:
   \[
   2xy - 2xz - 2y + 2z = 0
   \]
   \[
   2(xy - xz - y + z) = 0
   \]
   \[
   xy - xz - y + z = 0
   \]
   \[
   (y - z)(x - 1) = 0
   \]
   This implies either \(y = z\) or \(x = 1\).

4. **Subtract equation (1) from equation (3):**
   \[
   (2xy - x - y + z) - (2yz + x - y - z) = 0
   \]
   Simplifying, we get:
   \[
   2xy - 2yz - 2x + 2z = 0
   \]
   \[
   2(xy - yz - x + z) = 0
   \]
   \[
   xy - yz - x + z = 0
   \]
   \[
   (x - z)(y - 1) = 0
   \]
   This implies either \(x = z\) or \(y = 1\).

5. **Analyze the implications:**
   From the above steps, we have the following cases:
   - \(x = y\)
   - \(y = z\)
   - \(z = 1\)
   - \(x = 1\)
   - \(y = 1\)
   - \(x = z\)

   These cases imply that the variables \(x, y, z\) are not independent and must satisfy certain relationships. Therefore, it is impossible to have five distinct solutions.

6. **Find the values of \(a\) for four distinct integer solutions:**
   Consider the solutions \((1, 1, a)\), \((1, a, 1)\), \((a, 1, 1)\), and \((t, t, t)\), where \(t = \frac{1 \pm \sqrt{1 + 8a}}{4}\).

   For these to be distinct integer solutions, \(t\) must be an integer. This requires \(1 + 8a\) to be a perfect square, say \(k^2\), where \(k\) is an integer:
   \[
   1 + 8a = k^2
   \]
   \[
   8a = k^2 - 1
   \]
   \[
   a = \frac{k^2 - 1}{8}
   \]

   For \(a\) to be an integer, \(k^2 - 1\) must be divisible by 8. The possible values of \(k\) that satisfy this condition are \(k = 1, 3, 5, 7, \ldots\).

   Therefore, the values of \(a\) are:
   \[
   a = \frac{1^2 - 1}{8} = 0, \quad a = \frac{3^2 - 1}{8} = 1, \quad a = \frac{5^2 - 1}{8} = 3, \quad a = \frac{7^2 - 1}{8} = 6, \ldots
   \]

The final answer is \( \boxed{ a = 0, 1, 3, 6, \ldots } \)