Find all pairs of integers $a, b$ such that the following system of equations has a unique integral solution $(x , y , z )$ : 
 $\begin{cases}x + y = a - 1 \\
x(y + 1) - z^2 = b \end{cases}$
1. **Identify the system of equations:**
   \[
   \begin{cases}
   x + y = a - 1 \\
   x(y + 1) - z^2 = b 
   \end{cases}
   \]

2. **Analyze the symmetry in the solutions:**
   Notice that if \((x, y, z)\) is a solution, then \((y + 1, x - 1, z)\) and \((x, y, -z)\) are also solutions. For the system to have a unique solution, \(z\) must be zero, and \(y\) must be \(x - 1\).

3. **Assume the unique solution has \(z = 0\) and \(y = x - 1\):**
   Let the value of \(x\) in the unique solution be \(k\). Then:
   \[
   y = k - 1
   \]
   \[
   a = 2k
   \]
   \[
   b = k^2
   \]

4. **Verify that these values of \(a\) and \(b\) give a unique solution:**
   Substitute \(x = k\) and \(y = k - 1\) into the first equation:
   \[
   x + y = k + (k - 1) = 2k - 1
   \]
   This matches \(a - 1\), so \(a = 2k\).

5. **Substitute into the second equation:**
   \[
   x(y + 1) - z^2 = k((k - 1) + 1) - 0^2 = k \cdot k = k^2
   \]
   This matches \(b\), so \(b = k^2\).

6. **Rearrange the second equation to confirm the unique solution:**
   Substitute \(y + 1 = 2k - x\) into the second equation:
   \[
   x(2k - x) - z^2 = k^2
   \]
   Rearrange to:
   \[
   x(2k - x) = k^2 + z^2
   \]
   \[
   2kx - x^2 = k^2 + z^2
   \]
   \[
   x^2 - 2kx + k^2 + z^2 = 0
   \]
   \[
   (x - k)^2 + z^2 = 0
   \]
   Since \((x - k)^2 + z^2 = 0\), both terms must be zero:
   \[
   x = k
   \]
   \[
   z = 0
   \]
   Therefore, \(y = k - 1\).

7. **Conclusion:**
   The only possible values of \(a\) and \(b\) that give a unique solution are of the form \(a = 2k\) and \(b = k^2\) for some integer \(k\). This can be expressed as:
   \[
   b = \frac{a^2}{4}
   \]

The final answer is \( \boxed{ b = \frac{a^2}{4} } \).