Determine all solutions of the diophantine equation $a^2 = b \cdot (b + 7)$ in integers $a\ge 0$ and $b \ge 0$.

(W. Janous, Innsbruck)
1. We start with the given diophantine equation:
   \[
   a^2 = b \cdot (b + 7)
   \]
   We can rewrite this equation as a quadratic in \( b \):
   \[
   b^2 + 7b - a^2 = 0
   \]

2. To solve for \( b \), we use the quadratic formula:
   \[
   b = \frac{-7 \pm \sqrt{49 + 4a^2}}{2}
   \]
   For \( b \) to be an integer, the discriminant \( 49 + 4a^2 \) must be a perfect square. Let \( m \) be an integer such that:
   \[
   49 + 4a^2 = m^2
   \]

3. Rearrange the equation to:
   \[
   m^2 - 4a^2 = 49
   \]
   This is a difference of squares, which can be factored as:
   \[
   (m - 2a)(m + 2a) = 49
   \]

4. We now consider the factor pairs of 49:
   \[
   (1, 49), (-1, -49), (7, 7), (-7, -7)
   \]
   Since \( a \geq 0 \) and \( b \geq 0 \), we only consider the positive factor pairs:
   \[
   (1, 49) \quad \text{and} \quad (7, 7)
   \]

5. For the factor pair \( (1, 49) \):
   \[
   m - 2a = 1 \quad \text{and} \quad m + 2a = 49
   \]
   Adding these equations:
   \[
   2m = 50 \implies m = 25
   \]
   Subtracting these equations:
   \[
   4a = 48 \implies a = 12
   \]

6. For the factor pair \( (7, 7) \):
   \[
   m - 2a = 7 \quad \text{and} \quad m + 2a = 7
   \]
   Adding these equations:
   \[
   2m = 14 \implies m = 7
   \]
   Subtracting these equations:
   \[
   4a = 0 \implies a = 0
   \]

7. We now check the corresponding values of \( b \):
   - For \( a = 12 \):
     \[
     b = \frac{-7 + \sqrt{49 + 4 \cdot 12^2}}{2} = \frac{-7 + \sqrt{625}}{2} = \frac{-7 + 25}{2} = 9
     \]
   - For \( a = 0 \):
     \[
     b = \frac{-7 + \sqrt{49}}{2} = \frac{-7 + 7}{2} = 0
     \]

8. Therefore, the solutions in non-negative integers are:
   \[
   (a, b) = (12, 9) \quad \text{and} \quad (0, 0)
   \]

The final answer is \(\boxed{(12, 9) \text{ and } (0, 0)}\)