Let $f : Z_{\ge 0} \to Z_{\ge 0}$ satisfy the functional equation $$f(m^2 + n^2) =(f(m) - f(n))^2 + f(2mn)$$ for all nonnegative integers $m, n$. If $8f(0) + 9f(1) = 2006$, compute $f(0)$.
1. Given the functional equation:
   \[
   f(m^2 + n^2) = (f(m) - f(n))^2 + f(2mn)
   \]
   for all nonnegative integers \(m\) and \(n\), and the condition:
   \[
   8f(0) + 9f(1) = 2006
   \]

2. Let's start by setting \(m = 0\) and \(n = 1\):
   \[
   f(0^2 + 1^2) = (f(0) - f(1))^2 + f(2 \cdot 0 \cdot 1)
   \]
   Simplifying, we get:
   \[
   f(1) = (f(0) - f(1))^2 + f(0)
   \]

3. Let \(x = f(0)\) and \(y = f(1)\). Then the equation becomes:
   \[
   y = (x - y)^2 + x
   \]

4. Expanding and rearranging the equation:
   \[
   y = x^2 - 2xy + y^2 + x
   \]
   \[
   y - y^2 + 2xy - x^2 - x = 0
   \]
   \[
   y^2 - 2xy + y - x^2 - x = 0
   \]

5. This is a quadratic equation in \(y\):
   \[
   y^2 - (2x + 1)y + (x^2 + x) = 0
   \]

6. Solving this quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -(2x + 1)\), and \(c = x^2 + x\):
   \[
   y = \frac{2x + 1 \pm \sqrt{(2x + 1)^2 - 4(x^2 + x)}}{2}
   \]
   \[
   y = \frac{2x + 1 \pm \sqrt{4x^2 + 4x + 1 - 4x^2 - 4x}}{2}
   \]
   \[
   y = \frac{2x + 1 \pm \sqrt{1}}{2}
   \]
   \[
   y = \frac{2x + 1 \pm 1}{2}
   \]

7. This gives us two possible solutions:
   \[
   y = \frac{2x + 2}{2} = x + 1
   \]
   or
   \[
   y = \frac{2x}{2} = x
   \]

8. We have two cases:
   - \(f(1) = f(0) + 1\)
   - \(f(1) = f(0)\)

9. Given the condition \(8f(0) + 9f(1) = 2006\), let's analyze both cases:
   - If \(f(1) = f(0) + 1\):
     \[
     8f(0) + 9(f(0) + 1) = 2006
     \]
     \[
     8f(0) + 9f(0) + 9 = 2006
     \]
     \[
     17f(0) + 9 = 2006
     \]
     \[
     17f(0) = 1997
     \]
     \[
     f(0) = \frac{1997}{17} \approx 117.47
     \]
     Since \(f(0)\) must be an integer, this case is not possible.

   - If \(f(1) = f(0)\):
     \[
     8f(0) + 9f(0) = 2006
     \]
     \[
     17f(0) = 2006
     \]
     \[
     f(0) = \frac{2006}{17} = 118
     \]

10. Therefore, the only valid solution is:
    \[
    f(0) = 118
    \]

The final answer is \(\boxed{118}\)