Suppose $a, b$ are integers and $a+b$ is a root of $x^2 +ax+b = 0$. What is the maximum possible value of $b^2$?
1. We are given that \(a + b\) is a root of the quadratic equation \(x^2 + ax + b = 0\). Therefore, we can substitute \(x = a + b\) into the equation:
   \[
   (a + b)^2 + a(a + b) + b = 0
   \]
   Expanding and simplifying, we get:
   \[
   a^2 + 2ab + b^2 + a^2 + ab + b = 0
   \]
   \[
   2a^2 + 3ab + b^2 + b = 0
   \]

2. This is a quadratic equation in terms of \(a\). For this equation to have integer solutions, its discriminant must be a perfect square. The discriminant \(\Delta\) of the quadratic equation \(2a^2 + 3ab + b^2 + b = 0\) is given by:
   \[
   \Delta = (3b)^2 - 4 \cdot 2 \cdot (b^2 + b)
   \]
   \[
   \Delta = 9b^2 - 8(b^2 + b)
   \]
   \[
   \Delta = 9b^2 - 8b^2 - 8b
   \]
   \[
   \Delta = b^2 - 8b
   \]

3. For the discriminant to be a perfect square, we set:
   \[
   b^2 - 8b = k^2
   \]
   for some nonnegative integer \(k\). Rearranging, we get:
   \[
   (b - 4)^2 - k^2 = 16
   \]
   This can be factored as:
   \[
   (b - 4 - k)(b - 4 + k) = 16
   \]

4. We now solve for \(b\) and \(k\) by considering the factor pairs of 16:
   \[
   \begin{cases}
   b - 4 - k = -16 \\
   b - 4 + k = -1
   \end{cases}
   \]
   \[
   \begin{cases}
   b - 4 - k = -8 \\
   b - 4 + k = -2
   \end{cases}
   \]
   \[
   \begin{cases}
   b - 4 - k = -4 \\
   b - 4 + k = -4
   \end{cases}
   \]
   \[
   \begin{cases}
   b - 4 - k = 1 \\
   b - 4 + k = 16
   \end{cases}
   \]
   \[
   \begin{cases}
   b - 4 - k = 2 \\
   b - 4 + k = 8
   \end{cases}
   \]
   \[
   \begin{cases}
   b - 4 - k = 4 \\
   b - 4 + k = 4
   \end{cases}
   \]

5. Solving these systems, we get the following pairs \((b, k)\):
   \[
   \begin{cases}
   b - 4 - k = -8 \\
   b - 4 + k = -2
   \end{cases}
   \Rightarrow b = 0, k = 0
   \]
   \[
   \begin{cases}
   b - 4 - k = 1 \\
   b - 4 + k = 16
   \end{cases}
   \Rightarrow b = 9, k = 3
   \]
   \[
   \begin{cases}
   b - 4 - k = 2 \\
   b - 4 + k = 8
   \end{cases}
   \Rightarrow b = 8, k = 0
   \]
   \[
   \begin{cases}
   b - 4 - k = -4 \\
   b - 4 + k = -4
   \end{cases}
   \Rightarrow b = -1, k = 3
   \]

6. We now substitute these values of \(b\) back into the original quadratic equation to check for integer solutions for \(a\):
   - For \(b = -1\):
     \[
     2a^2 - 3a = 0 \Rightarrow a(2a - 3) = 0 \Rightarrow a = 0 \text{ or } a = \frac{3}{2} \text{ (not an integer)}
     \]
   - For \(b = 0\):
     \[
     2a^2 = 0 \Rightarrow a = 0
     \]
   - For \(b = 9\):
     \[
     2a^2 + 27a + 90 = 0 \Rightarrow a = -6 \text{ (solving the quadratic equation)}
     \]
   - For \(b = 8\):
     \[
     2a^2 + 24a + 72 = 0 \Rightarrow a = -6 \text{ (solving the quadratic equation)}
     \]

7. The possible solutions are \((a, b) = (0, -1), (0, 0), (-6, 9), (-6, 8)\). Among these, the maximum possible value of \(b^2\) is:
   \[
   9^2 = \boxed{81}
   \]