Suppose a, b are integers and a + b is a root of `x^2 + ax + b = 0`. What is the maximum possible values of `b^2` ?

Answer

To solve the problem, we need to find the maximum possible value of \( b^2 \) given that \( a + b \) is a root of the equation \( x^2 + ax + b = 0 \).

### Step-by-step Solution:

1. **Identify the given quadratic equation**:
The equation is \( x^2 + ax + b = 0 \).

2. **Substitute \( a + b \) as a root**:
Since \( a + b \) is a root, we substitute \( x = a + b \) into the equation:
\[ (a + b)^2 + a(a + b) + b = 0 \]

3. **Expand the equation**:
Expanding \( (a + b)^2 \):
\[ a^2 + 2ab + b^2 + a^2 + ab + b = 0 \]
Combine like terms:
\[ 2a^2 + 3ab + b^2 + b = 0 \]

4. **Rearrange the equation**:
Rearranging gives:
\[ b^2 + (3a + 1)b + 2a^2 = 0 \]
This is a quadratic equation in \( b \).

5. **Use the quadratic formula**:
The solutions for \( b \) can be found using the quadratic formula:
\[ b = \frac{-(3a + 1) \pm \sqrt{(3a + 1)^2 - 4 \cdot 1 \cdot 2a^2}}{2 \cdot 1} \]

6. **Simplify the discriminant**:
The discriminant \( D \) is:
\[ D = (3a + 1)^2 - 8a^2 \]
Simplifying this:
\[ D = 9a^2 + 6a + 1 - 8a^2 = a^2 + 6a + 1 \]

7. **Condition for \( b \) to be an integer**:
For \( b \) to be an integer, \( D \) must be a perfect square:
\[ a^2 + 6a + 1 = k^2 \quad \text{for some integer } k \]

8. **Rearranging gives a quadratic in \( a \)**:
Rearranging gives:
\[ a^2 + 6a + (1 - k^2) = 0 \]
The discriminant of this quadratic must also be a perfect square:
\[ 36 - 4(1 - k^2) = 4k^2 + 32 \]

9. **Setting \( 4k^2 + 32 \) to be a perfect square**:
Let \( m^2 = 4k^2 + 32 \) for some integer \( m \):
\[ m^2 - 4k^2 = 32 \quad \Rightarrow \quad (m - 2k)(m + 2k) = 32 \]

10. **Finding pairs of factors of 32**:
The pairs of factors of 32 are \( (1, 32), (2, 16), (4, 8) \) and their negatives. Solving for \( m \) and \( k \) gives possible values.

11. **Calculating values of \( b \)**:
Substitute back to find \( b \) for each valid \( a \) derived from the factor pairs.

12. **Finding maximum \( b^2 \)**:
After calculating possible values of \( b \) for valid integers \( a \), we find:
- For \( a = 0 \), \( b = 1 \) or \( -1 \) (values \( 0, 1 \)).
- For \( a = -6 \), \( b = 9 \) or \( 8 \) (values \( 8, 9 \)).

The maximum value of \( b \) is \( 9 \).

13. **Calculate \( b^2 \)**:
The maximum value of \( b^2 \):
\[ b^2 = 9^2 = 81 \]

### Conclusion:
The maximum possible value of \( b^2 \) is \( \boxed{81} \).