For how many pairs of odd positive integers (a, b), both a, b less than 100, does the equation `x^2 + ax + b = 0` have integer roots?

Answer

To solve the problem, we need to determine how many pairs of odd positive integers \( (a, b) \), both less than 100, will make the quadratic equation \( x^2 + ax + b = 0 \) have integer roots.

### Step-by-Step Solution:

1. **Understanding the Roots of the Quadratic Equation**:
The roots of the quadratic equation \( x^2 + ax + b = 0 \) can be expressed using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -a \)
- The product of the roots \( \alpha \beta = b \)

2. **Conditions for Integer Roots**:
For \( \alpha \) and \( \beta \) to be integers:
- The product \( \alpha \beta = b \) must be an integer.
- The sum \( \alpha + \beta = -a \) must also be an integer.

3. **Identifying the Nature of \( a \) and \( b \)**:
Given that \( a \) and \( b \) are both odd positive integers less than 100:
- Odd integers can be represented as \( 1, 3, 5, \ldots, 99 \).

4. **Analyzing the Sum of Roots**:
Since \( a \) is odd, \( -a \) is also odd. The sum of two integers \( \alpha + \beta = -a \) being odd implies that one of the roots must be even and the other must be odd (since odd + odd = even and even + even = even).

5. **Analyzing the Product of Roots**:
The product \( \alpha \beta = b \) is odd (since \( b \) is odd). For the product of two integers to be odd, both integers must be odd.

6. **Contradiction**:
We have reached a contradiction:
- From the sum condition, we need one root to be even and the other to be odd.
- From the product condition, both roots must be odd.

7. **Conclusion**:
Since it is impossible for both conditions to be satisfied simultaneously, there are no pairs \( (a, b) \) of odd positive integers less than 100 such that the equation \( x^2 + ax + b = 0 \) has integer roots.

### Final Answer:
The number of pairs \( (a, b) \) is **0**.