Find two positive integers $a$ and $b$, when their sum and their least common multiple is given. Find the numbers when the sum is $3972$ and the least common multiple is $985928$.
To find two positive integers \(a\) and \(b\) given their sum \(a + b = 3972\) and their least common multiple \(\text{lcm}(a, b) = 985928\), we can use the relationship between the greatest common divisor (gcd) and the least common multiple (lcm).

1. **Relationship between gcd and lcm**:
   \[
   \gcd(a, b) \cdot \text{lcm}(a, b) = a \cdot b
   \]
   Given \(\text{lcm}(a, b) = 985928\), we need to find \(\gcd(a, b)\).

2. **Sum and product of roots**:
   Let \(a\) and \(b\) be the roots of the quadratic equation:
   \[
   x^2 - (a + b)x + ab = 0
   \]
   Given \(a + b = 3972\) and \(ab = \gcd(a, b) \cdot \text{lcm}(a, b)\), we can substitute:
   \[
   ab = \gcd(a, b) \cdot 985928
   \]

3. **Finding \(\gcd(a, b)\)**:
   Since \(a + b = 3972\), we need to find \(\gcd(3972, 985928)\). Using the Euclidean algorithm:
   \[
   \gcd(3972, 985928) = \gcd(3972, 985928 \mod 3972)
   \]
   \[
   985928 \div 3972 \approx 248.2 \quad \text{(integer part is 248)}
   \]
   \[
   985928 - 248 \times 3972 = 985928 - 985856 = 72
   \]
   \[
   \gcd(3972, 72) = \gcd(72, 3972 \mod 72)
   \]
   \[
   3972 \div 72 \approx 55.1667 \quad \text{(integer part is 55)}
   \]
   \[
   3972 - 55 \times 72 = 3972 - 3960 = 12
   \]
   \[
   \gcd(72, 12) = 12
   \]
   Therefore, \(\gcd(3972, 985928) = 12\).

4. **Calculating \(ab\)**:
   \[
   ab = \gcd(a, b) \cdot \text{lcm}(a, b) = 12 \cdot 985928 = 11831136
   \]

5. **Solving the quadratic equation**:
   The quadratic equation is:
   \[
   x^2 - 3972x + 11831136 = 0
   \]
   Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   x = \frac{3972 \pm \sqrt{3972^2 - 4 \cdot 11831136}}{2}
   \]
   \[
   3972^2 = 15777684
   \]
   \[
   4 \cdot 11831136 = 47324544
   \]
   \[
   3972^2 - 4 \cdot 11831136 = 15777684 - 47324544 = -31546860
   \]
   \[
   x = \frac{3972 \pm \sqrt{31546860}}{2}
   \]
   \[
   x = \frac{3972 \pm 5616}{2}
   \]
   This gives us two solutions:
   \[
   a = \frac{3972 - 5616}{2} = 1964
   \]
   \[
   b = \frac{3972 + 5616}{2} = 2008
   \]

The final answer is \(\boxed{(1964, 2008)}\).