To solve the problem of finding the number of quadruplets of positive integers \((a, b, c, d)\) that satisfy the conditions \(1 \leq a \leq b \leq c \leq d\) and \(ab + cd = a + b + c + d + 3\), we can follow these steps:
### Step 1: Rewrite the equation
We start with the equation:
\[
ab + cd = a + b + c + d + 3
\]
Rearranging gives:
\[
ab + cd - a - b - c - d = 3
\]
### Step 2: Substitute values
To simplify the problem, we can assume \(a = x\), \(b = x + 1\), \(c = x + 2\), and \(d = x + 3\) for some positive integer \(x\). This satisfies the condition \(1 \leq a \leq b \leq c \leq d\).
### Step 3: Calculate \(ab\) and \(cd\)
Now, we calculate:
\[
ab = x(x + 1) = x^2 + x
\]
\[
cd = (x + 2)(x + 3) = x^2 + 5x + 6
\]
Thus, we have:
\[
ab + cd = (x^2 + x) + (x^2 + 5x + 6) = 2x^2 + 6x + 6
\]
### Step 4: Calculate \(a + b + c + d\)
Next, we calculate:
\[
a + b + c + d = x + (x + 1) + (x + 2) + (x + 3) = 4x + 6
\]
### Step 5: Set up the equation
Substituting these into our rearranged equation gives:
\[
2x^2 + 6x + 6 = 4x + 6 + 3
\]
This simplifies to:
\[
2x^2 + 6x + 6 = 4x + 9
\]
Rearranging yields:
\[
2x^2 + 2x - 3 = 0
\]
### Step 6: Solve the quadratic equation
Dividing the entire equation by 2 gives:
\[
x^2 + x - \frac{3}{2} = 0
\]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
x = \frac{-1 \pm \sqrt{1 + 6}}{2} = \frac{-1 \pm \sqrt{7}}{2}
\]
### Step 7: Determine the values of \(x\)
Calculating the roots:
\[
x_1 = \frac{-1 + \sqrt{7}}{2}, \quad x_2 = \frac{-1 - \sqrt{7}}{2}
\]
Since \(x\) must be a positive integer, we only consider \(x_1\):
\[
\sqrt{7} \approx 2.64575 \implies x_1 \approx \frac{-1 + 2.64575}{2} \approx 0.822875
\]
This value is not a positive integer.
### Step 8: Check for integer solutions
Since \(x\) must be a positive integer, we check for integer values of \(x\) starting from 1.
### Step 9: Testing integer values
1. For \(x = 1\):
- \(a = 1\), \(b = 2\), \(c = 3\), \(d = 4\)
- Check: \(1 \cdot 2 + 3 \cdot 4 = 2 + 12 = 14\) and \(1 + 2 + 3 + 4 + 3 = 13\) (not valid)
2. For \(x = 2\):
- \(a = 2\), \(b = 3\), \(c = 4\), \(d = 5\)
- Check: \(2 \cdot 3 + 4 \cdot 5 = 6 + 20 = 26\) and \(2 + 3 + 4 + 5 + 3 = 17\) (not valid)
3. For \(x = 3\):
- \(a = 3\), \(b = 4\), \(c = 5\), \(d = 6\)
- Check: \(3 \cdot 4 + 5 \cdot 6 = 12 + 30 = 42\) and \(3 + 4 + 5 + 6 + 3 = 21\) (not valid)
Continuing this process, we find that there are no integer solutions that satisfy the original equation.
### Conclusion
After testing various values of \(x\), we conclude that there are no quadruplets of positive integers \((a, b, c, d)\) that satisfy the given conditions.
