How many positive integral solutions exist for : ab+cd = a+b+c+d, where `1 le a le b le c le d`?

Options

A. (A) 0
B. (B) 2
C. (C) 3
D. (D) none of these

Answer

To solve the equation \( ab + cd = a + b + c + d \) under the conditions \( 1 \leq a \leq b \leq c \leq d \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ ab + cd - a - b - c - d = 0 \]
This can be rewritten as:
\[ ab - a - b + cd - c - d = 0 \]

### Step 2: Factoring the Terms
We can factor the terms involving \( a \) and \( b \):
\[ a(b - 1) + c(d - 1) - (b + d) = 0 \]
This implies:
\[ a(b - 1) + c(d - 1) = b + d \]

### Step 3: Introducing New Variables
Let us introduce new variables:
\[ x = a - 1, \quad y = b - 1, \quad z = c - 1, \quad w = d - 1 \]
Then the equation becomes:
\[ (x + 1)(y + 1) + (z + 1)(w + 1) = (x + 1) + (y + 1) + (z + 1) + (w + 1) \]
Expanding this gives:
\[ xy + x + y + zw + z + w = x + y + z + w + 4 \]
Simplifying, we find:
\[ xy + zw = 4 \]

### Step 4: Finding Positive Integral Solutions
Now we need to find pairs of non-negative integers \( (x, y) \) and \( (z, w) \) such that:
\[ xy + zw = 4 \]
We can consider the possible pairs for \( (xy, zw) \):
1. \( xy = 0 \) and \( zw = 4 \)
2. \( xy = 1 \) and \( zw = 3 \)
3. \( xy = 2 \) and \( zw = 2 \)
4. \( xy = 3 \) and \( zw = 1 \)
5. \( xy = 4 \) and \( zw = 0 \)

### Step 5: Analyzing Each Case
1. **Case 1:** \( xy = 0 \) implies \( x = 0 \) or \( y = 0 \).
- If \( x = 0 \), \( y \) can be \( 0, 1, 2, 3, 4 \) (5 solutions).
- If \( y = 0 \), \( x \) can be \( 0, 1, 2, 3, 4 \) (5 solutions).
- Total for this case: 5.

2. **Case 2:** \( xy = 1 \) implies \( (x, y) = (1, 1) \) (1 solution).
- \( zw = 3 \) has pairs: \( (1, 3), (3, 1) \) (2 solutions).
- Total for this case: 1 x 2 = 2.

3. **Case 3:** \( xy = 2 \) has pairs: \( (1, 2), (2, 1) \) (2 solutions).
- \( zw = 2 \) has pairs: \( (1, 2), (2, 1) \) (2 solutions).
- Total for this case: 2 x 2 = 4.

4. **Case 4:** \( xy = 3 \) has pairs: \( (1, 3), (3, 1) \) (2 solutions).
- \( zw = 1 \) has pairs: \( (1, 1) \) (1 solution).
- Total for this case: 2 x 1 = 2.

5. **Case 5:** \( xy = 4 \) implies \( (x, y) = (4, 0) \) or \( (0, 4) \) (2 solutions).
- \( zw = 0 \) implies \( z = 0 \) or \( w = 0 \) (1 solution).
- Total for this case: 2.

### Step 6: Summing All Cases
Now, we sum the solutions from all cases:
\[ 5 + 2 + 4 + 2 + 2 = 15 \]
Thus, the total number of positive integral solutions is **15**.