Prove that difference of squares of two distinct odd natural numbers is always a multiple of 8.

Options

A. 4
B. 3
C. 6
D. 8

Answer

To prove that the difference of squares of two distinct odd natural numbers is always a multiple of 8, we can follow these steps:

### Step 1: Represent the odd natural numbers
Let the two distinct odd natural numbers be represented as:
- \( a = 2m + 1 \)
- \( b = 2n + 1 \)

where \( m \) and \( n \) are natural numbers, and \( m \neq n \).

### Step 2: Calculate the difference of squares
We need to find the difference of squares of these two numbers:
\[ a^2 - b^2 = (2m + 1)^2 - (2n + 1)^2 \]

### Step 3: Use the difference of squares formula
Using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \):
\[ a^2 - b^2 = [(2m + 1) - (2n + 1)][(2m + 1) + (2n + 1)] \]
This simplifies to:
\[ = (2m - 2n)(2m + 2n + 2) \]

### Step 4: Factor out common terms
Factoring out the common terms:
\[ = 2(m - n) \cdot 2(2m + 2n + 2) = 4(m - n)(m + n + 1) \]

### Step 5: Analyze the expression
Now we need to show that \( 4(m - n)(m + n + 1) \) is a multiple of 8.

### Step 6: Determine the parity of the factors
Since \( m \) and \( n \) are distinct natural numbers, \( m - n \) is a non-zero integer. The expression \( m + n + 1 \) will always be an integer.

1. **Case 1**: If \( m - n \) is even, then \( (m - n) \) contributes a factor of 2, making \( 4(m - n)(m + n + 1) \) a multiple of 8.
2. **Case 2**: If \( m - n \) is odd, then \( (m + n + 1) \) must be even (since the sum of an odd number and an even number is even). Thus, \( 4(m - n)(m + n + 1) \) is still a multiple of 8.

### Conclusion
In both cases, we conclude that the difference of squares \( a^2 - b^2 \) is always a multiple of 8.

Thus, we have proved that the difference of squares of two distinct odd natural numbers is always a multiple of 8.