Which of the following numbers always divides the difference between the squares of two consecutive odd integers?

Options

A. 7
B. 3
C. 8
D. 6

Answer

To find which number always divides the difference between the squares of two consecutive odd integers, we can follow these steps:

### Step 1: Define Consecutive Odd Integers
Let the two consecutive odd integers be represented as \( n \) and \( n + 2 \), where \( n \) is an odd integer.

### Step 2: Calculate the Squares
Now, we calculate the squares of these integers:
- The square of the first odd integer \( n \) is \( n^2 \).
- The square of the second odd integer \( n + 2 \) is \( (n + 2)^2 \).

### Step 3: Find the Difference of Squares
Next, we find the difference between the squares:
\[ (n + 2)^2 - n^2 \]

### Step 4: Expand the Expression
Now, we expand the expression:
\[ (n + 2)^2 = n^2 + 4n + 4 \]
Thus,
\[ (n + 2)^2 - n^2 = (n^2 + 4n + 4) - n^2 = 4n + 4 \]

### Step 5: Factor the Result
We can factor out the common term:
\[ 4n + 4 = 4(n + 1) \]

### Step 6: Conclusion
From the factorization, we see that the difference between the squares of two consecutive odd integers is always divisible by 4. Therefore, the number that always divides the difference between the squares of two consecutive odd integers is **4**.