The difference between the squares of any two consecutive odd integers is always divisible by which of the following numbers?

Introduction / Context:
This question examines patterns in the squares of consecutive odd integers. Recognizing such patterns is important for quickly answering number theory and algebra questions in aptitude exams. The goal is to determine a fixed number that always divides the difference between the squares of consecutive odd integers.

Given Data / Assumptions:
We consider two consecutive odd integers.Let the smaller odd integer be n and the next odd integer be n + 2.We are interested in the difference between their squares: (n + 2)^2 − n^2.

Concept / Approach:
Using algebraic expansion, we can find a general expression for the difference of squares of consecutive odd integers. Then, by factoring and simplifying, we look for common factors that always appear. Once we have the simplified expression, we can determine which of the given options is always a divisor.

Step-by-Step Solution:
Step 1: Let the consecutive odd integers be n and n + 2, where n is odd.Step 2: Compute the difference of squares: (n + 2)^2 − n^2.Step 3: Expand (n + 2)^2: it equals n^2 + 4n + 4.Step 4: Subtract n^2: (n^2 + 4n + 4) − n^2 = 4n + 4.Step 5: Factor the expression: 4n + 4 = 4(n + 1).Step 6: Since n is odd, n + 1 is even, so write n + 1 = 2k for some integer k.Step 7: Then 4(n + 1) = 4 * 2k = 8k.Step 8: Therefore the difference of the squares is always a multiple of 8.

Verification / Alternative check:
Test with concrete pairs. For 3 and 5, the difference of squares is 5^2 − 3^2 = 25 − 9 = 16, which is divisible by 8. For 7 and 9, we get 81 − 49 = 32, also divisible by 8. For 11 and 13, the difference is 169 − 121 = 48, again a multiple of 8. These examples confirm the general algebraic result.

Why Other Options Are Wrong:
Although the expression 8k is also divisible by 2 and 4, the question asks for the number by which the difference is always divisible, and among the options provided, 8 is the strongest such divisor. The option 6 is not guaranteed, because 8k need not be a multiple of 6 for all integers k. Therefore 6 is not always a divisor, while 8 always is.

Common Pitfalls:
Some learners mistakenly consider consecutive integers instead of consecutive odd integers, which leads to a different algebraic form. Others stop at 4(n + 1) and choose 4 as the answer, forgetting that n + 1 is even and contributes another factor 2. Careful use of the fact that n is odd is essential.

Final Answer:
The difference between the squares of any two consecutive odd integers is always divisible by 8.