Consecutive Odd Squares — Difference Property Which number always divides the difference between the squares of two consecutive odd integers?

Introduction / Context:
This question probes number properties and the algebra of squares. For consecutive odd integers, their separation is 2. Squaring and subtracting reveals a fixed factorization that exposes guaranteed divisibility by a specific integer regardless of which odd numbers are chosen.

Given Data / Assumptions:

• Let the two consecutive odd integers be n and n+2.
• n is odd, so n+1 is even.
• We analyze (n+2)^2 - n^2.

Concept / Approach:
Use the identity a^2 - b^2 = (a - b)(a + b). Here, a - b = 2 and a + b = 2n + 2 = 2(n + 1). Hence the difference equals 2 * 2(n + 1) = 4(n + 1). Since n is odd, n + 1 is even, so write n + 1 = 2k. Then 4(n + 1) = 4 * 2k = 8k, which is divisible by 8 for all choices of n.

Step-by-Step Solution:
1) Compute difference: (n+2)^2 - n^2 = 4n + 4.2) Factor: 4n + 4 = 4(n + 1).3) Since n is odd, let n + 1 = 2k, giving 4 * 2k = 8k.4) Therefore the difference is always a multiple of 8.

Verification / Alternative check:
Try sample pairs: 3 and 5 → 25 - 9 = 16 (divisible by 8). 7 and 9 → 81 - 49 = 32 (divisible by 8). The pattern holds universally.

Why Other Options Are Wrong:
7 and 5 do not always divide the result; 3 and 6 divide some cases but not all. Only 8 is guaranteed for every pair of consecutive odd integers.

Common Pitfalls:
Forgetting that “consecutive odd” means a gap of 2; applying the result to consecutive integers (gap of 1) which yields a different divisibility pattern.

Final Answer:
8