The difference between the squares of two consecutive odd integers is always…

Concept

For two numbers a and b, the difference of their squares factors as a2 − b2 = (a − b)(a + b). When a and b are consecutive odd integers, they differ by exactly 2, and the sum of two consecutive odd integers is always a multiple of 4. In a question that asks what an expression is 'always divisible by' among several offered numeric options, the intended answer is the largest guaranteed common divisor — a smaller true divisor of that same guaranteed factor is not the exact answer being sought.

Applying it here

1. Let the two consecutive odd integers be 2n + 1 and 2n + 3, where n is an integer.

2. Their difference of squares is (2n + 3)2 − (2n + 1)2.

3. Using a2 − b2 = (a − b)(a + b): difference = [(2n + 3) − (2n + 1)] × [(2n + 3) + (2n + 1)] = 2 × (4n + 4).

4. Simplifying, 2 × 4(n + 1) = 8(n + 1).

5. Since n + 1 is always an integer, 8(n + 1) is always a multiple of 8 — and 8 is the largest number that is guaranteed to divide every such difference.

Cross-check

Testing with actual pairs confirms it: 52 − 32 = 16, 92 − 72 = 32, and 112 − 92 = 40 — every result is a multiple of 8. None of these results is divisible by 6 or by 12 (16 fails both tests), which rules those two options out. 4 also divides every one of these results, but only because 8 does — 4 is a weaker consequence of the same guaranteed factor, not the tightest one, so among the offered options 8 is the precise, largest guaranteed divisor and the intended answer.