Final answer:
The difference between the squares of two consecutive odd integers is calculated as 4(n+1); since n is an odd integer and n+1 is even, this difference is always divisible by 8.
Step-by-step explanation:
The question asks about the difference between the squares of two consecutive odd integers, and which number this difference is always divisible by. To solve this, let's denote the smaller odd integer as n, and since the integers are consecutive odd numbers, the next integer would be n + 2.
Now, the squares of these integers would be n2 and (n + 2)2 respectively. The difference between these squares is:
(n + 2)2 - n2 = (n2 + 4n + 4) - n2 = 4n + 4
The difference can be further simplified as 4(n + 1). Since n is an odd integer, n + 1 is an even integer, and thus, the difference is always a multiple of 8.
Therefore, the correct answer is D. 8. The difference between the squares of two consecutive odd integers is always divisible by 8.