The difference of the squares of two consecutive odd integers is divisible by which of the following integers

To solve this problem, let's first understand what is meant by the difference of the squares of two consecutive odd integers.

Let's assume the first odd integer is represented by the variable n. Since it is an odd integer, we can express it as 2k + 1, where k is an integer.

The next consecutive odd integer can be represented as n + 2. So, the second odd integer can be expressed as 2k + 1 + 2, which simplifies to 2k + 3.

Now, let's find the difference of the squares of these two consecutive odd integers.

The square of the first odd integer (n) is (2k + 1)^2, which simplifies to (4k^2 + 4k + 1).

The square of the second odd integer (n + 2) is (2k + 3)^2, which simplifies to (4k^2 + 12k + 9).

To find the difference, we subtract the second square from the first square:

(4k^2 + 4k + 1) - (4k^2 + 12k + 9)

Simplifying this expression, we get:

-8k - 8

Now, we need to determine which integers can divide this expression.

The expression -8k - 8 can be factored as -8(k + 1).

From this, we can see that -8 is divisible by 8, and (k + 1) is divisible by (k + 1) for any integer value of k.

Therefore, the difference of the squares of two consecutive odd integers is divisible by 8.