If n is an odd integer, then \(n^2 - 1\) is always divisible by:
Answer: D
The expression \(n^2 - 1\) can be factored as \((n-1)(n+1)\). Since n is an odd integer, both \(n-1\) and \(n+1\) are consecutive even integers. For example, if n=3, we get 2 and 4. If n=5, we get 4 and 6. The product of two consecutive even integers is always divisible by 8. This is because one of them must be a multiple of 4. Let the consecutive even integers be 2k and 2k+2. Their product is \(4k(k+1)\). Since either k or k+1 must be even, \(k(k+1)\) is always divisible by 2. Thus the product is divisible by \(4 \times 2 = 8\).