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For any integer $n$, the expression $n^2 - 1$ is always divisible by 8 when $n$ is
Aany integer
Ban even integer
Ca prime number
Dan odd integer
Answer & Solution
Correct answer: D. an odd integer
If $n$ is odd, write $n = 2k+1$. Then $n^2 - 1 = 4k(k+1)$. Since $k(k+1)$ is a product of two consecutive integers it is even, so $4k(k+1)$ is divisible by 8. For even $n$, $n^2-1$ is odd, so not divisible by 8.
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