n^{2}-1 is divisible by 8, if n is | vedclass

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(A) Let $a = n^{2} - 1$.Case $I$: If $n$ is an even integer,let $n = 2k$ for some integer $k$.Then $a = (2k)^{2} - 1 = 4k^{2} - 1$. This is always an

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an odd integer

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(A) Let $a = n^{2} - 1$.Case $I$: If $n$ is an even integer,let $n = 2k$ for some integer $k$.Then $a = (2k)^{2} - 1 = 4k^{2} - 1$. This is always an odd number,so it cannot be divisible by $8$.Case $II$: If $n$ is an odd integer,let $n = 2k + 1$ for some integer $k$.Then $a = (2k + 1)^{2} - 1 = (4k^{2} + 4k + 1) - 1 = 4k^{2} + 4k = 4k(k + 1)$.Since $k$ and $k + 1$ are consecutive integers,one of them must be even. Therefore,the product $k(k + 1)$ is divisible by $2$.Let $k(k + 1) = 2m$ for some integer $m$.Then $a = 4(2m) = 8m$.Since $a$ is a multiple of $8$,$n^{2} - 1$ is divisible by $8$ whenever $n$ is an odd integer.

$n^{2}-1$ is divisible by $8,$ if $n$ is

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