If n is an odd integer,then n^{2}-1 is divisi | vedclass

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(D) Any odd integer $n$ can be expressed in the form $2k+1$ where $k$ is an integer.Alternatively,any odd integer $n$ can be represented as $n = 4k \p

Vedclass · Accepted Answer

$8$

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(D) Any odd integer $n$ can be expressed in the form $2k+1$ where $k$ is an integer.Alternatively,any odd integer $n$ can be represented as $n = 4k \pm 1$ or simply by testing values.Let $n = 1$,then $n^{2}-1 = 1^{2}-1 = 0$ (divisible by $8$).Let $n = 3$,then $n^{2}-1 = 3^{2}-1 = 9-1 = 8$ (divisible by $8$).Let $n = 5$,then $n^{2}-1 = 5^{2}-1 = 25-1 = 24$ (divisible by $8$).General proof: Since $n$ is odd,$n = 2k+1$.$n^{2}-1 = (2k+1)^{2}-1 = 4k^{2}+4k+1-1 = 4k(k+1)$.Since $k(k+1)$ is the product of two consecutive integers,it is always even,i.e.,$k(k+1) = 2m$.Therefore,$n^{2}-1 = 4(2m) = 8m$.Thus,$n^{2}-1$ is always divisible by $8$.

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