MCQ
$n^2-1$ is divisible by 8 , if $n$ is
  • A
    an integer
  • B
    a natural number
  • an odd integer
  • D
    an even integer

Answer

Correct option: C.
an odd integer
(C) an odd integer
If $n$ is an even integer, then so is $n^2$ and hence $n^2-1$ is an odd integer. Consequently, $n^2-1$ cannot be divisible by 8 , if $n$ is an even integer. So, let $n$ be an odd integer. Then, $n=2 m+1$ for some integer $m$.
$
\therefore \quad n^2-1=(2 m+1)^2-1=4 m^2+4 m=4 m(m+1)
$
For any natural number $m, m(m+1)$ is an even natural number. Let $m=2 k$, where $k$ is a natural number.
$\therefore \quad n^2-1=8 k(2 k+1)$, which is divisible by 8 .
Hence, $n$ is an odd integer.

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