If n is an odd integer, then show that n2 – 1 is divisible by 8.
Let $\quad a=n^{2}-1 \quad \ldots$ (i)
Given that, $n$ is an odd integer.
$\therefore$ $n=1,3,5, \ldots$
From Eq. (i), at $n=1, a=(1)^{2}-1=1-1=0$,
which is divisible by $8 .$
From Eq. (i), at $n=3, a=(3)^{2}-1=9-1=8$,
which is divisible by 8 .
From Eq. (i), at $n=5, a=(5)^{2}-1=25-1=24=3 \times 8$,
which is divisible by 8 .
From Eq. (i), at $n=7, a=(7)^{2}-1=49-1=48=6 \times 8$,
which is divisible by 8 .
Hence, $\left(n^{2}-1\right)$ is divisible by 8, where $n$ is an odd integer.
Alternate Method
We know that an odd integer $n$ is of the from $(4 q+1)$ or $(4 q+3)$ for some integer $q$.
Case I When $n=4 q+1$
In this case, we have
$\left(n^{2}-1\right)=(4 q+1)^{2}-1$
$=16 q^{2}+1+8 q-1 \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
$=16 q^{2}+8 q=8 q(2 q+1)$
$=16 q^{2}+8 q=8 q(2 q+1)$
which is clearly, divisible by 8 .
Case II When $n=4 q+3$
In this case, we have
$\left(n^{2}-1\right)=(4 q+3)^{2}-1$
$=16 q^{2}+9+24 q-1 \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
$=16 q^{2}+24 q+8$
$=8\left(2 q^{2}+3 q+1\right)$
which is clearly divisible by 8 .
Hence, $\left(n^{2}-1\right)$ is divisible by 8 .
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