If n is an odd integer,
Question:

If n is an odd integer, then show that n2 – 1 is divisible by 8.

Solution:

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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