Question
If n is an odd integer, then show that $n^2 - 1$ is divisible by 8.
✓
Answer
Let $a = n ^2-1$.....(i) Given that, n is an odd integer.
$\therefore n=1,3,5$ 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 Answer
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 $\left(n^2-1\right)=(4 q+1)^2-1=16 q^2+1+8 q-1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]=16 q^2+8 q=8 q(2 q+1)=16 q^2+8 q=8 q(2 q+1)$
Which is clearly, divisible by 8 .
Case II:
When $n=4 q+3 \ln$ this case,
we have $\left(n_2-1\right)=(4 q+3)^2-1=16 q^2+9+24 q-1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]=16 q^2+24 q+8=8\left(2 q^2+3 q+1\right)$
which is clearly divisible by 8 . Hence, $\left(n^2-1\right)$ is divisible by 8 .
$\therefore n=1,3,5$ 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 Answer
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 $\left(n^2-1\right)=(4 q+1)^2-1=16 q^2+1+8 q-1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]=16 q^2+8 q=8 q(2 q+1)=16 q^2+8 q=8 q(2 q+1)$
Which is clearly, divisible by 8 .
Case II:
When $n=4 q+3 \ln$ this case,
we have $\left(n_2-1\right)=(4 q+3)^2-1=16 q^2+9+24 q-1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]=16 q^2+24 q+8=8\left(2 q^2+3 q+1\right)$
which is clearly divisible by 8 . Hence, $\left(n^2-1\right)$ is divisible by 8 .
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