If n is an odd positive integer, show that `(n^2-1)` is divisible by 8.

Explanation

Let `a=n^(2)-1` …(i)
Given that, n is an odd integer 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 divisibly by 8.
From Eq. (i), at n=5, a=`(5)^(2)-1=25-1=24=3 xx 8`
Which is divisible by 8
From Eq. (i), at n=7, a`=(7)^(2)-1=49-1=48=6xx8`
Which is divisible by 8.
Hence, `(n^(2)-1)` is divisibly by 8m where n is an odd integer.
Alternate Method
We know that that an odd integer n is of the from `(4q+1)or (4q+3)` for some integer q.
CASE I When m=4q+1
In this case, we have `(n^(2)-1)=(4q+1)^(2)-1`
`=16q^(2)+1+8q-1` `[therefore (a+b)^(2)=a^(2)+2ab+b^(2)]`
`=16q^(2)+8q=8q(2q+1)`
`=16(q^(2)+8q=8q(2q+1)`
Which is clearly, divisibly by 8.
CASE II When n=4p+3
In this case, we have
`(n^(2)-1)=(4q+3)^(2)-1`
`=16q^(2)+9+24q-1` `[therefore(a+b)^(2)=a^(2)+2ab+b^(2)]`
`=16q^(2)+24q+8`
`=8(2q^(2)+3q+1)`
which is clearly divisibly by 8.
Hence, `(n^(2)-1)` is divisibly by 8.