Question

show that n square, nsquare-1 divisible by 8

Answer 1

wd know that the smallest no whose square come under 8 is 3.
(n)square -1=3 square -1=8 (which is divisible by 8.
another case
n=8
(n)square =8 square
=64(which is divisible by 8
from these two cases we can say that n square, n square -1 are divisible by8.

Answer 2

hello friend
n^2 (n+1)(n-1)
Case 1 ..

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p. { n^2 = odd , no use in this situation }

Let n = 4p+ 1,

(n2 – 1) = (4p + 1)2 – 1 = 16p2 + 8p + 1 = 16p2 + 8p = 8p (2p + 1)
⇒ (n2 – 1) is divisible by 8.


Therefore, n2– 1 is divisible by 8 if n is an odd positive integer.

Case 2
If n is a even number
N= 2p
N^2 = 4p^2
{ N^2 -1 = odd number }

It shows that .. only odd value of n is divisible by 8 ..
Hope it helps you