Show that n2 1 is divisible by 8 when n is an odd natural n

Show that n^2 - 1 is divisible by 8 when n is an odd natural number.

Solution

show that n²-1 is divisible by 8 when n is an odd natural nmber

step(1): n=1
n^2-1 = 1-1 = 0 is divisible by 8
step(2): assume k is odd and k^2-1 is divisible by 8
odd numbers differ by 2, so we have to prove
(k+2)^2-1 is divisible by 8
(k+2)^2 - 1 = k^2 +4k +4 - 1 = (k^2-1)+4(k+1)
k^2-1 is divisible by 8
k is odd, k+1 is even => 4(k+1) is divisible by 8
sum of above 2 terms is also divisible by 8
=> (k+2)^2-1 is divisible
hence n^2-1 is divisible by 8 for all odd numbers