Question

if n is an odd integer then show that n square -
1 is divisible by 8 ​

Answer 1

Answer:

Step-by-step explanation:

the remainders when divisible by 8 is 1,2,3,4,5,6,7

odd integers are=1,3,5,7

n^2-1

n=1  1^2=1=0 it is divisible by 8

n=3   3^2=9-1=8 it is divisible by8

n=5    5^2-1=25-1=24 it is divisible by8

n=7   7^2-1=49-1=48 it is divisible by 8

;we can say n^2-1 is divisible by8

Answer 2

Hey

Let 'a' be any positive odd integer and b=4.

Every positive odd integer is of the form 4q+1 or 4q+3.

CASE 1:When n=4q+1

n^2-1=(n+1)(n-1)=[(4q+1)+1][(4q+1)-1]=(4q+2)(4q)=16q^2+8q=8q(2q+1)

Thus,divisible by 8.

CASE 2:When n=4q+3

n^2-1=(n+1)(n-1)=[(4q+3)+1][(4q+3)-1]=(4q+4)(4q+2)=4(q+1)2(2q+1)=8(q+1)(2q+1)

Thus,divisible by 8.

Hence,n^2-1 is divisible by 8.

I hope it helps you!!!