Question

if n^2-1 is divisible by 8 then what is n

Answer 1

If 'n' is an odd positive integer, show that (n2-1) Is divisible by 8??

 

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.

 

Let n = 4p+ 1,

 

(n2 – 1) = (4p + 1)2 – 1 = 16p2 + 8p + 1 = 16p2 + 8p = 8p (2p + 1)

⇒ (n2 – 1) is divisible by 8.

 

(n2 – 1) = (4p + 3)2 – 1 = 16p2 + 24p + 9 – 1 = 16p2 + 24p + 8 = 8(2p2 + 3p + 1)

⇒ n2– 1 is divisible by 8.

 

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

Answer 2

Answer:

Prove that n²-1is divisible by 8 ✓

Step-by-step explanation:

Let n=8

n²-1 +(n²-1)+8

n²(2n²-n²)+(n²+1)+8

n²(2n²+1)+8

n²(2n²+9)

n²(2n²/9*9/2n²)

1. n²-1 is divisible by 8 this equation is proved