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  • If n is an odd integer, then know that n square – 1 is divisible by 8.

If n is an odd integer, then show that n2 – 1 is divisible by 8.

Answers (1)

We can represent any odd positive integer in the form of (4q + 1) or (4q + 3) for some integer q.

Case 1:
If    n = 4q + 1
n2 – 1 = (4q + 1)2 –1
  n2– 1 = (4q)2 + (1)2 + 2(4q) (1) – 1
Using (a + b)2 = a2 + b2 + 2ab
n2 – 1 = 16q2 + 1 + 8q – 1
n2 – 1 = 16q2 + 8q = 8q (2q + 1)
Which is divisible by 8.

Case 2:
If n = 4q + 3
n2 – 1 = (4q + 3)2 – 1
n2 – 1 = (4q)2 + (3)2 + 2(4q) (3) – 1
(Using (a + b)2 = a2 + b2 + 2ab)
n2 – 1 =16q2 + 9 + 24q –1
 n2 – 1 = 16q2 + 8 + 24q = 8(2q2 + 1+3q)
Which is divisible by 8
Hence any odd positive integer n when written in n2 – 1 from is divisible by 8.

 

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