Show that (n^2-1) is divisible by 2 if n is an odd positive integer?

Asked by manhas | 8th Apr, 2017, 11:31: PM

Expert Answer:

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 2, since 8 is divisible by 2.
 
(n2 – 1) = (4p + 3)2 – 1 = 16p2 + 24p + 9 – 1 = 16p2 + 24p + 8 = 8(2p2 + 3p + 1)
⇒ (n– 1) is divisible by 2, since 8 is divisible by 2.
 
Therefore, n2– 1 is divisible by 2 if n is an odd positive integer.

Answered by Rebecca Fernandes | 27th Nov, 2017, 02:35: PM