Question

if n is a is an odd positive integer show that n square minus one is divisible by 8

Answer 1

Any odd positive integer n can be written in form of 4q + 1 or 4q + 3.

 

If n = 4q + 1, when n2 - 1 = (4q + 1)2 - 1 = 16q2 + 8q + 1 - 1 = 8q(2q + 1) which is divisible by 8.

If n = 4q + 3, when n2 - 1 = (4q + 3)2 - 1 = 16q2 + 24q + 9 - 1 = 8(2q2 + 3q + 1) which is divisible by 8.

 

So, it is clear that n2 - 1 is divisible by 8, if n is an odd positive integer.

Answer 2

Answer:

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

letn=4p+3n2−1=(4p+1)2−1=16p2+8p+1−1=8p(2p+1)⇒n2−1isdivisibleby8n2−1=(4p+3)2−1=16p2+24p+9−1=16p2+24p+8=8(2p2+3p+1)⇒n2−1isdivisibleby8

Therefore, n2−1 is divisible by 8 if n is an odd positive integer.

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