If n is an odd positive integer, show that (n^2-1) divisible by 8.

2
et n = 2m+1 where m=0,1,2,....

n^2=4m^2+1+4m

n^2-1=4m^2+4m
=4m(m+1)
Let K=m(m+1)
if m is odd number, odd*even will give u even number,
If m is even number even into odd will give u even number.

So K is always even no matter what is the value of m.

and even multiple of 4 will always divisble by 8

Answers

The Brainliest Answer!
2015-05-30T10:05:52+05:30
Thiz can b proved by taking any odd number let us take n=3 3sq-1/8=9-1/8=8/8=1 Let's take n=5 5sq-1/8=25-1/8=24/8=3 So nsq-1for any odd number iz divisible by 8
2 3 2
et n = 2m+1 where m=0,1,2,....

n^2=4m^2+1+4m

n^2-1=4m^2+4m
=4m(m+1)
Let K=m(m+1)
if m is odd number, odd*even will give u even number,
If m is even number even into odd will give u even number.

So K is always even no matter what is the value of m.

and even multiple of 4 will always divisble by 8
2015-05-30T10:17:09+05:30
Take any 2 nuiimebrs 
lets take 5
we get 25-1=24 which ius diuuvisible
also takje 13 we get 168 which ius divuisiuble
1 5 1