if n is an odd number integer then show that n square minus 1 is divisible by 8
Answers
Answered by
37
suppose let odd n=3
then its square=9-1= 8 which is divisible by 8..
then its square=9-1= 8 which is divisible by 8..
raj829:
thnxs
welcome
my pleasure;?(
Answered by
53
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 integer.
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 integer.
thnx
welcome
why only 4p+1 , why cant we take 2p+1
Srry don't know the exact ans
ur ans is right , but just pls remove my doubt
Note that n2−1=(n+1)(n−1).
If n is odd, then both n+1 and n-1 are even; further, exactly one of them is a multiple of 4. Therefore, their product is a multiple of 8.
If n is odd, then both n+1 and n-1 are even; further, exactly one of them is a multiple of 4. Therefore, their product is a multiple of 8.
but in 2p+1 , n is not negative
anyway, thnx
My pleasure
Similar questions