Question

n2– 1 is divisible by 8, if n is an odd positive integer. Prove it.

Answer 1

Hey i got the answer
$n {}^{2} - 1 \\ since \: you \: said \: that \: n \: is \: odd \: it \: can \: be \: written \: in \: a \: 2k + 1 \: form \\ \: = > (2k + 1) {}^{2} - 1 \\ = 4k {}^{2} + 4k + 1 - 1 \\ = 4k { {}}^{2} + 4k \\ = 4(k {}^{2} + k) \\ = 4(k(k + 1)) \\ now \: observe \: that \: k \: and \: (k + 1) \: are \: consecutive \: natural \: numbers. \\ \: hence \: k(k + 1) \: is \: divisible \: by \: 2. \\ this \: makes \: it \: an \: even \: number \: which \: can \: be \: expresses \: in \: a \: 2p \: form \\ = > 4(2p) \\ = > 8p$
Now you can easily see that 8p is divisible by 8. Hence proved..

Answer 2

Answer:

yes it is the odd integer

Step-by-step explanation:

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

let

n=4p+3n2 −1

=(4p+1)2 −1

=16p2 +8p+1−1

=8p(2p+1)

⇒n 2 −1isdivisibleby8

n2−1

=(4p+3)2 −1

=16p  2 +24p+9−1

=16p2 +24p+8

=8(2p2 +3p+1)

⇒n  2 −1isdivisibleby8

​

 

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