# Thread: help me with Inductive Proof #2

1. ## help me with Inductive Proof #2

$n^2-1$ is divisible by 8 whenever n is an odd positive integer

1) Basic step: P(1) = 1-1 =0 is divisible by 8.

2) inductive step:
hypothesis: $P(k): k^2-1$ is divisible by 8

conclusion: $P(k+2): (k+2)^2 -1$ is divisible by 8?
OR
$P(k+1): (k+1)^2 -1$ is divisible by 8? <- to make it odd

help me pls

2. Originally Posted by zpwnchen
$n^2-1$ is divisible by 8 whenever n is an odd positive integer

1) Basic step: P(1) = 1-1 =0 is divisible by 8.

2) inductive step:
hypothesis: $P(k): k^2-1$ is divisible by 8

conclusion: $P(k+2): (k+2)^2 -1$ is divisible by 8?
OR
$P(k+1): (k+1)^2 -1$ is divisible by 8? <- to make it odd

help me pls

It is P(k+2), of course: if k is an odd positive integer, the next ODD integer is k + 2, not k + 1.

Without induction: n^2 - 1 = (n-)(n+1). As n is odd, exactly one of n-1 or n+1 is divisible by 4 and the other factor is even ==> the whole thing is divisible by 4*2 = 8.

Tonio