- #1

r0bHadz

- 194

- 17

## Homework Statement

If n ∈ ℤ^+ and n is odd, prove 8|(n^2 -1)

## Homework Equations

## The Attempt at a Solution

n = 2b+1 for b ∈ ℤ^+ by the definition of positive odd int

n^2 = (2b+1)(2b+1) = 4b^2+4b+1 = 2(2b^2+2b) + 1

n^2 - 1 = 2(2b^2+2b) so n^2 - 1 will be even ∀n s.t n is odd

For n = 1, 8|0 is true, for n=3, 8|8 is true

If n ≥3, n^2 - 1 ≥ 8, and since n^2 -1 will always be even, n^2 -1 will always be able to be divided three times by 2, thus 8 | n^2 -1

I don't want an answer. If you post an answer please put a spoiler so I can know not to read it. Can someone just tell me if this proof is valid or not?

To me, it makes sense. Since n^2 - 1 will be greater than 8 for n>3, and 8 is 2^3, and since n^2 - 1 is an even number which means it is divisible by 2, since it is greater than 8 it will be divisible by 2 at least 3 times.