MATH: Help with integer proof

[madpersian]

Hey guys, just wanted to see if anyone had an idea about this proof:

"Prove that when n is odd, n^2 - 1 is divisible by 8."

Case 1: |n|=1. N^2 - 1 = 0 which is divisible by 8.

Case 2: |n| > 2. (n^2 - 1)=(n-1)(n+1). Since n is odd, n-1 and n+1 are divisible by 2. So (n-1)(n+1) = 2q2p, p,q belong to integers. So (n^2 - 1)=4k. Since n>=3, n^2-1>= 8. So k>=2.
So (n^2 - 1)=4k, k is an integer greater or equal to 2.

This is as far as I got. Not sure if I have the right approach. Any comments or suggestions would be appreciated.

[Kwijiboe]

Feed the cat less, and use a spray bottle.

[SDSD]

There are a few ways to do this. I'd say going with induction (as you seem to sort of are) might be a little tedious. You should try to prove this around the fact that, by the given, n^2-1=8l for some l_N(l in the set of natural numbers)

You seem to have some of the right ideas. First, as n is odd, n = 2k+1 for some k_N. Then with some basic algebra you can show that n^2-1=4(k)(k+1). From here you know that either k and k+1 are a pair of even or odd number. Use what you know about odd and even numbers here and the rest of the proof will fall into place.

[madpersian]

Thanks for the advice. My mistake was not thinking of odds as n=2k+1.

[Enedin]

Like SDSD said...

n²-1 = (2k+1)²-1
n²-1 = 4k²+4k
n²-1 = 4(k)(k+1)

k is any natural number, so either k or k+1 is even. Multiplying any natural number with an even number makes the outcome even, so (k)(k+1) will always be divisible by 2.

And because there's a 4 in that equation as well, 4*2 = 8, the entire thing is divisible by 8.

Put that in mathy terms and QED.

[madpersian]

Thanks a lot. I knew BG would deliver.