Divisibility Proof for Odd IntegersDate: 02/13/2002 at 02:16:47 From: Mike Smith Subject: Divisibility proof for odd integers I really like your site. I'm an applied math major and I enjoy reading your college and beyond section often. That said, here's my question: How can I prove that for all odd integers N, N^3 - N is divisible by 8? I know it's divisible by 4, I've gotten that far (using a 2k+1 sub. for N^3 - N and pulling out 4 as a factor) already. Where next? I want to say that >> if a number is divisible by 4 and it is divisible by 2, then it is divisible by 8. I know that this is incorrect. 28, voila. I'm only a beginner in the proofwriting realm, so any direction would be helpful. I'm sure it's something obvious that I'm overlooking... Thank you, Mike Date: 02/13/2002 at 02:48:02 From: Doctor Schwa Subject: Re: Divisibility proof for odd integers Hi Mike, Sounds like you're off to a great start on this problem! With your approach, N = 2k+1, you get (2k+1)^3 - (2k+1) = 8k^3 + 12k^2 + 4k, right? Now that is 4(2k^3 + 3k^2 + k), or, factoring, you have 4k(2k+1)(k+1) ... What I notice there is that you have both k and k+1 as factors, one of which must be even ... and thus the problem is solved. I took a slightly different approach from the beginning, with a little less messy algebra but slightly more subtle reasoning necessary: n^3 - n = n(n+1)(n-1). Now if n is odd, n+1 and n-1 are both even, and of any two consecutive even numbers like that, what can you conclude? Enjoy, - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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