Now by the induction hypothesis, (n^3 - n) is divisible by 3 and obviously, 3n(n+1) is divisible by 3. Therefore (n+1)^3 - (n+1) is also divisible by 3.
So it is not clear what you mean by yours:
``Your identity is far superior to what I was doing.''
On Oct 23, 2011, at 4:47 AM, Esther Brink wrote:
> Alexander, > Your identity is far superior to what I was doing. Thanks. > Ben > > CC: email@example.com; firstname.lastname@example.org > From: email@example.com > Subject: Re: Induction proof-need help. Thanks! > Date: Sat, 22 Oct 2011 23:56:15 -0400 > To: firstname.lastname@example.org > > Obviously, (n+1)^3 - (n+1) = (n+1)(n^2 - n) + 3n(n+1). The proof by > induction follows. > > On Oct 21, 2011, at 7:46 PM, Esther Brink wrote: > > Meo, > For n = 1, note n^3 - n = 1^3 - 1 = 0, which is divisible by 3. > Next suppose that k is an integer and the proposition "works". > Then k^3 - k is divisible by 3. > Notice that k^3 - k = k(k^2 - 1) = k(k-1)(k+1) = (k-1)k(k+1). > Therefore 3 must divide either k-1, k or k+1. > Now can we similarly divide (k+1)^3 - (k+1) into a product of > three expressions? If 3 divides either k or k+1, will it divide (k > +1)^3 - (k+1)? If not, what must 3 divide? Can we still conclude > that 3 divides (k+1)^3 - (k+1)? Why? > Thanks for providing a great question! > Ben > > > Date: Fri, 21 Oct 2011 16:46:04 -0400 > > From: email@example.com > > To: firstname.lastname@example.org > > Subject: Induction proof-need help. Thanks! > > > > Induction proof problem: write an induction proof for all natural > numbers for n that n^3-n is diviable with 3. > > > > Thanks for any help. > >