James Lobo	Posted on Monday, 05 May, 2003 - 03:22 pm: I have been stuck on this question Can you write down a careful proof for this statement: If n=m3 -m for some integer m, then n is a multiple of 6.
Peter Gyarmati	Posted on Monday, 05 May, 2003 - 03:58 pm: m3 -m = m(m2 -1) = m(m-1)(m+1), so n = (m-1)m(m+1). Here from the three factors at least one is even and one is divisible by three, thus n is divisible by six.
Michael Doré	Posted on Monday, 05 May, 2003 - 05:24 pm: More generally the product of k consecutive integers is a multiple of k!. This is a simple consequence of the fact the binomial coefficient C(n,k) is an integer.
Philip Ellison	Posted on Monday, 05 May, 2003 - 06:53 pm: A really nice way of proving that the binomial coefficient is always an integer is by considering the floor function and Legendre's formula. Not what was asked, but it interested me when I saw it!
David Loeffler	Posted on Monday, 05 May, 2003 - 07:26 pm: I would beg to differ on whether that constitutes a nice proof. The combinatorial interpretation of C(n,k) implies immediately that it's an integer, and you can just read off n! / k!(n-k)! by a simple counting argument. IMHO this is much nicer than messing around with floor functions. David
Philip Ellison	Posted on Monday, 05 May, 2003 - 09:02 pm: A matter of taste I suppose!
Demetres Christofides	Posted on Tuesday, 06 May, 2003 - 09:25 am: I agree 100% with David. By the way, I thought it was called De Polignac's formula. [I agree that this proof is interesting but it is not very nice] Demetres
Philip Ellison	Posted on Tuesday, 06 May, 2003 - 05:29 pm: Okay, interesting then! I meant it to be nice in that I liked the way it used the floor function (which I'd only just met) in conjunction with very basic number theory to obtain the result... despite the fact that it's clearly not the easiest/simplest way of establishing the result. Not sure about the name; perhaps it varies depending on which textbook you read!