Determine the probability that for a positive base ten integer N drawn at random between 2 and 201 inclusively, the number N³ - 1 is expressible in the form p*q*r, where p, q and r are three distinct positive integers such that p, q and r (in this order) corresponds to three consecutive terms of an arithmetic progression.

n^3 - 1 = (n-1)(n^2+n+1)
(q-a)(q)(q+a) = q(q^2-a^2)

Each is a linear term times a quadratic term so what if the linear terms were equal and the quadratic terms were equal?
q = n-1
q+1=n
Substituting into n^n+n+1 gives the equation
q^2 + 3q + 3 = q^2 - a^2
3q + 3 = -a^2

Unfortunately this has no real solutions for positive q.

It does give plenty of negative solutions though:
(-3)^3 - 1 = -1*-4*-7
(-12)^3 - 1 = -7*-13*-19
etc...

Posted by Jer on 2010-06-01 15:14:59