Determine the probability that for a positive base ten integer N drawn at random between 2 and 201 inclusively, the number N^{3}  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 = (n1)(n^2+n+1)
(qa)(q)(q+a) = q(q^2a^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 = n1
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 20100601 15:14:59 