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.

One striking thing about the solutions is that so many of them have p=2. Directly substituting p=2 into the equations makes r=2q-2 and then N^3=(2q-1)^2. These must both equal a number to the sixth power, call it k. 2q-1 is odd, therefore k and N are odd. Then for __every odd integer k__, there is a solution __N=k^2, p=2, q=(k^3+1)/2, r=k^3-1__

Note: this does not cover all solutions, just ones with p=2.

*Edited on ***June 1, 2010, 8:23 pm**