Determine the probability that for a base ten positive integer N drawn at random between 2 and 101 inclusively, the number N^{3} + 1 is expressible in the form p*q*r, where p, q, r are three distinct positive integers such that p, q and r (in this order) corresponds to three consecutive terms of an arithmetic sequence.
Disclosure: I figured this out using Charlies solution.
Instead of using p,q,r call the constant difference d so the numbers become pd, p, p+d
If d is of the form 3x and q is of the form 3x^2 + 1
then
(pd)(p)(p+d)=(3x23x+1)(3x^2+1)(3x^2+3x+1)
=(9x^43x^2+1)(3x^2+1)
=27x^6 + 1
=(3x^2)^3 + 1
so N can be written in the form 3x^2
which for x={1,2,3,4,5}
Gives the solutions N={3,12,27,48,75} as found by Charlie.
Unfortunately this does not prove that there are not other solutions.

Posted by Jer
on 20100311 22:52:58 