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Cubic and Consecutive Concern II (Posted on 2010-05-31) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

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re: Infinite family of solutions | Comment 9 of 10 |
(In reply to Infinite family of solutions by Brian Smith)

Yes, because pq then has a value of k^3+1, and (k^3+1)*(k^3-1) = (k^2)^3-1. Does this assist with explaining the outliers, though?

I also noticed that since (n^2+n+1) divides q(q^2-x^2), and (n-1) also divides q(q^2-x^2), at least 1 of p,q,r must have at least one of the members of the series: 3,7,13,21,31,43,57...(Sloane A002061) as a factor  and at least one of the series: 3,7,13,31,43,73,157...Sloane: A002383) as a prime factor. In addition, one of p,q,r must have one of 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397...(Sloane:A003215) as a factor.

The notes to A002061 include this, which might be helpful:

a(n+1) = n^2 + n + 1. a(n+1)*a(n)=(n^6-1)/(n^2-1)=n^4+n^2+1=a(n^2+1) - a product of two consecutive numbers from this sequence belongs to this sequence too. (a(n+1)+a(n))/2=n^2+1. (a(n+1)-a(n))/2=n. a((a(n+1)+a(n))/2)=a(n+1)*a(n).

Is it also correct that the whole of the expression N^3-1 cannot be of the form 4k+2, given that it is divisible by a difference of squares?

Lastly, since every factor of p,q,r, is a factor of N^3-1, can we also infer that no factor of n is also a factor of p,q,or r?

Any more views?



Edited on June 3, 2010, 5:43 am
  Posted by broll on 2010-06-02 10:03:22

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