(In reply to
re: Possible solution...2 hints by Ady TZIDON)
Ady,
I think this works for both regular and irregular cases. As to regular cases, we need only observe that in the base case:
A001653 1, 5, 29, 169, 985, 5741, ...
A002315 1, 7, 41, 239, 1393, 8119,...
all solutions for b and c are either (6k+1) or (6k1):
(6a+1)^2(6b+1)^2 = 12(ab)(3a+3b+1)
(6a+1)^2(6b1)^2 = 12(a+b)(3a3b+1)
(6a1)^2(6b+1)^2 = 12(a+b)(3a3b1)
(6a1)^2(6b1)^2 = 12(ab)(3a+3b1)
If a,b are both even, or both odd, then 2 divides (a+b), or 2 divides (ab). If a,b are of opposite parities, then 2 divides (3a3b+1).
The 'irregular cases' that bothered me earlier are specifically those {1,7,17,31,49,71...} where a is itself of the form 2*n^2  1. This allows for a lot of variations on the basic theme, since if a is an integer of the form 2r(r+2)+1, then b can be of the form 2r(Xr+P)+Y, and c of the form 2r(Pr+4n)+Q, where X,Y are in Sloane A001653 {1, 5, 29, 169, 985,...} and P,Q are in Sloane A001333 {1, 3, 7, 17, 41, 99, 239, ...} but the same principle applies; all solutions for b and c are again either (6k+1) or (6k1), hence by the same approach as above, 24 divides all differences, d.
Edited on March 14, 2013, 6:08 am

Posted by broll
on 20130314 05:55:29 