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UN cargo (Posted on 2013-03-13) Difficulty: 4 of 5
A triplet (1,25,49) represents 3 perfect squares forming an arithmetic progression.

1. Provide few additional samples like the above.
2. What can be said about the possible values of d (the constant
difference between the adjacent members of an arithmetic progression)?
3. How relates the title of my post to its contents?

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.5000 (2 votes)

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Some Thoughts re(2): Possible solution...2 hints | Comment 6 of 9 |
(In reply to re: Possible solution...2 hints by Ady TZIDON)


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 (6k-1):

(6a+1)^2-(6b+1)^2 = 12(a-b)(3a+3b+1) 
(6a+1)^2-(6b-1)^2 = 12(a+b)(3a-3b+1) 
(6a-1)^2-(6b+1)^2 = 12(a+b)(3a-3b-1)  
(6a-1)^2-(6b-1)^2 = 12(a-b)(3a+3b-1)  
If a,b are both even, or both odd, then 2 divides (a+b), or 2 divides (a-b). If a,b are of opposite parities, then 2 divides (3a-3b+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 (6k-1), hence by the same approach as above, 24 divides all differences, d.





Edited on March 14, 2013, 6:08 am
  Posted by broll on 2013-03-14 05:55:29

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