Find all sets of four natural numbers such that the square of each of them, when added to the sum of the other three, again yields a perfect square.

(In reply to a start by xdog)

I think I agree that for (a,a,a,a), only a = 1 is a solution.

1² + (1 + 1 + 1) = 4 = 2² 

For (a,a,b,b) [with the assumption that sets (a,b,a,b); (a,b,b,a); (b,a,a,b); (b,a,b,a) and (b,b,a,a) are counted as the same set as (a,a,b,b)], there is (6,6,11,11) as a solution.

6² + (6 + 11 + 11) = 64 = 8² 
11² + (11 + 6 + 6) = 144 = 12²

Posted by Dej Mar on 2008-12-11 22:48:58