Find four distinct positive integers such that the sum of any three of them is a perfect square.

(In reply to re: Number of Solutions (spoilers present) by Charlie)

Charlie's observation is dead-on; if we want to look at lists and count totals of solutions, we want to index by the maximum sum, not the minimum. Thus, the lists are actually exhaustive. I don't know why I was listing from the smallest :-(

So the question turns to: When can we pick 4 distinct naturals and know that every combo of the form a^2+b^2+c^2-2d^2, is divisible by 3 and not negative? Notice that Ptolemy's suggested form meets this criteria, and as a matter of fact, picking any four naturals that are all divisible by 3 will generate a solution quad (sans negative anwers). 

The negativity issue is really not too restrictive. It is not hard at all to show that Robert's constraints are positive for all Ptolemy quads with largest square of 33^2 and greater (hint: this boils down to looking at the smallest valued Robert constraint and finding the roots of a quadratic.) Thus we have an infinite number of solutions. But are there an infinite number (or any) non-Ptolemy quads?

Wow, I have two perfect cardinals (birds) sitting outside my window.

Posted by owl on 2005-02-24 14:32:18