Find three positive rational numbers such that their sum is a square, and the sum of any pair exceeds the third by a square.

Classical Rules: Let a "square" be any number that is the square of a rational number.

(In reply to Super Simple Method for Finding Solutions! by tomarken)

Here is a method that works when you select two even squares:

Pick any two even square numbers, A and B.

Find ((A+B-4)/4)^2.  Call this number C.  Then:

x = (A+B)/2
y = (A+C)/2
z = (B+C)/2

For example:

A = 16, B = 36

C = [(16+36-4)/4]^2 = 144

x = (16+36)/2 = 26
y = (16+144)/2 = 80
z = (36+144)/2 = 90

 

I don't think this could ever work if you select two odd squares to begin with, as their sum will always be an even number that is not divisible by 4.  The distance between squares is always either odd or divisible by 4, so choosing two odd values for A and B won't lead to a solution.

Edited on April 21, 2006, 1:21 pm

Posted by tomarken on 2006-04-18 14:04:05