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.
Here is a really simple method to find solutions to this problem. It won't find all of them, of course, but it will find some:
Pick any odd square number, A.
Find ((A1)/2)^2. Call this C.
You can then use these two to find numbers x, y, and z that will satisfy the puzzle:
x = A/2
y = C/2
z = x + y
For example:
A = 25
Then, C = ((251)/2)^2 = 144.
x = 25/2 = 12.5
y = 144/2 = 72
z = 12.5 + 72 = 84.5
Try this to find more solutions:
Pick any two square numbers, A and B, such that their sum is odd (this is actually the same as above, except in that method, the second square "B" is always zero, so I just left it out).
Find ((A+B1)/2)^2. Call this C.
Then:
x = (A+B)/2
y = (A+C)/2
z = (B+C)/2
For example:
A = 4, B = 9
C = ((4 + 9  1)/2)^2 = 36
x = (4+9)/2 = 6.5
y = (4+36)/2 = 20
z = (9+36)/2 = 22.5

Posted by tomarken
on 20060418 12:02:35 