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 ((A-1)/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 = ((25-1)/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+B-1)/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 2006-04-18 12:02:35