Determine all possible positive integer triplets (x, y, z) with y ≥ z satisfying the following system of equations :

                  x² = 2(x+y+z)

              x³- y³ - z³ = 3xyz

y³+z³=(y+z)³-3yz(y+z)
x³-3yzx=(y+z)³-3yz(y+z)
(x-y-z)(x²+y²+z²+xy+yz+zx)=0
x=y+z or 1/2{(x+y)²+(y+z)²+(z+x)²}=0 => x=y=z=0.
Substitute x=y+z in eq(1)
x²=2(x+x) => x=0 or x=4
y+z=0 => y=-z
y+z=4 => y=4-z
Solutions: (0,c,-c) and (4,b,4-b) where c≥0 and b≥2.
As the solutions should have positive integrs only.
The solutions will be (4,3,1) and (4,2,2).

Edited on August 31, 2007, 6:11 am

Posted by Praneeth on 2007-08-24 15:23:57