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

The right side of equation (2) is positive, so is the left side: Therefore x>y>=z, hence

x^2 < 2(x+x+x) = 6x Therefore x<=5

The right side of equation (1) is even, so is the left. Therefore x is even. Combining with x<=5, we see that x must be 2 or 4.

The remaining candidate space is quite small and quickly harvested by hand. The solutions are (4,3,1) and (4,2,2)

Posted by FrankM on 2008-01-19 21:51:13