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

(In reply to re: Solution (little thing to add) by Daniel)

Zero is neither positive nor negative, therefore (0,0,0), (4,0,0) and (4,4,0) are not valid solutions for the given problem. Of the given FIVE solutions, TWO are positive integer triplets -- (4,2,2) and (4,3,1). 

Posted by Dej Mar on 2007-08-25 09:11:05