Determine whether there exist real numbers x, y and z satisfying the following system of equations:
x²+4yz+2z = 0; x+2xy+2z² = 0; 2xz+y²+y+1 = 0
x^2+2z(1+2y) = 0 (1)
2z^2+x(1+2y) = 0 (2)
y^2+y+1+2xz = 0 (3)
Multiply (1) by x to get
x^3+2xz(1+2y) = 0 (4)
Multiply (2) by 2z to get
4z^32xz(1+2y) = 0 (5)
Adding (4) and (5) gives
x^3 = 4z^3 (6)
For x and z real, (6) implies
xz >= 0 (7)
Solving (3) for y gives
y = (1+sqrt(14(1+2xz))/2 (8)
For y real, (8) implies
xy < 0
This conflicts with (7).
Therefore, there are no real
x, y, and z satisfying (1),
(2), and (3).

Posted by Bractals
on 20070210 12:17:44 