System of Equations

Basic Algebra Time

Let's see it Veeky Forums :^)

Impossible to solve, proven by Mr Wiles haha :>>>>>)

Link to the proof?

No integer solutions, since [math]x + y + z \equiv x^2 + y^2 + z^2 \mod 2[/math]

But it doesn't have to be integer solutions

Nah, but it's still something. I'm thinking about the general solutions.

But 3 equations with three unknowns should be able to be solved

Okay, it's solvable but I don't think it's pretty:
Write [math]y = 1 - x - z[/math]
Now [math]x^2 + (x+z - 1)^2 + z^2 = 34[/math] and [math]x^3 + (x+z-1)^2 + z^3 = 97[/math]
Set [math]u = x+z, v = xz[/math]
The system becomes [math]2u^2 - 2u - 2v = 34[/math] and [math]u^3 +u^2 - 3uv - 2u = 96[/math]
Substituting v in the second equation, we find a third degree equation in u, which is solvable using Cardan's formulas. That gives us 3 values of u, and hence 3 corresponding values of v, and hence, solving each quadratic equation [math]x^2 - ux + v = 0[/math], we get 6 values of (x,y), and 6 corresponding values of z. But I'm not doing it by hand.

It can be solved.
Solve the first for z => z=1-x-y, substitute into the second and third: Solve the second (quadratic) for x, substitute into the third. That gives you a cubic in y:
4*y^3 - 4*y^2 - 99*y - 93 = 0
That can be solved (it has 3 real roots), but it's ugly. Numerically, the solutions are
-1.02544845449982
-3.855398344452967
5.880846798952788 [*]
Back-substitute into the second and first to get x and z. The third value marked [*] results in complex x, the other two give x, z as
-2.917748692529978, 4.943197147029798
0.5108714460169772, 4.34452689843599

IOW:
x = -2.917748692529978 y = - 1.02544845449982 z = 4.943197147029798
x = 4.943197147029798 y = - 1.02544845449982 z = -2.917748692529977
x = 0.5108714460169772 y = - 3.855398344452967 z = 4.34452689843599
x = 4.34452689843599 y = - 3.855398344452967 z = 0.5108714460169774

Only if the system is linear
[math]X^2 + Y^2 + Z^2 =34[/math] is not linear