## A non linear system: x+y+z=2, x^2+y^2+z^2=14, x^4+y^4+z^4=98

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PoCTo
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### A non linear system: x+y+z=2, x^2+y^2+z^2=14, x^4+y^4+z^4=98

$\begin{cases} x+y+z=2 \\
x^2+y^2+z^2=14 \\
x^4+y^4+z^4=98 \end{cases}$

Maybe some ideas, tricks?

Honeysuckle588
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### Re: A non linear system: x+y+z=2, x^2+y^2+z^2=14, x^4+y^4+z^4=98

Well...graphing it showed me there are 6 points that solve the system. Can you first determine the intersection of the plane and the sphere? It's a circle, and if you can get the equation for that you can then work on finding the intersection of the circle with the cube-like 4th degree.

stapel_eliz
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Sorry to resurrect such an old thread, but questions of this sort do arise from time to time, and this one's been bugging me for a while....

We have the following relations:

. . . . .$(x\, +\, y\, +\, z)^2\, =\, x^2\, +\, y^2\, +\, z^2\, +\, 2(xy\, +\, yz\, +\, xz)$

Then:

. . . . .$-2(xy\, +\, yz\, +\, xz)\, =\, x^2\, +\, y^2\, +\, z^2\, -\, (x\, +\, y\, +\, z)^2\,$

And plugging in gives us:

. . . . .$-2(xy\, +\, yz\, +\, xz)\, =\, 14\, -\, (2)^2\, =\, 10$

. . . . .$xy\, +\, yz\, +\, xz\, =\, -5$

If one does the long multiplication, one will find that:

. . . . .$(xy\, +\, yz\, +\, xz)^2$

. . . . . . . . .$=\, (x^2y^2\, +\, y^2z^2\, +\, x^2z^2)\, +\, 2xyz(x\, +\, y\, +\, z)$

Then plugging in gives us:

. . . . .$(-5)^2\, =\, (x^2y^2\, +\, y^2z^2\, +\, x^2z^2)\, +\, 2xyz(2)$

. . . . .$25\, =\, (x^2y^2\, +\, y^2z^2\, +\, x^2z^2)\, +\, 4xyz$

Restating:

. . . . .$-2(xy\, +\, yz\, +\, xz)\, =\, x^2\, +\, y^2\, +\, z^2\, -\, (x\, +\, y\, +\, z)^2\,$

...as:

. . . . .$x^2\, +\, y^2\, +\, z^2\, -\, 2(xy\, +\, yz\, +\, xz)\, =\, (x\, +\, y\, +\, z)^2$

...and squaring, we get:

. . . . .$(x\, +\, y\, +\, z)^4\,$

. . . . . . . . .$=\, (x^4\, +\, y^4\, +\, z^4)\,$

. . . . . . . . . . .$+\, 2(x^2y^2\, +\, y^2z^2\, +\, x^2z^2)$

. . . . . . . . . . . . .$-\, 4(x^2\, +\, y^2\, +\, z^2)(xy\, +\, yz\, +\, xz)$

. . . . . . . . . . . . . . .$+\, 4(xy\, +\, yz\, +\, xz)^2$

Substitution yields:

. . . . .$2^4\, =\, 98\, +\, 2(25\, -\, 4xyz)\, -\, 4(14)(-5)\, +\, 4(-5)^2$

. . . . .$-512\, =\, -8xyz$

. . . . .$64\, =\, xyz$

But this says that:

. . . . .$x^2y^2\, +\, y^2z^2\, +\, x^2z^2\, =\, 25\, -\, 4(64)\, =\, -231$

...which clearly is not a possible value for the sum of squares of real numbers. Now what...?

PoCTo
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### Re: A non linear system: x+y+z=2, x^2+y^2+z^2=14, x^4+y^4+z^4=98

It's too late to understand where is the mistake, I can only write my own solution to this now
It's similar to yours, I think

Let:
$a = x+y+z = 2$
$b = x^2+y^2+z^2 = 14$
$c = x^4+y^4+z^4 = 98$

Then:
$d = \frac{a^2 - b}{2}=xy+yz+xz= -5$
$e = \frac{b^2 - c}{2}=x^2 y^2+y^2 z^2 +x^2 z^2= 49$
$f = \frac{d^2 - e}{2}=x^2 yz+xy^2 z+xyz^2 = -12$
$g = \frac{f}{a}=xyz = -6$ (After this we can get new roots, so we have to check them after solving)

So we know:
$x+y+z=2$, $xy+yz+xz=-5$, $xyz=-6$

We know, that these formulas are cube polynomial's coefficients.
If $f$ is the polynom:
$(f-x)(f-y)(f-z)=f^3-(x+y+z)f^2+(xy+xz+yz)f-xyz=f^3-2f^2 - 5f +6$, so $x,\ y,\ z$ are roots of this polynom

Now we can use some formulas (http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function) or find trivial solutions, they are: $\{3,-2,1\}$

Then we must understand that all variables are equivalent, have the same rights, so if we swap some of them, we will get a new answer.
$[3,-2,1];\ [3,1,-2];\ [-2,1,3];\ [-2,3,1];\ [1,-2,3];\ [1,3,-2]$