System of simultaneous equations $x^n+y^n+z^n=3$ ($n$=1,2,5)

Question

Does someone know the solution of the following system of simultaneous equations ($x$, $y$, $z$ are real numbers):

$$ x+y+z=3 $$ $$ x^2+y^2+z^2=3 $$ $$ x^5+y^5+z^5=3 $$

Presented at one of math competitions, don't remember when and where.

Answer 1

$$ x+y+z=3$$ Is a plane which contains the point $(1,1,1)$ and has a normal vector $(1,1,1)$. In other words it is a tangent plane to the sphere at the point $(1,1,1)$. This will then be the only point of intersection between the plane and the sphere; no other point can be a solution to the system as a whole.

Since the point $(1,1,1)$ also solves the third equation it turns out to be a solution of the system.

Answer 2

Here is the algebraic solution. Consider z as parameter, so we have

$$ x+y=3-z $$ $$ x^2+y^2=3-z^2 $$

Subtracting second equation from first squared gives: $$ 2xy=6-6z+2z^2 $$

or $$ xy=z^2-3z+3 $$

Roots of the system

$$ x+y=3-z $$ $$ xy=z^2-3z+3 $$

are the roots of polynomial

$$ t^2-(3-z)t+(z^2-3z+3) $$

which has real roots if $$ (3-z)^2-4(z^2-3z+3) \ge 0 $$

Simplifying the LHS expression gives:

$$ -3(z-1)^2 \ge 0 $$

or $$ (z-1)^2 \le 0 $$

which may happen only if $z=1$.

Substituting $z=1$ gives $x=y=1$