Solve this equations system: $x^{5} + y^{5} + z^{5} = 1, x + y + z = 1$

Question

Solve this equations system: \begin{cases} x^{5} + y^{5} + z^{5} = 1 \\ x + y + z = 1 \end{cases}

Sketch of solutions:

If x,y,z is solution of this equations system then $$(x+y+z)^{5} - (x^{5} + y^{5}+z^{5}) = 0$$ when $x+y= 0$, then equation system is true for any z.

So we have $$(x+y+z)^5 - (x^{5}+y^{5}+z^{5})= A(x+y)(y+z)(x+y)(x^{2}+xy+xz+y^{2}+yz+z^{2})=$$

for $x=1, y=1, z=0$ we have $$2^{5}-1-1 = 6A$$ $$ A = 5 $$ So we have $5(x+y)(y+z)(x+z)(x^{2}+xy+xz+y^{2}+yz+z^{2})=0$

So the solutions: 1. $$x+y= 0, z = 1 $$ 2. $$y+z= 0, x = 1 $$ 3. $$x+z= 0, y = 1 $$ 4. $$x^{2}+xy+xz+y^{2}+yz+z^{2} =$$ $$ \frac{1}{2}(x^{2}+2xy+y^{2}+x^{2}+2xy+z^{2}+y^{2}+2yz+z^{2}$$ $$ \frac{1}{2}((x+y)^{2}+(y+z)^{2}+(x+z)^{2})==0$$ this imply that (0,0,0) is solution. It's wrong because we have $0=1$

Do you see mistake?

Solutions from Mathematica: $\left\{\{x\to 1,z\to -y\},\{y\to 1,z\to -x\},\{y\to -x,z\to 1\},\left\{y\to \frac{1}{2} \left(-\sqrt{-3 x^2+2 x-3}-x+1\right),z\to \frac{1}{2} \left(\sqrt{-3 x^2+2 x-3}-x+1\right)\right\},\left\{y\to \frac{1}{2} \left(\sqrt{-3 x^2+2 x-3}-x+1\right),z\to \frac{1}{2} \left(-\sqrt{-3 x^2+2 x-3}-x+1\right)\right\}\right\}$

Answer 1

Your answer of $0$ is not a mistake because it is a solution of $(x+y+z)^{5} - (x^{5} + y^{5}+z^{5}) = 0$ which is the equation you were solving.

It just happens not to be a solution which makes each of $(x+y+z)^{5}$ and $(x^{5} + y^{5}+z^{5})$ equal to $1$.

Answer 2

If we have $x^5+y^5=A$ and $x+y=B$ we have $$ A = x^5+(B-x)^5 $$ such that $$ x = \frac{B}{2}\pm\frac{\sqrt{5} \sqrt{\pm 2 \sqrt{5} B \sqrt{B \left(4 A+B^5\right)}-5 B^4}}{10 B},\quad y = \frac{B}{2}\mp\frac{\sqrt{5} \sqrt{\pm 2 \sqrt{5} B \sqrt{B \left(4 A+B^5\right)}-5 B^4}}{10 B}$$ because $(B/2+z)^5+(B/2-z)^5$ is a biquadratic polynomial. Now you may just set $A=1-z^5, B=1-z$ and notice that the intersection (in $\mathbb{R}^3$) of $x+y+z=1$ and $x^5+y^5+z^5=1$ is just given by three lines: