<s>[INST] Find all $x,\ y,\ z$ that satisfy: $(x+y-z)^{5}+(x+z-y)^{5}+(z+y-x)^{5}=0$ 
Find all integers $x,\ y,\ z$ that satisfy:
  $$(x+y-z)^{5}+(x+z-y)^{5}+(z+y-x)^{5}=0$$

I  guessed that, considering the symmetry and the similar case $\sum x_{i}^{2n}=0$, that these three expressions must be equal to zero, which leads to $x=y=z=0$. This leaves two other cases:
i. $x+y-z<0,\ x+z-y<0,\ z+y-x>0$
ii. $x+y-z<0,\ x+z-y>0,\ z+y-x>0$
But this didn't help much. My other attempt was considering this equation mod $2x,\ 2y,\ 2z$.
Note. This problem states that $x,y,z$ are integers, but if you have solved the case that $x,\ y,\ z$ are real, please share your solution.
 [/INST] Since $x, y, z$ are integers so are $a=x+y-z, b=x+z-y, c=z+y-x$.
We know from Fermat's last theorem that $a^5+b^5+c^5=0$ is impossible for non-zero integers $a, b, c$.
This means that $x+y-z, x+z-y$ or $z+y-x$ must be zero.
If $x+y-z=0$ then your equation gives $x+z-y=x-y-z$ which means that $z=0$ and $x=-y$.
If $x+z-y=0$ then your equation gives $x+y-z=x-y-z$ which means that $y=0$ and $x=-z$.
If $z+y-x=0$ then your equation gives $x+y-z=y-x-z$ which means that $x=0$ and $y=-z$.
So, a general solution is $(x, y, z)=(-a, a, 0)$ or $(x, y, z)=(-a, 0, a)$ or $(x, y, z)=(0, -a, a)$ for any integer $a$.
If you want real solutions, then there are many of them. Simply choose $y=z=1$ and solve for $x$. You can find that $(x, y, z)=(-\frac{2}{\sqrt[5]{2}-1}, 1, 1)$
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