Vtyjkyuk

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.

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 $(x, y, z)=(-frac2sqrt[5]2-1, 1, 1)$