Find all the integer solutions $(x, y, z)$ of the equation

\[
(x+y+z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
\]
We directly check the identity

$$
(x+y+z)^{5}-(-x+y+z)^{5}-(x-y+z)^{5}-(x+y-z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$

Therefore, if integers $x, y$ and $z$ satisfy the equation from the statement, we then have

$$
(-x+y+z)^{5}+(x-y+z)^{5}+(x+y-z)^{5}=0
$$

By Fermat's theorem at least one of the parenthesis equals 0. Let, w.l.o.g., $x=y+z$. Then the previous equation reduces to $(2 z)^{5}+(2 y)^{5}=0$, which is equivalent to $y=-z$. Therefore, the solution set of the proposed equation is

$$
(x, y, z) \in\{(0, t,-t),(t, 0,-t),(t,-t, 0): t \in \mathbb{Z}\}
$$

Remark(PSC): Due to the volume of calculations that are involved in the solution of this problem, a more simplified version can be:

Find all the integer solutions $(x, y, z)$ of the equation

$$
(x+y+z)^{3}=24 x y z
$$

The solution would be similar to the one proposed in the original problem using the identity

$$
(x+y+z)^{3}-(-x+y+z)^{3}-(x-y+z)^{3}-(x+y-z)^{3}=24 x y z
$$