908*. Does the equation

$$
x^{2}+y^{2}+z^{2}=2 x y z
$$

have a solution in non-negative integers?
$\triangle$ By the condition, the sum $x^{2}+y^{2}+z^{2}$ is divisible by 2. Therefore, among the numbers $x, y$, and $z$, either one is even and two are odd, or all of them are even. Let's consider both possibilities.

1) Suppose one of the numbers $x, y, z$ is even, and two are odd.

But in this case, the right-hand side of the equation is divisible by 4, while the left-hand side is not (think about why). Therefore, there are no solutions here.

2) Suppose each of the numbers $x, y$, and $z$ is even:

$$
x=2 x_{1}, \quad y=2 y_{1}, \quad z=2 z_{1}
$$

where $x_{1}, y_{1}$, and $z_{1}$ are non-negative integers. We get:
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$$
4 x_{1}^{2}+4 y_{1}^{2}+4 z_{1}^{2}=16 x_{1} y_{1} z_{1}, \quad x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=4 x_{1} y_{1} z_{1}
$$

For the last equation, there can be the same two possibilities. The first one, as before, is ruled out, and in the second case, the numbers $x_{1}, y_{1}, z_{1}$ are divisible by 2, and therefore, the numbers $x, y$, and $z$ are already divisible by $2^{2}$.

Further, we obtain that each of the numbers $x, y$, and $z$ is divisible by $2^{3}$, then by $2^{4}$, and generally, by any power of 2 with any natural exponent. The latter is possible only if $x=y=z=0$.

Answer: It has, and indeed the unique solution is $(0, 0, 0)$.