5. Find all positive integers $n$ such that the ternary polynomial
$$
\begin{array}{l} 
P_{n}(x, y, z) \\
=(x-y)^{2 n}(y-z)^{2 n}+(y-z)^{2 n}(z-x)^{2 n}+ \\
(z-x)^{2 n}(x-y)^{2 n}
\end{array}
$$

divides the ternary polynomial
$$
\begin{array}{l}
Q_{n}(x, y, z) \\
=\left[(x-y)^{2 n}+(y-z)^{2 n}+(z-x)^{2 n}\right]^{2 n} .
\end{array}
$$
5. The only positive integer that satisfies the condition is $n=1$, and it is easy to verify that $Q_{1}=4 P_{1}$.
Below is the proof: For $n \geqslant 2, P_{n} \times Q_{n}$.
Assume $P_{n} \mid Q_{n}$, then set
$$
Q_{n}(x, y, z)=R_{n}(x, y, z) P_{n}(x, y, z) .
$$

Define $p_{n}(x)=P_{n}(x, 0,-1)$,
$$
q_{n}(x)=Q_{n}(x, 0,-1) \text {. }
$$

Then $q_{n}=R_{n}(x, 0,-1) p_{n}$.
$p_{n}$ and $q_{n}$, as polynomials in $x$ with real coefficients, have $p_{n} \mid q_{n}$. In particular, $p_{n}$ and $q_{n}$ are both polynomials with integer coefficients, and $p_{n}$ is monic.

Therefore, $R_{n}(x, 0,-1)$ is a polynomial in $x$ with integer coefficients. In particular, when $x$ takes integer values, it also takes integer values.

For any integer $a$, $p_{n}(a)$ and $q_{n}(a)$, as integers, have $p_{n}(a) \mid q_{n}(a)$.
Let $a=1$. Then
$$
\left(2^{2 n+1}+1\right) \mid\left(2^{2 n}+2\right)^{2 n} \text {. }
$$

Let $p$ be a prime such that $p \mid\left(2^{2 n+1}+1\right)$. Then $p \mid\left(2^{2 n}+2\right)$.
Thus, $0 \equiv 2^{2 n+1}+1-2\left(2^{2 n}+2\right) \equiv-3(\bmod p)$.
Therefore, $p=3$. This indicates that $2^{2 n+1}+1=3^{k}$, where $k$ is some positive integer.
Since the left side is congruent to 1 modulo 4, $k$ must be even.
Let $k=2 m$. Then
$$
2^{2 n+1}=\left(3^{m}+1\right)\left(3^{m}-1\right) \text {. }
$$

Therefore, $3^{m}+1$ and $3^{m}-1$ are both powers of 2.
Thus, $m=1, n=1$.
In conclusion, $n=1$ is the only positive integer that satisfies the condition.