5. For every non-negative integer $n$, define $\operatorname{rad}(n)$ as follows: if $n=0$ or 1, then $\operatorname{rad}(n)=1$; if the prime factors of $n$ are $p_{1}, p_{2}, \cdots, p_{k}\left(p_{1}<p_{2}<\cdots<p_{k}\right)$, then $\operatorname{rad}(n)=p_{1} p_{2} \cdots p_{k}$. Find all non-negative integer coefficient polynomials $f(x)$ such that for each non-negative integer $n$, we have
$$
\operatorname{rad}(f(n)) \mid \operatorname{rad}\left(f\left(n^{\operatorname{rad}(n)}\right)\right).
$$
5. First, we explain: $f(x)=a x^{m}(a, m \in \mathbf{N})$ satisfies the condition.
If $f(x)$ is the zero polynomial, the conclusion is obviously true.
If $f(x)$ has at least one positive coefficient, then $f(1)>0$.
Let $p$ be a prime. Then
$$
\begin{array}{l}
f(n) \equiv 0(\bmod p) \\
\Rightarrow f\left(n^{\operatorname{rad}(n)}\right) \equiv 0(\bmod p) .
\end{array}
$$

Since for all positive integers $k$ we have
$$
\operatorname{rad}\left(n^{\operatorname{rad}(n)^{k}}\right)=\operatorname{rad}(n),
$$

repeatedly applying equation (1) yields that if $p \mid f(n)$, then for all positive integers $k$ we have
$$
f\left(n^{\operatorname{rad}(n)^{k}}\right) \equiv 0(\bmod p) .
$$

Construct a prime $p$ and a positive integer $n$ such that
$$
(p-1) \mid n \text {, and } p \mid f(n) \text {. }
$$

At this point, for sufficiently large $k$ we have
$$
(p-1) \mid \operatorname{rad}(n)^{k} \text {. }
$$

Thus, if $(p, n)=1$, by Fermat's Little Theorem we know
$$
n^{\operatorname{rad}(n)^{k}} \equiv 1(\bmod p) \text {. }
$$

Therefore, $f(1) \equiv f\left(n^{\operatorname{rad}(n)^{k}}\right) \equiv 0(\bmod p)$.
Assume $f(x)=g(x) x^{m}$, where $g(0) \neq 0$.
Let $t$ be a positive integer, and $p$ be any prime factor of $g(-t)$, and $n=(p-1) t$. Then
$$
\begin{array}{l}
(p-1) \mid n, \\
f(n)=f((p-1) t) \equiv f(-t) \equiv 0(\bmod p) .
\end{array}
$$

Thus, either $(p, n)>1$ or equation (2) holds.
If $(p,(p-1) t)>1$, then $p \mid t$.
Hence $g(0) \equiv g(-t) \equiv 0(\bmod p)$, which indicates that $p \mid g(0)$.
In summary, for each prime factor $p$ of $g(-t)$, we have
$$
p \mid g(0) f(1) \neq 0 \text {. }
$$

Therefore, when $t$ takes all positive integers, the set of prime factors of $g(-t)$ is finite. Thus, $g(x)$ is a constant polynomial.
$$
\text { Hence, } f(x)=a x^{m} \text {. }
$$