Let $n$ be a natural number, with the prime factorisation
    \[ n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r} \] where $p_1, \ldots, p_r$ are distinct primes, and $e_i$ is a natural number. Define
    \[ rad(n) = p_1p_2 \cdots p_r \] to be the product of all distinct prime factors of $n$. Determine all polynomials $P(x)$ with rational coefficients such that there exists infinitely many naturals $n$ satisfying $P(n) = rad(n)$.
1. **Understanding the Problem:**
   We need to find all polynomials \( P(x) \) with rational coefficients such that there exist infinitely many natural numbers \( n \) satisfying \( P(n) = \text{rad}(n) \). Here, \(\text{rad}(n)\) is defined as the product of all distinct prime factors of \( n \).

2. **Analyzing the Range of \( \text{rad}(n) \):**
   Since \( \text{rad}(n) \) is the product of distinct primes dividing \( n \), we have:
   \[
   0 < \text{rad}(n) \leq n
   \]
   for all \( n \in \mathbb{N} \). This implies:
   \[
   0 < P(n) \leq n
   \]
   for infinitely many \( n \).

3. **Degree of \( P(x) \):**
   Given that \( P(n) \leq n \) for infinitely many \( n \), we can infer:
   \[
   \lim_{n \to \infty} \frac{P(n)}{n} \leq 1
   \]
   This implies that the degree of \( P(x) \) must be at most 1, i.e., \( \deg P(x) \leq 1 \).

4. **Case 1: \( P(x) \) is a constant polynomial:**
   Let \( P(x) = c \) where \( c \in \mathbb{Q} \). For \( P(n) = \text{rad}(n) \) to hold for infinitely many \( n \), \( c \) must be a square-free integer (since \( \text{rad}(n) \) is always square-free). Therefore, \( P(x) = c \) where \( c \) is a square-free integer.

5. **Case 2: \( P(x) \) is a linear polynomial:**
   Let \( P(x) = \frac{ax + b}{c} \) for some \( a, b, c \in \mathbb{Z} \) with \( a, c > 0 \). We need to check if this form can satisfy \( P(n) = \text{rad}(n) \) for infinitely many \( n \).

   - **Subcase 2.1: \( b \neq 0 \):**
     If \( b \neq 0 \), then for \( P(n) = \text{rad}(n) \) to hold for infinitely many \( n \), we would have:
     \[
     \text{rad}(n) \mid b
     \]
     for infinitely many \( n \). This is impossible because \( \text{rad}(n) \) can take arbitrarily large values, leading to a contradiction.

   - **Subcase 2.2: \( b = 0 \):**
     If \( b = 0 \), then \( P(x) = \frac{ax}{c} \). For \( P(n) = \text{rad}(n) \) to hold for infinitely many \( n \), we need:
     \[
     an = c \cdot \text{rad}(n)
     \]
     This implies:
     \[
     \frac{n}{\text{rad}(n)} = \frac{c}{a}
     \]
     Since \( \frac{n}{\text{rad}(n)} \) is an integer, \( c \) must be a multiple of \( a \). We can then write \( c = ka \) for some \( k \in \mathbb{N} \), and thus:
     \[
     P(x) = \frac{x}{k}
     \]

6. **Conclusion:**
   The polynomials \( P(x) \) that satisfy the given condition are of the form \( P(x) = \frac{x}{k} \) where \( k \in \mathbb{N} \) and \( P(x) = c \) where \( c \) is a square-free integer.

The final answer is \( \boxed{ P(x) = \frac{x}{k} } \) where \( k \in \mathbb{N} \) or \( P(x) = c \) where \( c \) is a square-free integer.