Let $p$ and $q$ be prime numbers. The sequence $(x_n)$ is defined by $x_1 = 1$, $x_2 = p$ and $x_{n+1} = px_n - qx_{n-1}$ for all $n \geq 2$.
Given that there is some $k$ such that $x_{3k} = -3$, find $p$ and $q$.
1. **Initial Conditions and Recurrence Relation:**
   The sequence \((x_n)\) is defined by:
   \[
   x_1 = 1, \quad x_2 = p, \quad x_{n+1} = px_n - qx_{n-1} \quad \text{for all } n \geq 2.
   \]
   We are given that \(x_{3k} = -3\) for some \(k\).

2. **Considering Modulo 2:**
   If both \(p\) and \(q\) are odd primes, then modulo 2, the sequence behaves as follows:
   \[
   x_1 \equiv 1 \pmod{2}, \quad x_2 \equiv 1 \pmod{2}, \quad x_3 \equiv p \cdot 1 - q \cdot 1 \equiv 0 \pmod{2}.
   \]
   This pattern repeats every three terms:
   \[
   x_4 \equiv p \cdot 0 - q \cdot 1 \equiv 1 \pmod{2}, \quad x_5 \equiv p \cdot 1 - q \cdot 0 \equiv 1 \pmod{2}, \quad x_6 \equiv p \cdot 1 - q \cdot 1 \equiv 0 \pmod{2}.
   \]
   Thus, \(x_{3k} \equiv 0 \pmod{2}\), which contradicts \(x_{3k} = -3\). Therefore, either \(p = 2\) or \(q = 2\).

3. **Case Analysis:**
   - **Case 1: \(p = 2\):**
     \[
     x_3 = 2^2 - q = 4 - q.
     \]
     Given \(x_{3k} = -3\), we need \(x_3 = -3\):
     \[
     4 - q = -3 \implies q = 7.
     \]
     Therefore, \(p = 2\) and \(q = 7\).

   - **Case 2: \(q = 2\):**
     \[
     x_3 = p^2 - 2.
     \]
     Given \(x_{3k} = -3\), we need \(x_3 = -3\):
     \[
     p^2 - 2 = -3 \implies p^2 = -1.
     \]
     This is not possible since \(p\) is a prime number and \(p^2\) cannot be negative.

4. **Verification:**
   To verify, we substitute \(p = 2\) and \(q = 7\) back into the sequence:
   \[
   x_1 = 1, \quad x_2 = 2, \quad x_3 = 2^2 - 7 = 4 - 7 = -3.
   \]
   This satisfies the condition \(x_{3k} = -3\).

The final answer is \( \boxed{ p = 2 } \) and \(q = 7\).