Find all primes $ p,q $ such that $ \alpha^{3pq} -\alpha \equiv 0 \pmod {3pq} $ for all integers $ \alpha $.
1. We start by noting that \(3, p, q\) are distinct primes. We need to find all primes \(p\) and \(q\) such that \(\alpha^{3pq} - \alpha \equiv 0 \pmod{3pq}\) for all integers \(\alpha\).

2. Assume \(p = q\). Then we have \(n^{3p^2} \equiv n \pmod{p^2}\) for all \(n \in \mathbb{Z}\). If we take \(n = p\), we get:
   \[
   p^{3p^2} \equiv p \pmod{p^2} \implies p^{3p^2 - 1} \equiv 1 \pmod{p^2}
   \]
   This implies:
   \[
   p^{3p^2 - 1} \equiv 1 \pmod{p}
   \]
   Since \(p^{3p^2 - 1} \equiv 0 \pmod{p}\), we get \(0 \equiv 1 \pmod{p}\), which is a contradiction. Therefore, \(p \neq q\).

3. Let \(g_1\) and \(g_2\) be primitive roots modulo \(p\) and \(q\) respectively. Then:
   \[
   g_1^{3pq} \equiv g_1 \pmod{p} \implies p - 1 \mid 3pq - 1
   \]
   Similarly:
   \[
   g_2^{3pq} \equiv g_2 \pmod{q} \implies q - 1 \mid 3pq - 1
   \]
   Thus, we have:
   \[
   p - 1 \mid 3q - 1 \quad \text{and} \quad q - 1 \mid 3p - 1
   \]

4. Let \(3q - 1 = x(p - 1)\) and \(3p - 1 = y(q - 1)\) for some \(x, y \in \mathbb{N}\). Notice that \(1 \leq x \leq 3\).

5. If \(x = 1\), we have:
   \[
   3q - 1 = p - 1 \implies 3q = p
   \]
   This is absurd because \(p\) and \(q\) are distinct primes.

6. If \(x = 3\), we have:
   \[
   3q - 1 = 3p - 3 \implies 3q = 3p - 2 \implies q = p - \frac{2}{3}
   \]
   Reducing modulo 3 gives us \(2 \equiv 0 \pmod{3}\), which is another contradiction.

7. Therefore, \(x = 2\). Now we have:
   \[
   3q - 1 = 2(p - 1) \implies 3q - 1 = 2p - 2 \implies 3q = 2p - 1
   \]
   and:
   \[
   3p - 1 = y(q - 1)
   \]

8. Solving for \(q\) in the system of equations gives us:
   \[
   q = \frac{2p - 1}{3}
   \]
   Substituting \(q\) into the second equation:
   \[
   3p - 1 = y\left(\frac{2p - 1}{3} - 1\right)
   \]
   Simplifying, we get:
   \[
   3p - 1 = y\left(\frac{2p - 1 - 3}{3}\right) \implies 3p - 1 = y\left(\frac{2p - 4}{3}\right) \implies 9p - 3 = 2yp - 4y
   \]
   Rearranging, we get:
   \[
   9p - 2yp = 4y - 3 \implies p(9 - 2y) = 4y - 3
   \]

9. Solving for \(p\), we get:
   \[
   p = \frac{4y - 3}{9 - 2y}
   \]
   For \(p\) to be an integer, \(9 - 2y\) must divide \(4y - 3\). Testing values of \(y\), we find that \(y = 5\) and \(y = 7\) work.

10. If \(y = 5\), then:
    \[
    p = \frac{4(5) - 3}{9 - 2(5)} = \frac{20 - 3}{9 - 10} = \frac{17}{-1} = -17
    \]
    which is not a prime.

11. If \(y = 7\), then:
    \[
    p = \frac{4(7) - 3}{9 - 2(7)} = \frac{28 - 3}{9 - 14} = \frac{25}{-5} = -5
    \]
    which is also not a prime.

12. Therefore, the only valid solution is when \(p = 11\) and \(q = 17\).

The final answer is \( \boxed{ (p, q) = (11, 17) } \)