Find all primes $p,q,r$ such that $qr-1$ is divisible by $p$, $pr-1$ is divisible by $q$, $pq-1$ is divisible by $r$.
1. We are given three primes \( p, q, r \) such that:
   \[
   qr - 1 \text{ is divisible by } p, \quad pr - 1 \text{ is divisible by } q, \quad pq - 1 \text{ is divisible by } r
   \]
   We need to find all such primes.

2. Assume \( p \leq q \leq r \). This assumption does not lose generality because the problem is symmetric in \( p, q, r \).

3. From the given conditions, we have:
   \[
   p \mid qr - 1, \quad q \mid pr - 1, \quad r \mid pq - 1
   \]

4. Consider the product of the three conditions:
   \[
   p \cdot q \cdot r \mid (qr - 1)(pr - 1)(pq - 1)
   \]

5. Expanding the right-hand side:
   \[
   (qr - 1)(pr - 1)(pq - 1) = qr \cdot pr \cdot pq - qr \cdot pr - pr \cdot pq - pq \cdot qr + qr + pr + pq - 1
   \]

6. Simplifying, we get:
   \[
   (qr - 1)(pr - 1)(pq - 1) = pqr \cdot pqr - pqr \cdot (qr + pr + pq) + (qr + pr + pq) - 1
   \]
   \[
   = (pqr)^2 - pqr \cdot (qr + pr + pq) + (qr + pr + pq) - 1
   \]

7. Since \( p \cdot q \cdot r \mid (pqr)^2 - pqr \cdot (qr + pr + pq) + (qr + pr + pq) - 1 \), we have:
   \[
   pqr \mid (qr + pr + pq) - 1
   \]

8. Let \( k \) be an integer such that:
   \[
   pqr \cdot k = qr + pr + pq - 1
   \]

9. Rearranging, we get:
   \[
   1 + k \cdot pqr = qr + pr + pq
   \]

10. Dividing both sides by \( pqr \):
    \[
    1 + k = \frac{qr}{pqr} + \frac{pr}{pqr} + \frac{pq}{pqr}
    \]
    \[
    1 + k = \frac{1}{p} + \frac{1}{q} + \frac{1}{r}
    \]

11. Since \( k \) is a positive integer, we have:
    \[
    k = \frac{1}{p} + \frac{1}{q} + \frac{1}{r} - 1
    \]

12. Given that \( p, q, r \) are primes, we need to check possible values for \( p, q, r \).

13. If \( p > 3 \), then \( \frac{1}{p} < \frac{1}{3} \). Similarly, \( \frac{1}{q} < \frac{1}{3} \) and \( \frac{1}{r} < \frac{1}{3} \). Thus:
    \[
    \frac{1}{p} + \frac{1}{q} + \frac{1}{r} < 1
    \]
    This contradicts \( k \geq 1 \). Hence, \( p, q, r \leq 3 \).

14. Since \( p, q, r \) are primes, the possible values are 2 and 3. We need to check combinations of these values.

15. If \( p = 2 \):
    \[
    \frac{1}{2} + \frac{1}{q} + \frac{1}{r} = 1 + k
    \]
    \[
    \frac{1}{q} + \frac{1}{r} = 1 - \frac{1}{2} + k = \frac{1}{2} + k
    \]

16. If \( q = 3 \):
    \[
    \frac{1}{3} + \frac{1}{r} = \frac{1}{2} + k
    \]
    \[
    \frac{1}{r} = \frac{1}{2} - \frac{1}{3} + k = \frac{1}{6} + k
    \]

17. Since \( r \) is a prime, the only possible value for \( r \) is 5:
    \[
    \frac{1}{5} = \frac{1}{6} + k
    \]
    \[
    k = \frac{1}{5} - \frac{1}{6} = \frac{6 - 5}{30} = \frac{1}{30}
    \]
    This is not an integer, so \( k = 0 \).

18. Therefore, the only solution is \( (p, q, r) = (2, 3, 5) \) and its permutations.

The final answer is \(\boxed{(2, 3, 5)}\) and its permutations.