Find all primes $p,q, r$ such that $\frac{p^{2q}+q^{2p}}{p^3-pq+q^3} = r$.

Titu Andreescu, Mathematics Department, College of Texas, USA
We are given the Diophantine equation:
\[
\frac{p^{2q} + q^{2p}}{p^3 - pq + q^3} = r
\]
where \( p, q, \) and \( r \) are primes. We need to find all such primes.

1. **Assume \( p = q \):**
   \[
   \frac{p^{2p} + p^{2p}}{p^3 - p^2 + p^3} = r \implies \frac{2p^{2p}}{2p^3 - p^2} = r \implies \frac{2p^{2p}}{p^2(2p - 1)} = r \implies \frac{2p^{2p-2}}{2p - 1} = r
   \]
   Since \( \gcd(2p, 2p-1) = 1 \), we must have \( 2p - 1 = 1 \), which implies \( p = 1 \). However, 1 is not a prime number, so \( p \neq q \).

2. **Assume \( p = 2 \):**
   \[
   \frac{2^{2q} + q^4}{2^3 - 2q + q^3} = r \implies \frac{4^q + q^4}{8 - 2q + q^3} = r
   \]
   - **Check \( q = 5 \):**
     \[
     \frac{4^5 + 5^4}{8 - 2 \cdot 5 + 5^3} = \frac{1024 + 625}{8 - 10 + 125} = \frac{1649}{123}
     \]
     Since \( 3 \mid 123 \) and \( 4^q + q^4 \equiv 2 \pmod{3} \), this is impossible. Hence, \( q \neq 5 \).

   - **Using Fermat's Little Theorem:**
     \[
     4^q + q^4 \equiv (-1)^q + 1 = -1 + 1 = 0 \pmod{5}
     \]
     This implies:
     \[
     5 \mid r(q^3 - 2q + 8)
     \]
     Since \( q^3 - 2q + 3 \equiv 2, 3, 4 \pmod{5} \), we must have \( 5 \mid r \). Thus, \( r = 5 \).

   - **Check \( q = 3 \):**
     \[
     \frac{4^3 + 3^4}{8 - 2 \cdot 3 + 3^3} = \frac{64 + 81}{8 - 6 + 27} = \frac{145}{29} = 5
     \]
     This is valid, so \( (p, q, r) = (2, 3, 5) \) is a solution.

3. **Assume \( p \neq 2 \):**
   Since \( p \) and \( q \) are both odd primes:
   \[
   p^{2q} + q^{2p} \text{ is even and } p^3 - pq + q^3 \text{ is odd}
   \]
   This implies \( r \) must be even, so \( r = 2 \). However:
   \[
   2(p^3 - pq + q^3) = p^{2q} + q^{2p} > 2p^q \cdot q^p > 2(pq)^3
   \]
   This leads to a contradiction:
   \[
   (p^3 - 1)(q^3 - 1) < 1 - pq < 0
   \]
   Hence, there are no solutions when \( p \) and \( q \) are both odd primes.

Conclusion:
The only solutions of the equation in primes are \( (p, q, r) = (2, 3, 5) \) and \( (p, q, r) = (3, 2, 5) \).

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