Find all ordered triplets $(p,q,r)$ of positive integers such that $p$ and $q$ are two (not necessarily distinct) primes, $r$ is even, and 
\[p^3+q^2=4r^2+45r+103.\]
1. Given the equation \( p^3 + q^2 = 4r^2 + 45r + 103 \), where \( p \) and \( q \) are primes and \( r \) is an even positive integer.
2. Since \( r \) is even, let \( r = 2k \) for some integer \( k \). Substituting \( r = 2k \) into the equation, we get:
   \[
   p^3 + q^2 = 4(2k)^2 + 45(2k) + 103 = 16k^2 + 90k + 103
   \]
3. We need to consider the parity of \( p \) and \( q \). Since \( p \) and \( q \) are primes, one of them must be 2 (the only even prime) to ensure the left-hand side is even. Let's consider the cases where \( p = 2 \) or \( q = 2 \).

### Case 1: \( q = 2 \)
4. Substituting \( q = 2 \) into the equation, we get:
   \[
   p^3 + 2^2 = 4r^2 + 45r + 103 \implies p^3 + 4 = 4r^2 + 45r + 103
   \]
   \[
   p^3 = 4r^2 + 45r + 99
   \]
5. Notice that \( p^3 = 4r^2 + 45r + 99 \) can be factored as:
   \[
   p^3 = (r+3)(4r + 33)
   \]
6. Since \( p \) is a prime, \( p^3 \) has exactly 4 divisors: \( 1, p, p^2, p^3 \). Therefore, \( r+3 \) and \( 4r+33 \) must be among these divisors. Given \( 1 < r+3 < 4r+33 \), we have:
   \[
   r+3 = p \quad \text{and} \quad 4r+33 = p^2
   \]
7. Solving the system of equations:
   \[
   r + 3 = p \implies r = p - 3
   \]
   \[
   4(p-3) + 33 = p^2 \implies 4p - 12 + 33 = p^2 \implies p^2 - 4p - 21 = 0
   \]
8. Solving the quadratic equation:
   \[
   p^2 - 4p - 21 = 0 \implies p = \frac{4 \pm \sqrt{16 + 84}}{2} = \frac{4 \pm 10}{2}
   \]
   \[
   p = 7 \quad \text{or} \quad p = -3 \quad (\text{not a prime})
   \]
9. Thus, \( p = 7 \) and:
   \[
   r = 7 - 3 = 4
   \]
10. Substituting \( p = 7 \) and \( r = 4 \) back into the original equation to verify:
    \[
    7^3 + 2^2 = 4(4)^2 + 45(4) + 103 \implies 343 + 4 = 64 + 180 + 103 \implies 347 = 347
    \]
    The solution is verified.

### Case 2: \( p = 2 \)
11. Substituting \( p = 2 \) into the equation, we get:
    \[
    2^3 + q^2 = 4r^2 + 45r + 103 \implies 8 + q^2 = 4r^2 + 45r + 103
    \]
    \[
    q^2 = 4r^2 + 45r + 95
    \]
12. For \( r \) to be an integer, the discriminant of the quadratic equation \( 4r^2 + 45r + (95 - q^2) = 0 \) must be a perfect square:
    \[
    \Delta = 45^2 - 16(95 - q^2) = n^2
    \]
    \[
    2025 - 1520 + 16q^2 = n^2 \implies 505 + 16q^2 = n^2
    \]
13. This implies:
    \[
    (n + 4q)(n - 4q) = 505
    \]
14. Solving for \( n \) and \( q \):
    \[
    n + 4q = 505 \quad \text{and} \quad n - 4q = 1 \implies 2n = 506 \implies n = 253 \implies 8q = 504 \implies q = 63 \quad (\text{not a prime})
    \]
    \[
    n + 4q = 101 \quad \text{and} \quad n - 4q = 5 \implies 2n = 106 \implies n = 53 \implies 8q = 96 \implies q = 12 \quad (\text{not a prime})
    \]
15. Since \( q \) is not a prime in both cases, there are no solutions for \( p = 2 \).

Conclusion:
The only solution is \( (p, q, r) = (7, 2, 4) \).

The final answer is \( \boxed{ (7, 2, 4) } \)