Find all pairs $(a,b)$ of prime positive integers $a,b$ such that number $A=3a^2b+16ab^2$ equals to a square of an integer.
To find all pairs \((a, b)\) of prime positive integers such that \(A = 3a^2b + 16ab^2\) is a perfect square, we start by analyzing the given expression.

1. **Factor the expression**:
   \[
   A = 3a^2b + 16ab^2 = ab(3a + 16b)
   \]
   For \(A\) to be a perfect square, \(ab(3a + 16b)\) must be a perfect square.

2. **Case 1: \(a = b\)**:
   \[
   A = ab(3a + 16b) = a^2(3a + 16a) = a^2 \cdot 19a = 19a^3
   \]
   For \(19a^3\) to be a perfect square, \(a\) must be a perfect square. Since \(a\) is a prime number, the only prime that is a perfect square is \(a = 19\). Thus, \(a = b = 19\).

3. **Case 2: \(a \neq b\)**:
   Since \(ab(3a + 16b)\) is a perfect square and \(a\) and \(b\) are distinct primes, we need \(ab\) and \(3a + 16b\) to be structured such that their product is a perfect square. This implies:
   \[
   a \mid 3a + 16b \quad \text{and} \quad b \mid 3a + 16b
   \]
   Since \(a\) and \(b\) are primes, we can deduce:
   \[
   a \mid 16b \quad \text{and} \quad b \mid 3a
   \]
   Given that \(a\) and \(b\) are distinct primes, the only possible values are:
   \[
   a = 2 \quad \text{and} \quad b = 3
   \]
   Let's verify:
   \[
   A = 3(2)^2(3) + 16(2)(3)^2 = 3 \cdot 4 \cdot 3 + 16 \cdot 2 \cdot 9 = 36 + 288 = 324 = 18^2
   \]
   Thus, \((a, b) = (2, 3)\) is a valid solution.

Therefore, the pairs \((a, b)\) that satisfy the condition are \((19, 19)\) and \((2, 3)\).

The final answer is \(\boxed{(19, 19) \text{ and } (2, 3)}\).