Let $x_1,x_2,x_3, \cdots$ be a sequence defined by the following recurrence relation:
\[ \begin{cases}x_{1}&= 4\\ x_{n+1}&= x_{1}x_{2}x_{3}\cdots x_{n}+5\text{ for }n\ge 1\end{cases} \]
The first few terms of the sequence are $x_1=4,x_2=9,x_3=41 \cdots$

Find all pairs of positive integers $\{a,b\}$ such that $x_a x_b$ is a perfect square.
1. **Define the sequence and initial conditions:**
   The sequence \( \{x_n\} \) is defined by:
   \[
   \begin{cases}
   x_1 = 4 \\
   x_{n+1} = x_1 x_2 x_3 \cdots x_n + 5 \text{ for } n \geq 1
   \end{cases}
   \]
   The first few terms are given as \( x_1 = 4 \), \( x_2 = 9 \), and \( x_3 = 41 \).

2. **Assume \( a < b \) and express \( x_b \):**
   \[
   x_b = x_1 x_2 \cdots x_{b-1} + 5
   \]
   Suppose there is a prime \( p \) that divides both \( x_a \) and \( x_b \). Then \( p \) must divide \( 5 \) because:
   \[
   x_b - x_1 x_2 \cdots x_{b-1} = 5
   \]
   This implies \( p = 5 \).

3. **Check if \( p = 5 \) divides any \( x_i \):**
   If \( p = 5 \) divides \( x_a \), then it must divide some \( x_c \) where \( c < a \), and so on, until it divides \( x_1 \) or \( x_2 \). However, \( x_1 = 4 \) and \( x_2 = 9 \), neither of which is divisible by 5. Therefore, \( p \neq 5 \).

4. **Conclude that \( \gcd(x_a, x_b) = 1 \):**
   Since no prime \( p \) divides both \( x_a \) and \( x_b \), we have:
   \[
   \gcd(x_a, x_b) = 1
   \]
   This implies that \( x_a \) and \( x_b \) must individually be perfect squares for their product to be a perfect square.

5. **Check the congruence modulo 3 for \( x_n \) where \( n > 2 \):**
   For \( n > 2 \), we observe that:
   \[
   x_1 \equiv 1 \pmod{3}, \quad x_2 \equiv 0 \pmod{3}, \quad x_3 \equiv 2 \pmod{3}
   \]
   We need to show that \( x_n \equiv 2 \pmod{3} \) for \( n > 2 \). Assume \( x_n \equiv 2 \pmod{3} \) for some \( n \geq 3 \). Then:
   \[
   x_{n+1} = x_1 x_2 \cdots x_n + 5 \equiv 1 \cdot 0 \cdot 2 \cdots 2 + 5 \equiv 0 + 5 \equiv 2 \pmod{3}
   \]
   By induction, \( x_n \equiv 2 \pmod{3} \) for all \( n > 2 \).

6. **Conclude that \( x_a \) and \( x_b \) are not perfect squares for \( a, b > 2 \):**
   Since \( x_n \equiv 2 \pmod{3} \) for \( n > 2 \), \( x_a \) and \( x_b \) cannot be perfect squares for \( a, b > 2 \).

7. **Identify the possible pairs \( (a, b) \):**
   The only possibilities are \( (a, b) = (1, 2) \) or \( a = b \). For \( a = 1 \) and \( b = 2 \):
   \[
   x_1 x_2 = 4 \times 9 = 36
   \]
   which is a perfect square.

The final answer is \( \boxed{ \{(1, 2), (2, 1)\} } \) and \(\{(a, a)\}\) for any \(a\).