Find all sets of positive integers $\{x_1, x_2, \dots, x_{20}\}$ such that $$x_{i+2}^2=lcm(x_{i+1}, x_{i})+lcm(x_{i}, x_{i-1})$$ for $i=1, 2, \dots, 20$ where $x_0=x_{20}, x_{21}=x_1, x_{22}=x_2$.
1. **Claim:** For every \( i \), \(\gcd(x_{i+1}, x_i) \geq 2\).

   - To prove this, assume there exists an \( i \) such that \( x_i \) is divisible by a prime \( p \). If no such \( i \) exists, then \( x_i = 1 \) for all \( i \), which does not satisfy the given equation.
   - Given the equation \( x_{i+2}^2 = \text{lcm}(x_{i+1}, x_i) + \text{lcm}(x_i, x_{i-1}) \), we can deduce that \( p \mid x_i \) implies \( p \mid x_{i+2} \).
   - From the equation \( x_{i+3}^2 = \text{lcm}(x_{i+2}, x_{i+1}) + \text{lcm}(x_{i+1}, x_i) \), we get \( (x_{i+3} - x_{i+2})(x_{i+3} + x_{i+2}) = \text{lcm}(x_{i+2}, x_{i+1}) - \text{lcm}(x_i, x_{i-1}) \).
   - Since \( p \mid x_i \), it follows that \( p \mid x_{i+3} \). By induction, every \( x_i \) is divisible by \( p \). Therefore, for every \( i \), \(\gcd(x_{i+1}, x_i) \geq 2\).

2. **Summing the Equations:**
   - Summing all the given equations, we obtain:
     \[
     2 \sum_{i=1}^{20} \text{lcm}(x_{i+1}, x_i) = \sum_{i=1}^{20} x_i^2
     \]
   - Using the inequality \(\sum_{i=1}^{20} x_i^2 \geq \sum_{i=1}^{20} x_i x_{i+1}\), where equality holds if and only if \( x_i = x_{i+1} \) for all \( i \).

3. **Rewriting the LCM:**
   - Rewrite \(\text{lcm}(x_{i+1}, x_i) = \frac{x_i x_{i+1}}{\gcd(x_{i+1}, x_i)}\).
   - Since \(\gcd(x_{i+1}, x_i) \geq 2\), we conclude that the inequality must hold, and therefore all integers must be equal.

4. **Solving for \( x_i \):**
   - Given \( x_i = x_{i+1} \) for all \( i \), we substitute into the original equation:
     \[
     x_i^2 = 2x_i
     \]
   - Solving \( x_i^2 = 2x_i \) gives \( x_i (x_i - 2) = 0 \), so \( x_i = 2 \).

Conclusion:
\[
\boxed{x_i = 2 \text{ for all } i \in \{1, 2, \ldots, 20\}}
\]