For every natural number $ n$, find all polynomials $ x^2\plus{}ax\plus{}b$, where $ a^2 \ge 4b$, that divide $ x^{2n}\plus{}ax^n\plus{}b$.
To solve the problem, we need to find all polynomials \( x^2 + ax + b \) that divide \( x^{2n} + ax^n + b \) for every natural number \( n \), given that \( a^2 \ge 4b \).

1. **Case \( n = 1 \):**
   \[
   x^2 + ax + b \text{ divides } x^2 + ax + b
   \]
   This is trivially true for any \( a \) and \( b \).

2. **Case \( b = 0 \):**
   \[
   x^2 + ax \text{ divides } x^{2n} + ax^n
   \]
   Factoring out \( x \) from both polynomials:
   \[
   x(x + a) \text{ divides } x^n(x^n + a)
   \]
   For this to hold, \( x + a \) must divide \( x^n + a \). This implies:
   \[
   x^n + a = (x + a)Q(x)
   \]
   Evaluating at \( x = -a \):
   \[
   (-a)^n + a = 0 \implies a^{n+1} = 0
   \]
   Therefore, \( a = 0 \).

3. **Case \( b \neq 0 \):**
   Let \( x_1 \) and \( x_2 \) be the roots of \( x^2 + ax + b \). Then:
   \[
   x^2 + ax + b = (x - x_1)(x - x_2)
   \]
   Since \( x^2 + ax + b \) divides \( x^{2n} + ax^n + b \), the roots \( x_1 \) and \( x_2 \) must satisfy:
   \[
   x_1^n = x_1 \quad \text{and} \quad x_2^n = x_2
   \]
   This implies \( x_1 \) and \( x_2 \) are roots of unity, specifically \( x_1, x_2 \in \{1, -1\} \).

4. **Subcase \( x_1 = x_2 = 1 \):**
   \[
   (x - 1)^2 \text{ divides } (x^n - 1)^2
   \]
   This implies:
   \[
   a = -2, \quad b = 1
   \]

5. **Subcase \( x_1 = x_2 = -1 \):**
   \[
   (x + 1)^2 \text{ divides } (x^n + 1)^2
   \]
   This implies:
   \[
   a = 2, \quad b = 1
   \]

6. **Subcase \( x_1 = 1, x_2 = -1 \):**
   \[
   (x - 1)(x + 1) = x^2 - 1 \text{ divides } x^{2n} - 1
   \]
   This implies:
   \[
   a = 0, \quad b = -1
   \]

Therefore, the polynomials \( x^2 + ax + b \) that divide \( x^{2n} + ax^n + b \) for every natural number \( n \) are:
\[
x^2 - 2x + 1, \quad x^2 + 2x + 1, \quad x^2 - 1
\]

The final answer is \( \boxed{ x^2 - 2x + 1, x^2 + 2x + 1, x^2 - 1 } \).