Find all second degree polynomial $d(x)=x^{2}+ax+b$ with integer coefficients, so that there exists an integer coefficient polynomial $p(x)$ and a non-zero integer coefficient polynomial $q(x)$ that satisfy: \[\left( p(x) \right)^{2}-d(x) \left( q(x) \right)^{2}=1, \quad \forall x \in \mathbb R.\]
To find all second-degree polynomials \( d(x) = x^2 + ax + b \) with integer coefficients such that there exist integer coefficient polynomials \( p(x) \) and a non-zero integer coefficient polynomial \( q(x) \) satisfying:
\[
(p(x))^2 - d(x) (q(x))^2 = 1, \quad \forall x \in \mathbb{R},
\]
we will proceed as follows:

1. **Analyzing the given equation:**
   \[
   (p(x))^2 - (x^2 + ax + b)(q(x))^2 = 1.
   \]
   This is a form of the Pell's equation in polynomial form.

2. **Considering the degrees of \( p(x) \) and \( q(x) \):**
   - Let \( \deg(p(x)) = m \) and \( \deg(q(x)) = n \).
   - The left-hand side of the equation has degree \( 2m \) and the right-hand side is a constant (degree 0).
   - Therefore, \( 2m = 2n + 2 \) must hold, implying \( m = n + 1 \).

3. **Case analysis based on the parity of \( a \):**
   - **Case 1: \( a \) is odd.**
     - If \( a \) is odd, then \( 4b = a^2 - 1 \).
     - This is derived from the fact that the discriminant of the quadratic \( x^2 + ax + b \) must be a perfect square for the Pell's equation to have integer solutions.
     - Therefore, \( b = \frac{a^2 - 1}{4} \).

   - **Case 2: \( a \) is even.**
     - If \( a = 2r \) for some integer \( r \), then \( b = \frac{a^2}{4} + k \) where \( k \mid 2 \).
     - This is because the discriminant must again be a perfect square.

4. **Constructing the polynomials \( p(x) \) and \( q(x) \):**
   - For \( a \) even, let \( a = 2r \) and \( d(x) = x^2 + 2rx + b \).
   - Let \( f(x) = p(x + r) \) and \( g(x) = q(x + r) \).
   - Then we have:
     \[
     (p(x))^2 - (x^2 + k)(q(x))^2 = 1.
     \]
   - Let \( R(x) = p(x) - xq(x) \). Then:
     \[
     R(x)(2xq(x) + R(x)) = 1 + k(q(x))^2.
     \]

5. **Finding minimal solutions:**
   - If \( \deg(p) = 2 \), then the only solution for \( q(x) \) in \( \mathbb{Q}[x] \) is \( q(x) = \frac{2}{k}x \) and \( p(x) = 1 + \frac{2}{k}x^2 \).
   - If \( k = -m^2 \) for some \( m \in \mathbb{Q}^+ \), then the minimal solution is \( p(x) = mx \) and \( q(x) = m \).

6. **General solution construction:**
   - For \( k \neq -m^2 \), the minimal solution is:
     \[
     p_n(x) = 2p_0p_{n-1}(x) - p_{n-2}(x), \quad q_n(x) = 2p_0q_{n-1}(x) - q_{n-2}(x),
     \]
     where \( p_0 = 1 \), \( q_0 = 0 \), \( q_1(x) = \frac{2}{k}x \), and \( p_1(x) = 1 + \frac{2}{k}x^2 \).

7. **Conclusion:**
   - For \( k \mid 2 \), the solution exists in \( \mathbb{Z}[x] \).
   - For \( k = -1 \), the solution is \( p(x) = x \) and \( q(x) = 1 \).

The final answer is \( \boxed{ d(x) = x^2 + ax + b } \) where \( b = \frac{a^2 - 1}{4} \) if \( a \) is odd, and \( b = \frac{a^2}{4} + k \) with \( k \mid 2 \) if \( a \) is even.