Find all pair of constants $(a,b)$ such that there exists real-coefficient polynomial $p(x)$ and $q(x)$ that satisfies the condition below.

[b]Condition[/b]: $\forall x\in \mathbb R,$ $ $ $p(x^2)q(x+1)-p(x+1)q(x^2)=x^2+ax+b$
To find all pairs of constants \((a, b)\) such that there exist real-coefficient polynomials \(p(x)\) and \(q(x)\) satisfying the condition:
\[
\forall x \in \mathbb{R}, \quad p(x^2)q(x+1) - p(x+1)q(x^2) = x^2 + ax + b,
\]
we proceed as follows:

1. **Substitute specific values of \(x\)**:
   Let's consider the roots of the quadratic equation \(x^2 - x - 1 = 0\). The roots are:
   \[
   \alpha = \frac{1 + \sqrt{5}}{2} \quad \text{and} \quad \beta = \frac{1 - \sqrt{5}}{2}.
   \]
   These roots satisfy \(\alpha^2 = \alpha + 1\) and \(\beta^2 = \beta + 1\).

2. **Evaluate the given condition at \(x = \alpha\)**:
   \[
   p(\alpha^2)q(\alpha + 1) - p(\alpha + 1)q(\alpha^2) = \alpha^2 + a\alpha + b.
   \]
   Using \(\alpha^2 = \alpha + 1\) and \(\alpha + 1 = \alpha^2\), we get:
   \[
   p(\alpha + 1)q(\alpha^2) - p(\alpha^2)q(\alpha + 1) = \alpha^2 + a\alpha + b.
   \]
   Since \(\alpha + 1 = \alpha^2\) and \(\alpha^2 = \alpha + 1\), the left-hand side becomes zero:
   \[
   0 = \alpha^2 + a\alpha + b.
   \]

3. **Evaluate the given condition at \(x = \beta\)**:
   Similarly, for \(\beta\):
   \[
   p(\beta^2)q(\beta + 1) - p(\beta + 1)q(\beta^2) = \beta^2 + a\beta + b.
   \]
   Using \(\beta^2 = \beta + 1\) and \(\beta + 1 = \beta^2\), we get:
   \[
   p(\beta + 1)q(\beta^2) - p(\beta^2)q(\beta + 1) = \beta^2 + a\beta + b.
   \]
   Since \(\beta + 1 = \beta^2\) and \(\beta^2 = \beta + 1\), the left-hand side becomes zero:
   \[
   0 = \beta^2 + a\beta + b.
   \]

4. **Solve the system of equations**:
   We now have two equations:
   \[
   \alpha^2 + a\alpha + b = 0 \quad \text{and} \quad \beta^2 + a\beta + b = 0.
   \]
   Substituting \(\alpha = \frac{1 + \sqrt{5}}{2}\) and \(\beta = \frac{1 - \sqrt{5}}{2}\), we get:
   \[
   \left(\frac{1 + \sqrt{5}}{2}\right)^2 + a\left(\frac{1 + \sqrt{5}}{2}\right) + b = 0,
   \]
   \[
   \left(\frac{1 - \sqrt{5}}{2}\right)^2 + a\left(\frac{1 - \sqrt{5}}{2}\right) + b = 0.
   \]

5. **Simplify the equations**:
   \[
   \frac{1 + 2\sqrt{5} + 5}{4} + a\frac{1 + \sqrt{5}}{2} + b = 0,
   \]
   \[
   \frac{1 - 2\sqrt{5} + 5}{4} + a\frac{1 - \sqrt{5}}{2} + b = 0.
   \]
   Simplifying further:
   \[
   \frac{6 + 2\sqrt{5}}{4} + a\frac{1 + \sqrt{5}}{2} + b = 0,
   \]
   \[
   \frac{6 - 2\sqrt{5}}{4} + a\frac{1 - \sqrt{5}}{2} + b = 0.
   \]
   \[
   \frac{3 + \sqrt{5}}{2} + a\frac{1 + \sqrt{5}}{2} + b = 0,
   \]
   \[
   \frac{3 - \sqrt{5}}{2} + a\frac{1 - \sqrt{5}}{2} + b = 0.
   \]

6. **Combine and solve for \(a\) and \(b\)**:
   Adding the two equations:
   \[
   \left(\frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2}\right) + a\left(\frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2}\right) + 2b = 0,
   \]
   \[
   3 + a + 2b = 0.
   \]
   Therefore:
   \[
   a + 2b = -3.
   \]

   Subtracting the two equations:
   \[
   \left(\frac{3 + \sqrt{5}}{2} - \frac{3 - \sqrt{5}}{2}\right) + a\left(\frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2}\right) = 0,
   \]
   \[
   \sqrt{5} + a\sqrt{5} = 0,
   \]
   \[
   (1 + a)\sqrt{5} = 0.
   \]
   Since \(\sqrt{5} \neq 0\), we have:
   \[
   a = -1.
   \]
   Substituting \(a = -1\) into \(a + 2b = -3\):
   \[
   -1 + 2b = -3,
   \]
   \[
   2b = -2,
   \]
   \[
   b = -1.
   \]

The final answer is \( \boxed{ (a, b) = (-1, -1) } \).