Determine all pairs $(a, b)$ of real numbers for which there exists a unique symmetric $2\times 2$ matrix $M$ with real entries satisfying $\mathrm{trace}(M)=a$ and $\mathrm{det}(M)=b$.

(Proposed by Stephan Wagner, Stellenbosch University)
1. Consider a symmetric \(2 \times 2\) matrix \(M\) with real entries. The general form of such a matrix is:
   \[
   M = \begin{bmatrix} x & y \\ y & z \end{bmatrix}
   \]
   where \(x, y, z\) are real numbers.

2. The trace of \(M\) is given by:
   \[
   \text{trace}(M) = x + z = a
   \]

3. The determinant of \(M\) is given by:
   \[
   \text{det}(M) = xz - y^2 = b
   \]

4. We need to find the conditions under which there exists a unique matrix \(M\) satisfying the given trace and determinant conditions. To do this, we consider the characteristic polynomial of \(M\):
   \[
   \lambda^2 - (x+z)\lambda + (xz - y^2) = 0
   \]
   Substituting the given conditions, we get:
   \[
   \lambda^2 - a\lambda + b = 0
   \]

5. The roots of this quadratic equation are:
   \[
   \lambda = \frac{a \pm \sqrt{a^2 - 4b}}{2}
   \]
   These roots correspond to the eigenvalues of the matrix \(M\).

6. For \(M\) to be unique, the eigenvalues must be equal, which implies that the discriminant of the quadratic equation must be zero:
   \[
   a^2 - 4b = 0
   \]
   Solving for \(b\), we get:
   \[
   b = \frac{a^2}{4}
   \]

7. Substituting \(b = \frac{a^2}{4}\) back into the expressions for the eigenvalues, we get:
   \[
   \lambda = \frac{a \pm 0}{2} = \frac{a}{2}
   \]
   Thus, the eigenvalues are both \(\frac{a}{2}\).

8. Since the eigenvalues are equal, the matrix \(M\) must be of the form:
   \[
   M = \begin{bmatrix} \frac{a}{2} & y \\ y & \frac{a}{2} \end{bmatrix}
   \]

9. The determinant condition \(xz - y^2 = b\) becomes:
   \[
   \left(\frac{a}{2}\right)^2 - y^2 = \frac{a^2}{4} - y^2 = b
   \]
   Substituting \(b = \frac{a^2}{4}\), we get:
   \[
   \frac{a^2}{4} - y^2 = \frac{a^2}{4}
   \]
   which simplifies to:
   \[
   y^2 = 0 \implies y = 0
   \]

10. Therefore, the unique matrix \(M\) is:
    \[
    M = \begin{bmatrix} \frac{a}{2} & 0 \\ 0 & \frac{a}{2} \end{bmatrix}
    \]

11. The pairs \((a, b)\) for which there exists a unique symmetric \(2 \times 2\) matrix \(M\) are given by:
    \[
    (a, b) = (2t, t^2) \quad \text{for} \quad t \in \mathbb{R}
    \]

The final answer is \( \boxed{ (a, b) = (2t, t^2) \quad \text{for} \quad t \in \mathbb{R} } \).