Solution:
Let $z_{k}, 1 \leq k \leq 4$, be the roots of the given equation. Using Vieta's formulas we obtain
$$z_{1}+z_{2}+z_{3}+z_{4}=a \text{ and } z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}=a^{2}$$
Set $u_{k}=\frac{2 z_{k}}{a}=x_{k}+i y_{k}, 1 \leq k \leq 4$, where $x_{k}, y_{k} \in \mathbb{R}$. Then $u_{k}^{2}=x_{k}^{2}-y_{k}^{2}+2 i x_{k} y_{k}$ and we get
$$\begin{aligned} & x_{1}+x_{2}+x_{3}+x_{4}=2 \\ & x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2} \geq 4 \end{aligned}$$
On the other hand,
$$\left|a-z_{k}\right| \geq\left|z_{k}\right| \Longleftrightarrow\left|2-u_{k}\right| \geq\left|u_{k}\right| \Longleftrightarrow\left(2-x_{k}\right)^{2}+y_{k}^{2} \geq x_{k}^{2}+y_{k}^{2} \Longleftrightarrow x_{k} \leq 1$$
Now (1) implies $x_{k} \geq-1$, i.e. $x_{k}^{2} \leq 1$, which together with (2) gives $x_{k}= \pm 1$, $y_{k}=0$.
Hence (1) shows that three of the numbers $x_{k}$ are equal to 1 and the fourth one is $-1$. We can assume that $z_{1}=z_{2}=z_{3}=-z_{4}$. Then $z_{1} z_{2} z_{3} z_{4}=-1$ implies $z_{1}= \pm 1, \pm i$, which gives
$$(a, b)=(2,-2),(-2,2),(2 i, 2 i),(-2 i,-2 i)$$