We know very well that static electric fields have scalar potential fields, and magnetic fields can't have scalar fields if there is free current, as $$\nabla\times \vec{B}=\mu_0\vec J$$ $$\Rightarrow \int\nabla\times \vec{B}\cdot d\vec{a}=\oint\vec{B}\cdot d\vec{l}=\int\mu_0\vec J\cdot d\vec{a}$$Since the surface integral of $\vec{J}$ does not have to be zero, that means if we summing up $\vec{B}\cdot{d\vec{l}}$ from point $a$ to point $b$; the result depends not only on end points $a,b$ but also on how you sum it up. Therefore, we do not have a scalar potential for static magnetic field, which is defined as $$\int_{a}^{b}\vec{B}\cdot{d\vec{l}}$$ that is required to be independent of path chosen.
But I have just wondered: can we give static electric field a vector potential field? How? If we can't, how?
My attempt:
$$\nabla\times (\nabla\times \vec A_m)=0$$ $$-(\partial_x^2+\partial_y^2+\partial_z^2)\vec A+ \begin{bmatrix} \partial_x \\ \partial_y \\ \partial_z\\ \end{bmatrix} \ \nabla \cdot \vec A=0$$
Then we have this for the $x$ direction:
$$(\partial_x^2+\partial_y^2+\partial_z^2) A_x = \partial_x \nabla \cdot \vec A$$
I do not know how to solve it, except for requiring $\nabla \cdot \vec A=0$, we have $$(\partial_x^2+\partial_y^2+\partial_z^2) \begin{bmatrix} A_x \\ A_y \\ A_z\\ \end{bmatrix} \ =0 $$
Then I can go no further. (I have looked up the three dimensional Laplace partial equation, but I am not sure what does the solution means, especially it will appear in each terms of the vector)
Furthermore: can we define a vector potential field for a non-static eletirc field? $^1$
$1:$ I think it has something to do with the Maxwell equations that having each other partial t E/M field in the "curl M/E equations", but I am not quite sure about that.