Why do we need both Divergence and Curl to define a vector field? I was reading Classical Electrodynamics by J.D.Jacskon (section 1.5) where he said:

Perhaps some readers know that a vector field can be specified almost completely if its divergence and curl are given everywhere in space

with a comment

(almost)-Up to the gradient of a scalar function that satisfies the Laplace equation.

why is this so
if we want to know $\vec{E}$ in 3-D space we really just want to find out is $E_x,E_y$ and $E_z$ (in cartesian coordinates) which are just 3 scalar functions.
If we know the curl is (let's say) zero, that is $$\nabla \times \vec{E}=0$$
in cartesian coordinates that gives us three equations
$$\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}=0$$
$$\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}=0$$
$$\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}=0$$
these are three equations for 3 unknown quantities, that should give us enough* information about these three functions, it obviously isn't so hence we need to know the divergence as well, my question is why?
is there some theorem that proves this (I suspect there is $:)$ )
this is obviously a mathematical question although it is physics motivated
so I hope it is ok that I post it here.
(*up to maybe a constant because all the equations involve first derivatives, I can live with that)
 A: If
$\nabla \times \vec E = 0, \tag 1$
we know from a standard result that 
$\vec E = \nabla \phi \tag 2$
for some scalar function $\phi$.
If the divergence of $\vec E$ is also a known function $\rho$,
$\nabla \cdot \vec E = \rho, \tag 3$
then combining (2) and (3) we obtain
$\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \vec E = \rho, \tag 4$
which can in principle be solved for $\phi$ (assuming we admit appropiate boundary conditions on $\phi$; $\phi(x) \to 0$ sufficiently fast as $\vert x \vert \to \infty$ is often used; Jackson's book covers this, mos' likely); thus, we may discover $\vec E$ from (2).  
We observe that such a solution is not unique; indeed, let $\psi$ be any harmonic function,
$\nabla \cdot \nabla \psi = \nabla^2 \psi = 0; \tag 5$
then
$\nabla^2(\phi + \psi) = \nabla^2 \phi + \nabla^2 \psi = \rho + 0 = \rho, \tag 6$
and if
$\vec E = \nabla (\phi + \psi), \tag 7$
we still obtain
$\nabla \times \vec E = \nabla \times \nabla (\phi + \psi) = \nabla \times \nabla \phi + \nabla \times \nabla \times \psi = 0, \tag 8$
since the curl of any gradient vanishes.  Also,
$\nabla \cdot \vec E = \nabla \cdot (\nabla \phi + \nabla \psi) = \nabla^2 \phi + \nabla ^2 \psi = \nabla^2 \phi = \rho; \tag 9$
these last two equations show that we may transform any solution according to
$\phi \to \phi + \psi, \tag 9$
$\vec E = \nabla \phi \to \nabla \phi + \nabla \psi, \tag{10}$
$\psi$ as in (5), and preserve the divergence and curl of $\vec E$; so any solution is not unique; uniqueness may be attained by specifying appropriate boundary conditions on $\phi$ and $\psi$ which can then become unambiguously determined.
The above discussion addresses the relatively simple case (1), (3); we can, in fact, also address the significant generalization
$\nabla \times \vec E = \vec F, \; \nabla \cdot \vec E = \rho, \tag{11}$
where $\vec F$ is a pre-specified vector field; the situation is more complicated since we may no longer assume $\vec E$ is a gradient as in (1)-(2).  In this case we 
instead invoke the vector calculus identity
$\nabla \times (\nabla \times \vec A) = \nabla (\nabla \cdot \vec A) - \nabla^2 \vec A, \tag{12}$
where the Laplacian operator $\nabla^2$ occurring on the right-hand side is understood to act component-wise on $\vec A$; thus we have
$\nabla \times \vec E = \vec F \tag{13}$
leading to
$\nabla \times (\nabla \times \vec E) = \nabla \times \vec F, \tag{14}$
which we trasform according to (12):
$\nabla (\nabla \cdot \vec E) - \nabla^2(\vec E) = \nabla \times \vec F; \tag{15}$
by virtue of (3) we write this as
$\nabla \rho - \nabla^2 \vec E = \nabla \times \vec F, \tag{16}$
whence we find
$\nabla^2 \vec E = \nabla \rho - \nabla \times \vec F, \tag{17}$
which we may in principle solve component-wise for $\vec E$; for example,
$\nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \left ( \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \right) = \dfrac{\partial \rho}{\partial x} - 
\nabla^2 E_x = \dfrac{\partial \rho}{\partial x} -  \dfrac{\partial F_z}{\partial y} + \dfrac{\partial F_y}{\partial z}; \tag{18}$
of course, we need boundary conditions and perhaps certain restrictions on $\rho$ and $\vec F$; but I'll leave it to folks like David Jackson to explain such matters.
Apparently solutions to (11), (18) are still not unique; it is still true that adjusting $\vec E$ by the gradient of a harmonic function $\psi$ gives rise to another solution; of course, we may determine such $\psi$ uniquely via appropriate boundary conditions, and then a certain uniqueness is attained for $\vec E$.
