Solutions of EM wave equations that are not solutions of Maxwell equations? Maxwell equation in absence of charges and currents are 
$$\nabla \cdot \bf{E} = 0 \\  \nabla\cdot B=0  \\  \nabla \times E=-\frac{\partial B}{\partial t}  \\\nabla \times B=\mu \epsilon \frac{\partial E}{\partial t}$$
Wave equation is  $$\nabla ^2 \bf{E}=\mu \epsilon \frac{\partial^2 E}{\partial t^2}\tag{1}$$
How can I prove that, given a solution $\bf{E}$ that satisfies both Maxwell equations and $(1)$, then the vector $\bf{E+E'}$ where $\bf{}{}E'$ is any vector such that $\nabla \times \bf{}{}E'=0$ (and not necessarily a solution of Maxwell equations) is a solution of $(1)$?
 A: You cannot prove it, since it isn't true.
The curl of your E-field would be unchanged
$$\nabla \times (\vec{E} + \vec{E}^{\prime}) = \nabla \times \vec{E} + \nabla \times \vec{E}^{\prime} = \nabla \times \vec{E}$$
The left hand side of the wave equation is the Laplacian of the E-field 
$$\nabla^2 (\vec{E} + \vec{E}^{\prime}) = \nabla (\nabla \cdot (\vec{E} + \vec{E}^{\prime})) - \nabla \times (\nabla \times (\vec{E} + \vec{E}^{\prime})) $$
$$\nabla^2 (\vec{E} + \vec{E}^{\prime}) = \nabla(\nabla \cdot \vec{E}^{\prime})-\nabla \times (\nabla \times \vec{E}) $$
$$ \nabla^2 (\vec{E} + \vec{E}^{\prime}) = \nabla(\nabla \cdot \vec{E}^{\prime}) + \nabla^2 \vec{E} - \nabla (\nabla \cdot \vec{E}) = \nabla^2 \vec{E} + \nabla(\nabla \cdot \vec{E}^{\prime})$$
The right hand side of the wave equation becomes
$$
\mu \epsilon \frac{\partial^2 (\vec{E}+\vec{E}^{\prime})}{\partial t^2} = \mu \epsilon \frac{\partial^2 \vec{E}}{\partial t^2} +  \mu \epsilon \frac{\partial^2 \vec{E}^{\prime}}{\partial t^2}   $$
Thus the only way that the vacuum wave equation is still satisfied is if
$$\nabla (\nabla \cdot \vec{E}^{\prime} )= \mu \epsilon \frac{\partial^2 \vec{E}^{\prime}}{\partial t^2} $$
This is not a general identity! For example, $\vec{E}^{\prime}= \vec{E}_0 \sin \omega t$ has a zero curl, a zero divergence, but a non-zero second time derivative. However, if the additional field is a solution to Maxwell's vacuum equations with zero curl, then this condition is automatically satisfied.
A: I write a complement to the answer by Rob Jeffries. A possibility to add such a vector $\vec{E}'$ to $\vec{E}$ means that outside charges the total field is not obligatory transversal, but may contain (and contains) a longitudinal field. Normally it happens next to the radiating system of charges (a "near" field). Only far-far away the field becomes "radiated" and transversal.
From mathematical point of view, the equations are not sufficient for determining the field. One needs boundary conditions too.
