The electric field $\mathbf{E}$ and the magnetic induction $\mathbf{B}$ can be parameterized in terms of potentials $V$ and $\mathbf{A}$: $$ \mathbf{E}=-\nabla V-\frac{\partial \mathbf{A}}{\partial t},\quad \mathbf{B}=\nabla\times \mathbf{A}.$$ This parameterization is not unique, as we can find a scalar function $\theta$ and define a couple $(\tilde{V},\mathbf{\tilde{A}})$ via $\tilde{V} = V-\partial \theta/\partial t$ and $\mathbf{\tilde{A}}=A+\nabla \theta$ $(*)$. Then both $(V,\mathbf{A})$ and $(\tilde{V},\mathbf{\tilde{A}})$ will give rise to the same $(\mathbf{E},\mathbf{B})$.
Via Maxwell's equations we can find a coupled system of differential equations for $V$ and $\mathbf{A}$: $$ \begin{cases} \square \mathbf{A} = -\mu \mathbf{J}+\nabla\left( \nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t}\right) \\ \square V = -\frac{\rho}{\varepsilon} -\frac{\partial}{\partial t}\left( \nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t}\right)\end{cases},\quad \square=\nabla^2-\frac{\partial^2}{\partial t^2}.$$
These can be made independent by considering the Lorenz-gauge, in which we set $\nabla\cdot \mathbf{A}+\varepsilon\mu \frac{\partial V}{\partial t} = 0$. How can one explicitly show that for each $(V,\mathbf{A})$ there is $(\tilde{V},\mathbf{\tilde{A}})$ (i.e. give rise to the same fields) such that this couple will satisfy the Lorenz-gauge condition. Is it enough to consider the expressions $(*)$ and deduce that both couples of potentials need to satisfy the Lorenz-gauge condition, resulting in the condition $\square \theta = 0$, i.e. we can always choose a scalar function $\theta$ for which $\square \theta=0$ and consider a new potentials via $(*)$?
Thanks in advance.