Let $U$ solve the Neumann${}$ problem${}$ for laplace's equation on a${}$ domain $\Omega$. Show that $U+c$ also solves this problem for any $c\in\Bbb R$.
What is being asked of me? Does this mean that:
1) $\frac{\partial U}{\partial \mathbf n}=f$
2) $\nabla^2 U =0$
Where $\partial \Omega$ is the boundary and $f:\partial \Omega \to \Bbb C$ and $U:\Omega\cup \partial\Omega \to \Bbb C$
Does this mean I am being asked to show that $\frac{\partial U}{\partial \mathbf n} = f\implies\frac{\partial (U+c)}{\partial \mathbf n}=f $ and $\nabla^2 U =0\implies \nabla^2 (U+c)=0$?
Where $\nabla^2$ is the Laplacian operator and $U$ is a harmonic function on $\Omega$
Yes, all you need to do is to show that adding a constant affects neither the PDE nor the boundary condition.
This is the kind of question that are posed not to make you sweat, but to make you aware of an important fact: unlike the Dirichlet problem and mixed Dirichlet-Neumann problems, the pure Neumann problem never has a unique solution.