a PDE problem involving divergence 
Consider the PDE on a bounded smooth domain $\Omega \subset R^n$, $\triangle u(x)= 0$ on $\Omega$, $\frac{\partial u(x)}{\partial n}|_{\partial \Omega}=g(x)$. Prove that if it admits a smooth solution u, we have $\int_{\partial \Omega} g d \sigma=0$

I know that $\triangle u=div(\triangledown u)$. Can someone give me a hint to do this?

We have
$\nabla \cdot \nabla u = \triangle u = 0 \tag 1$
on $\Omega$.  Thus the divergence theorem yields
$\displaystyle \int_{\partial \Omega} \nabla u \cdot \mathbf n \; dS = \int_\Omega \nabla \cdot \nabla u; dV = \int_\Omega 0 \; dV = 0, \tag 2$
where $\mathbf n$ is the outward-pointing unit normal vector field on $\partial \Omega$ and $dS$ is the volume element on $\partial \Omega$; but on $\partial \Omega$ we have
$\nabla u \cdot \mathbf n = \dfrac{\partial u}{\partial n} = g = 0, \tag 3$
whence
$\displaystyle \int_{\partial \Omega} g \; dS = 0. \tag 4$
