On critical points of harmonic functions

Question

Let $u \in C^{\infty}(\mathbb R^3)$ be harmonic. Suppose that $u$ has no critical points outside the unit ball but that it has at least one critical point inside the unit ball.

Does it follow that $u$ is a polynomial?

Answer 1

This is not true in dimension 2. Function $f(z)=ze^z$ is entire, and $f'(z)=0$ at one point, $z=-1$. It follows that the function $$u(x,y)=\mathrm{Re}f(x+iy)=e^x(x\cos y-y\sin y)$$ is harmonic, has one critical point $(x,y)=(-1,0)$, and evidently not a polynomial.

Now one can construct a similar example in even dimension, for example: $$v(x_1,x_2,x_3,x_4)=u(x_1,x_2)+u(x_3,x_4).$$ But for $n=3$ I could not figure out.