How can we characterize polynomials in $\mathbb{R}^2$ that are harmonic How can we characterize polynomials $p(x,y)$ in  $\mathbb{R}^2$ (in two variables) that are harmonic (that is $\Delta p(x,y) = 0$)?
 A: This is a much-studied question.
A: (Since the Laplacian in two variables can be written as $\partial\circ\overline{\partial}$, for example), all harmonic polynomials in $x,y$ are linear combinations of $(x+iy)^n$ and $(x-iy)^n$, for non-negative integers $n$.
EDIT: one way to prove this is to use $z=x+iy$ and $\overline{z}=x-iy$, and observe that every (complex) polynomial in $x,y$ can be rewritten as such in $z$ and $\overline{z}$. Let $\partial=\partial/\partial z$ and $\overline{\partial}=\partial/\partial \overline{z}$. Now, of course, it's not clear what these are, except as expressing them in terms of $x,y$. Indeed. Ok, but it is not hard to prove the key lemma, that $\overline{\partial}z=0$ and $\partial \overline{z}=0$. Also, check that the Laplacian is $\partial \circ \overline{\partial}=\overline{\partial}\circ \partial$. Also, check that the obviously-different monomials in $z$ and $\overline{z}$ are linearly independent. Then show that $\partial(P(z,\overline{z})=0$ for a polynomial in these two occurs if and only if $z$ does not appear. And similarly for $\overline{z}$. 
Then annihilation by the Laplacian is $\partial(\overline{\partial}P)=0$, so $\overline{\partial}P$ must have no $z$'s in it. Thus, $P$ itself must have only monomials that purely involve $z$, or purely involve $\overline{z}$.
(Yes, this argument is an algebraic emulation of some things about holomorphic functions...)
